This section studies how the distribution of a random variable changes when the variable is transfomred in a deterministic way. If you are a new student of probability, you should skip the technical details.
As usual, our starting point is a random experiment, modeled by a probability space \((\Omega, \ms F, \P)\). So to review, \(\Omega\) is the set of outcomes, \(\ms F\) is the collection of events, and \(\P\) is the probability measure on the sample space \((\Omega, \ms F)\). Suppose now that \((S, \ms S)\) and \((T, \ms T)\) are measurable spaces and that \(X\) is a random variable for the experiment, with values in \(S\). If \(r: S \to T\) is measurable function, then \(Y = r(X)\) is a new random variable with values in \(T\). If the distribution of \(X\) is known, how do we find the distribution of \(Y\)? This is a very basic and important question. In applications, random variables are variables or measurements of interest in the experiment, and transformations of such variables are ubiquitous. In a superficial sense, the solution is easy, but first recall that the inverse image of \(B \in \ms T\) under \(r\) is \(r^{-1}(B) = \{x \in S: r(x) \in B\} \in \ms S\).
\(\P(Y \in B) = \P\left[X \in r^{-1}(B)\right]\) for \(B \in \ms T\).
\( \P(Y \in B) = \P\left[r(X) \in B\right] = \P\left[X \in r^{-1}(B)\right] \) for \( B \in \ms T \). This is just the probability version of the general theorem that a measurable function on a measure space produces a new measure on the range space.
Suppose now that \(\mu\) and \(\nu\) are \(\sigma\)-finite positive measures for \((S, \ms S)\) and \((T, \ms T)\). Frequently the distribution of \(X\) is known through its probability density function \(f\), and we would similarly like to find the probability density function of \(Y\). This is a difficult problem in general, because as we will see, even simple transformations of variables with simple distributions can lead to variables with complex distributions. We will solve the problem in various special cases. As always, density functions are with respect to the associated reference measures, and the two most important special cases are discrete measure spaces and Euclidean measure spaces as described in the introduction.
When the transformed variable \(Y\) has a discrete distribution, the probability density function of \(Y\) can be computed using basic rules of probability.
Suppose that \(Y\) has a discrete distribution, and that \(X\) has probability density function \(f\). Then \(Y\) has probability density function \(g\) given by \[g(y) = \int_{r^{-1}(y)} f(x) \, d\mu(x), \quad y \in T\]
The hypothesis means that \(T\) is countable, \(\ms T\) is the collection of all subsets of \(T\), and the reference measure is counting measure \(\#\). So by definition of the densities of \(Y\) and \(X\), \[g(y) = \P(Y = y) = \P[X \in r^{-1}(y)] = \int_{r^{-1}(y)} f(x) \, d\mu(x), \quad y \in T\]
Here are two special cases for the distribution of \(X\).
As a special case, if \(X\) also has a discrete distribution, then the integral in is a sum and \[ g(y) = \sum_{x \in r^{-1}\{y\}} f(x), \quad y \in T \]
As another special case, suppose that \(X\) has a continuous distribution on an \(n\)-dimensional Euclidean space for some \(n \in \N_+\). Then the integral in is the Lebesgue integral and \[ g(y) = \int_{r^{-1}\{y\}} f(x) \, dx, \quad y \in T \]
The assumptin that \(X\) has a density function actually means that \(X\) is absolutely continuous with respect to Lebesgue measure. Recall also that the Lebesgue integral agrees with the ordinary Riemann integral of calculus for the density functions that usually occur in applications.
So the main problem is often computing the inverse images \(r^{-1}\{y\}\) for \(y \in T\). The formulas in the discrete in and the Euclidean case in cases are not worth memorizing explicitly; it's usually better to just work each problem from scratch. The main step is to write the event \(\{Y = y\}\) in terms of \(X\), and then find the probability of this event using the probability density function of \( X \).
Suppose now that \(T \in \ms R^n\) for some \(n \in \N_+\). In some cases, the distribution function of \(Y\) can be found by using basic rules of probability. Then the probability density function of \(Y\), if it exists, can be found form the distribution function. This general method is referred to, appropriately enough, as the distribution function method. Here is the case with \(n = 1\).
Suppose that \(Y\) is real valued. The distribution function \(G\) of \(Y\) is given by
\[ G(y) = \int_{r^{-1}(-\infty, y]} f(x) \, d\mu(x), \quad y \in \R \]Again, this follows from the definition of \(f\) as a density of \(X\). For \( y \in \R \), \[ G(y) = \P(Y \le y) = \P\left[r(X) \in (-\infty, y]\right] = \P\left[X \in r^{-1}(-\infty, y]\right] = \int_{r^{-1}(-\infty, y]} f(x) \, d\mu(x) \]
As in the discrete case, the formula above not much help, and it's usually better to work each problem from scratch. The main step is to write the event \(\{Y \le y\}\) in terms of \(X\), and then find the probability of this event using the probability density function of \( X \).
Suppose now that \(S, \, T \in \ms R^n\) for some \(n \in \N_+\), and suppose that \(X\) has a continuous distribution probability density function \(f\). When the transformation \(r: S \to T\) is one-to-one and smooth, there is a formula for the probability density function of \(Y\) directly in terms of \(f\). This is known as the change of variables formula. Note that since \(r\) is one-to-one, it has an inverse function \(r^{-1}\). We will explore the one-dimensional case first, where the concepts and formulas are simplest. Thus, suppose that \(S \subseteq \R\) is an interval, and that \(X\) has a continuous distribution on \(S\) with probability density function \(f\). Suppose \(Y = r(X)\) where \(r\) is a one-to-one, continuously differentiable function from \(S\) onto an interval \(T \subseteq R\). Clearly \(r\) is either strictly increasing or strictly decreasing.
If \(r\) is strictly increasing then \(Y\) has a continuous distribution on \(T\) with density function \(g\) given by \[ g(y) = f\left[r^{-1}(y)\right] \frac{d}{dy} r^{-1}(y), \quad y \in T \]
Let \(G\) denote the distribution function of \(Y\). Then \[ G(y) = \P(Y \le y) = \P[r(X) \le y] = \P\left[X \le r^{-1}(y)\right] = F\left[r^{-1}(y)\right], \quad y \in T \] Note that the inquality is preserved since \( r \) is increasing. Taking derivatives with respect to \( y \) and using the chain rule gives the result. Recall that \( F^\prime = f \) at the points of continuity of \(f\).
If \(r\) is strictly decreasing then \(Y\) has a continuous distribution on \(T\) with density function \(g\) given by \[ g(y) = -f\left[r^{-1}(y)\right] \frac{d}{dy} r^{-1}(y), \quad y \in T \]
Again let \(G\) denote the distribution function of \(Y\). Then \[ G(y) = \P(Y \le y) = \P[r(X) \le y] = \P\left[X \ge r^{-1}(y)\right] = 1 - F\left[r^{-1}(y)\right], \quad y \in T \] Note that the inquality is reversed since \( r \) is decreasing. Taking derivatives with respect to \( y \) and using the chain rule gives the result. Recall again that \( F^\prime = f \) at the points of continuity of \(f\).
The formulas for the probability density functions in the increasing case in propostion and the decreasing case in can be combined:
Under the assumptions on \(r\), the probability density function \(g\) of \(Y\) is given by \[ g(y) = f\left[ r^{-1}(y) \right] \left| \frac{d}{dy} r^{-1}(y) \right|, \quad y \in T \]
Letting \(x = r^{-1}(y)\), the change of variables formula can be written more compactly as \( g(y) = f(x) |dx / dy| \) or even in differential form as \( g(y) |dy| = f(x) |dx| \). Although succinct and easy to remember, the formulas are a bit less clear. It must be understood that \(x\) on the right should be written in terms of \(y\) via the inverse function. The images below give a graphical interpretation of the formula in the two cases where \(r\) is increasing and where \(r\) is decreasing.
The generalization of this result to dimension \(n \in \N_+\) is basically a theorem in multivariate calculus. First we need some notation. Suppose \(S\) and \(T\) are open subsets of \(\R^n\) and that \( r: S \to T \) is a one-to-one, continuously differentiable fucntion from \(S\) onto \(T\). We will use the notation \(x = (x_1, x_2, \ldots, x_n) \in S\) and \(y = (y_1, y_2, \ldots, y_n) \in T\). The first derivative of the inverse function \(x = r^{-1}(y)\) is the \(n \times n\) matrix \( dx / dy \) of first partial derivatives: \( (dx / dy)_{i j} = \partial x_i / \partial y_j \). The Jacobian (named in honor of Karl Gustav Jacobi) of the inverse function is the determinant of the first derivative matrix \( \det (dx / dy) \). With this compact notation, the multivariate change of variables formula is easy to state.
Suppose that \(X\) has a continuous distribution on \(S\) with probability density function \(f\). Then \(Y = r(X)\) has a continuous distribution on \(T\) with density function \(g\) given by \[ g(y) = f(x) \left| \det \left( \frac{dx}{dy} \right) \right|, \quad y \in T \]
The result follows from the multivariate change of variables formula in calculus. The probability distribution of \(Y\) is given by \[\P(Y \in B) = \P[r(X) \in B] = \P[X \in r^{-1}(B)] = \int_{r^{-1}(B)} f(x) \, dx, \quad B \in \ms T\] This is just a special case of , and only requires that \(r\) be measurable. The next step is to use the change of variables \(x = r^{-1}(y)\), \(dx = \left|\det \left( dx / dy \right)\right|\, dy\). Under the assumptions on \(r\), \[\P(Y \in B) = \int_B f[r^{-1}(y)] \left|\det \left( \frac{dx}{dy} \right)\right|\, dy\] So it follows that \(g\) defined in the theorem is a density function for \(Y\).
The Jacobian is the infinitesimal scale factor that describes how \(n\)-dimensional volume changes under the transformation, as illustrated in the graphic below.
Linear transformations (or more technically affine transformations) are among the most common and important transformations. Moreover, this type of transformation leads to simple applications of the change of variable theorems. Suppose first that \(X\) has a continuous distribution on an interval \(S \subseteq \R\) with probability density function \(f\). Let \(Y = a + b \, X\) where \(a \in \R\) and \(b \in \R \setminus\{0\}\). Note that \(Y\) has values in the interval \(T = \{y = a + b x: x \in S\}\).
\(Y\) has a continuous distribution on \(T\) with probability density function \( g \) given by \[ g(y) = \frac{1}{\left|b\right|} f\left(\frac{y - a}{b}\right), \quad y \in T \]
When \(b \gt 0\) (which is often the case in applications), this transformation is known as a location-scale transformation; \(a\) is the location parameter and \(b\) is the scale parameter. Scale transformations arise naturally when physical units are changed (from feet to meters, for example). Location transformations arise naturally when the physical reference point is changed (measuring time relative to 9:00 AM as opposed to 8:00 AM, for example). The change of temperature measurement from Fahrenheit to Celsius is a location and scale transformation.
The multivariate version of this result has a simple and elegant form when the linear transformation is expressed in matrix-vector form. Thus suppose that \(X\) has a continuous distribution on \(S \in \ms R^n\) with probability density function \(f\). Let \(Y = a + B X\) where \(a \in \R^n\) and \(B\) is an invertible \(n \times n\) matrix. Note that \(Y\) has values in \(T = \{a + B x: x \in S\} \in \ms R^n\).
\(Y\) has a continuous distribution on \(T\) with probability density function \(g\) given by \[ g(y) = \frac{1}{\left| \det(B)\right|} f\left[ B^{-1}(y - a) \right], \quad y \in T \]
Simple addition of independent real-valued random variables is perhaps the most important of all transformations. Indeed, much of classical probability theory is concerned with sums of independent variables, in particular the law of large numbers, and the central limit theorem. In addition, aside from myriad applications, in many cases a variable with a special distribution can be decomposed inot a sum of independent random variables with simpler distributions. We will consider sums of independent variables in terms of their probability distributions, distribution functions, and density functions. Each case leads to a type of convolution.
Suppose that \(X\) and \(Y\) are independent, real-value random variables with distributions \(P\) and \(Q\) respectively. Then the distribution of \(Z = X + Y\) is the convolution \(P * Q\) of \(P\) and \(Q\) defined by \[(P * Q)(A) = \int_\R Q(A - x) \, dP(x), \quad A \in \ms R\]
Since \(X\) and \(Y\) are independent, the distribution of \((X, Y)\) on \(\R^2\) is the product probability measure \(P \times Q\). \begin{align*} \P(Z \in A) & = \P(X + Y \in A) = \int_{\R^2} \bs 1(x + y \in A) \, d(P \times Q)(x, y) = \int_{\R^2} \bs 1(y \in A - x) \, d(P \times Q)(x, y) \\ & = \int_\R \int_\R \bs 1(y \in A - x) \, dQ(y) \, dP(x) = \int_R Q(A - x) \, dP(x), \quad A \in \ms R \end{align*} Writing the double integral as an iterated integral is justified by Fubini's theorem. A more elegant and intutive proof can be given in terms of expected value.
Since addition of independent random variables is trivially commutative and assoicative, the same is true of the convolution of probability measures on \(\R\).
Suppose that \(P\), \(Q\), and \(R\) are probability measures on \(\R\). Then
We can restate in terms of distribution functions.
Suppose that \(X\) and \(Y\) are independent, real-valued random variables with distribution functions \(F\) and \(G\) respectively. The distribution function of \(Z = X + Y\) is the convoluation \(F * G\) of \(F\) and \(G\), defined by \[ (F * G)(z) = \int_\R G(z - x) \, dF(x), \quad z \in \R \]
Suppose that \(x, \, z \in \R\). If \(A = (-\infty, z]\) then \(A - x = (-\infty, z - x]\). Hence from the first convolution result in , \[ \P(Z \le z) = \P(X + Y \le z) = \int_\R Q(-\infty, z - x] \, dP(x) = \int_\R G(z - x) \, dF(x), \quad z \in \R \] The last equation is simply notation: \(P\) is the Lebesgue-Stieljes integral associated with the distribution function \(F\), so an integral with respect to \(F\) means the same thing as an integral with respect ot \(P\).
Convolution of probability distribution functions is commutative and associative.
Suppose that \(F\), \(G\), and \(H\) are distribution functions on \(\R\). Then
If the independent random variables \(X\) and \(Y\) are of the same type (either discrete or continuous) then we can find the density function of \(X + Y\) in terms of the density functions of \(X\) and \(Y\).
Suppose that \(X\) and \(Y\) are independent, real-valued random variables.
Both results follow from the previous convolution results, but it's also instructive to give direct proofs.
The formulas are simple and elegant, but gloss over domain complications. Suppose that \(X\) and \(Y\) have support sets \(S\) and \(T\), respectively. In part (a), as noted in the proof, \(S\) and \(T\) are countable and the sum is actually over the countable set \(\{x \in S: z - x \in T\}\). In part (b), \(S\) and \(T\) are typically intervals, and again, the integration is over the set \(\{x \in S: z - x \in T\}\). Determining this set for each possible value \(z\) of \(Z\) is often the most challenging part. However, when the variables are nonnegative, the complications are much reduced.
Suppose again that \( X \) and \( Y \) are independent random variables.
Convolution of density functions, like the two other forms of convolution, is commutative and associative.
Suppose that \(f\), \(g\), and \(h\) are probability density functions of the same type. Then
We have now seen three types of convolution—one for probability measures, one for distribution functions, and one for probabiity density functions. One of the takeaways is that convolution means different things depending on the type of objects involved. Moreover, convolution is a very important mathematical operation that occurs in areas of mathematics outside of probability.
Suppose that \(X\) is a real-valued random variable and that \(\bs X = (X_1, X_2, \ldots)\) is a sequence of independent copies of \(X\). Let \(Y_n = X_1 + X_2 + \cdots + X_n\) for \(n \in \N_+\).
In statistical terms, \( \bs X \) corresponds to sampling from the distribution of \(X\). When appropriately scaled and centered, the distribution of \(Y_n\) converges to the standard normal distribution as \(n \to \infty\). The precise statement of this result is the central limit theorem, one of the fundamental theorems of probability.
While not as important as sums, products and quotients of independent, real-valued random variables also occur frequently. We will limit our discussion to positive random variables with continuous distributions, the most important case.
Suppose that \( X \) and \( Y \) are independent and have continuous distributions on \( (0, \infty) \) with probability density functions \( f \) and \( g \), respectively.
First recall that \((X, Y)\) has probability density function \((x, y) \mapsto f(x) g(y)\) on \((0, \infty)^2\). We introduce the auxiliary variable \( U = X \) so that we have bivariate transformations and can use our change of variables formula .
If \( X \) is supported on a subset \( S \subseteq (0, \infty) \) and \( Y \) is supported on a subset \( T \subseteq (0, \infty) \), then for a given \( v \in (0, \infty) \), the integral in (a) is over \( \{x \in S: v / x \in T\} \), and for a given \( w \in (0, \infty) \), the integral in (b) is over \( \{x \in S: w x \in T\} \). As with convolution, determining the domain of integration is often the most challenging step.
Suppose that \(n \in \N_+\) and that \((X_1, X_2, \ldots, X_n)\) is a sequence of independent real-valued random variables. The minimum and maximum transformations \[U = \min\{X_1, X_2, \ldots, X_n\}, \quad V = \max\{X_1, X_2, \ldots, X_n\} \] are very important in a number of applications. For example, recall that in the standard model of structural reliability, a system consists of \(n\) components that operate independently. Suppose that \(X_i\) represents the lifetime of component \(i \in \{1, 2, \ldots, n\}\). Then \(U\) is the lifetime of the series system which operates if and only if each component is operating. Similarly, \(V\) is the lifetime of the parallel system which operates if and only if at least one component is operating.
A particularly important special case occurs when the random variables are identically distributed, in addition to being independent. In this case, the sequence of variables is a random sample of size \(n\) from the common distribution. The minimum and maximum variables are the extreme examples of order statistics. Our first result is in terms of distributon functions, and so applies without regard to the type of distributions.
Suppose that \((X_1, X_2, \ldots, X_n)\) is a sequence of indendent real-valued random variables and that \(X_i\) has distribution function \(F_i\) for \(i \in \{1, 2, \ldots, n\}\).
From part (a), note that the product of \(n\) distribution functions is another distribution function. From part (b), the product of \(n\) right-tail distribution functions is a right-tail distribution function. In the reliability setting, where the random variables are nonnegative, the last statement means that the product of \(n\) reliability functions is another reliability function. If \(X_i\) has a continuous distribution with probability density function \(f_i\) for each \(i \in \{1, 2, \ldots, n\}\), then \(U\) and \(V\) also have continuous distributions, and their probability density functions can be obtained by differentiating the distribution functions in parts (a) and (b) of . The computations are straightforward using the product rule for derivatives, but the results are a bit of a mess.
The formulas in are particularly nice when the random variables are identically distributed, in addition to being independent
Suppose that \((X_1, X_2, \ldots, X_n)\) is a sequence of independent real-valued random variables, with common distribution function \(F\).
In particular, it follows that a positive integer power of a distribution function is a distribution function. More generally, it's easy to see that every positive power of a distribution function is a distribution function. How could we construct a non-integer power of a distribution function in a probabilistic way?
Suppose that \((X_1, X_2, \ldots, X_n)\) is a sequence of independent real-valued random variables, with a common continuous distribution that has probability density function \(f\).
For our next discussion, we will consider transformations that correspond to common distance-angle based coordinate systems—polar coordinates in the plane, and cylindrical and spherical coordinates in 3-dimensional space. First, for \( (x, y) \in \R^2 \), let \( (r, \theta) \) denote the standard polar coordinates corresponding to the Cartesian coordinates \((x, y)\), so that \( r \in [0, \infty) \) is the radial distance and \( \theta \in [0, 2 \pi) \) is the polar angle.
It's best to give the inverse transformation: \( x = r \cos \theta \), \( y = r \sin \theta \). As we all know from calculus, the Jacobian of the transformation is \( r \). Hence the following result is an immediate consequence of our change of variables :
Suppose that \( (X, Y) \) has a continuous distribution on \( \R^2 \) with probability density function \( f \), and that \( (R, \Theta) \) are the polar coordinates of \( (X, Y) \). Then \( (R, \Theta) \) has probability density function \( g \) given by \[ g(r, \theta) = f(r \cos \theta , r \sin \theta ) r, \quad (r, \theta) \in [0, \infty) \times [0, 2 \pi) \]
Next, for \( (x, y, z) \in \R^3 \), let \( (r, \theta, z) \) denote the standard cylindrical coordinates, so that \( (r, \theta) \) are the standard polar coordinates of \( (x, y) \) as above, and coordinate \( z \) is left unchanged. Given our previous result, the one for cylindrical coordinates should come as no surprise.
Suppose that \( (X, Y, Z) \) has a continuous distribution on \( \R^3 \) with probability density function \( f \), and that \( (R, \Theta, Z) \) are the cylindrical coordinates of \( (X, Y, Z) \). Then \( (R, \Theta, Z) \) has probability density function \( g \) given by \[ g(r, \theta, z) = f(r \cos \theta , r \sin \theta , z) r, \quad (r, \theta, z) \in [0, \infty) \times [0, 2 \pi) \times \R \]
Finally, for \( (x, y, z) \in \R^3 \), let \( (r, \theta, \phi) \) denote the standard spherical coordinates corresponding to the Cartesian coordinates \((x, y, z)\), so that \( r \in [0, \infty) \) is the radial distance, \( \theta \in [0, 2 \pi) \) is the azimuth angle, and \( \phi \in [0, \pi] \) is the polar angle. (In spite of our use of the word standard, different notations and conventions are used in different subjects.)
Once again, it's best to give the inverse transformation: \( x = r \sin \phi \cos \theta \), \( y = r \sin \phi \sin \theta \), \( z = r \cos \phi \). As we remember from calculus, the absolute value of the Jacobian is \( r^2 \sin \phi \). Hence the following result is an immediate consequence of the change of variables theorem in :
Suppose that \( (X, Y, Z) \) has a continuous distribution on \( \R^3 \) with probability density function \( f \), and that \( (R, \Theta, \Phi) \) are the spherical coordinates of \( (X, Y, Z) \). Then \( (R, \Theta, \Phi) \) has probability density function \( g \) given by \[ g(r, \theta, \phi) = f(r \sin \phi \cos \theta , r \sin \phi \sin \theta , r \cos \phi) r^2 \sin \phi, \quad (r, \theta, \phi) \in [0, \infty) \times [0, 2 \pi) \times [0, \pi] \]
Our next discussion concerns the sign and absolute value of a real-valued random variable.
Suppose that \(X\) has a continuous distribution on \(\R\) with distribution function \(F\) and probability density function \(f\).
Recall that the sign function on \( \R \) (not to be confused, of course, with the sine function) is defined as follows: \[ \sgn(x) = \begin{cases} -1, & x \lt 0 \\ 0, & x = 0 \\ 1, & x \gt 0 \end{cases} \]
Suppose again that \( X \) has a continuous distribution on \( \R \) with distribution function \( F \) and probability density function \( f \), and suppose in addition that the distribution of \( X \) is symmetric about 0. Then
This subsection contains computational exercises, many of which involve special parametric families of distributions. It is always interesting when a random variable from one parametric family can be transformed into a variable from another family. It is also interesting when a parametric family is closed or invariant under some transformation on the variables in the family. Often, such properties are what make the parametric families special in the first place. Please note these properties when they occur.
Recall that a standard die is an ordinary 6-sided die, with faces labeled from 1 to 6 (usually in the form of dots). A fair die is one in which the faces are equally likely. An ace-six flat die is a standard die in which faces 1 and 6 occur with probability \(\frac{1}{4}\) each and the other faces with probability \(\frac{1}{8}\) each.
Suppose that two six-sided dice are rolled and the sequence of scores \((X_1, X_2)\) is recorded. Find the probability density function of \(Y = X_1 + X_2\), the sum of the scores, in each of the following cases:
Let \(Y = X_1 + X_2\) denote the sum of the scores.
| \(y\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\P(Y = y)\) | \(\frac{1}{36}\) | \(\frac{2}{36}\) | \(\frac{3}{36}\) | \(\frac{4}{36}\) | \(\frac{5}{36}\) | \(\frac{6}{36}\) | \(\frac{5}{36}\) | \(\frac{4}{36}\) | \(\frac{3}{36}\) | \(\frac{2}{36}\) | \(\frac{1}{36}\) |
| \(y\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\P(Y = y)\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{5}{64}\) | \(\frac{3}{32}\) | \(\frac{7}{64}\) | \(\frac{3}{16}\) | \(\frac{7}{64}\) | \(\frac{3}{32}\) | \(\frac{3}{32}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) |
| \(y\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\P(Y = y)\) | \(\frac{2}{48}\) | \(\frac{3}{48}\) | \(\frac{4}{48}\) | \(\frac{5}{48}\) | \(\frac{6}{48}\) | \(\frac{8}{48}\) | \(\frac{6}{48}\) | \(\frac{5}{48}\) | \(\frac{4}{48}\) | \(\frac{3}{48}\) | \(\frac{2}{48}\) |
In the dice experiment, select two dice and select the sum random variable. Run the simulation 1000 times and compare the empirical density function to the probability density function for each of the following cases:
Suppose that \(n\) standard, fair dice are rolled. Find the probability density function of the following variables:
Let \(U\) denote the minimum score and \(V\) the maximum score.
In the dice experiment, select fair dice and select each of the following random variables. Vary \(n\) with the scrollbar and note the shape of the density function. With \(n = 4\), run the simulation 1000 times and note the agreement between the empirical density function and the probability density function.
Recall again that for \(n \in \N_+\), \(\R^n\) is given the Borel \(\sigma\)-algebra \(\ms R^n\) and that \(\lambda^n\) denotes Lebesgue measure on \(\R^n\). So \(\lambda_1\) is length on sets in \(\ms R_1\), \(\lambda_2\) is area for sets in \(\ms R_2\), \(\lambda_3\) is volume for sets in \(\ms R_3\), and in general, \(\lambda^n\) is \(n\)-dimensional volume for sets in \(\ms R^n\). If \(S \in \ms R^n\) with \(0 \lt \lambda^n(S) \lt \infty\) then the uniform distribution on \( S \) is the continuous distribution with constant probability density function \(f\) defined by \( f(x) = 1 \big/ \lambda^n(S) \) for \( x \in S \).
Let \(Y = X^2\). Find the probability density function of \(Y\) and sketch the graph in each of the following cases:
Compare the distributions in . In part (c), note that even a simple transformation of a simple distribution can produce a complicated distribution. In this particular case, the complexity is caused by the fact that \(x \mapsto x^2\) is one-to-one on part of the domain \(\{0\} \cup (1, 3]\) and two-to-one on the other part \([-1, 1] \setminus \{0\}\).
On the other hand, the uniform distribution is preserved under a linear transformation of the random variable.
Suppose that \(S \in \ms R^n\) with \(0 \lt \lambda^n(S) \lt \infty\) and that \(X\) has the uniform distribution on \(S\). Let \(Y = a + B X\), where \(a \in \R^n\) and \(B\) is an invertible \(n \times n\) matrix. Then \(Y\) is uniformly distributed on \(T = \{a + B x: x \in S\}\).
For the following three exercises, recall that the standard uniform distribution is the uniform distribution on the interval \( [0, 1] \).
Suppose that \(X\) and \(Y\) are independent and that each has the standard uniform distribution. Let \(U = X + Y\), \(V = X - Y\), \( W = X Y \), \( Z = Y / X \). Find the probability density function of each of the follow:
Suppose that \(X\), \(Y\), and \(Z\) are independent, and that each has the standard uniform distribution. Find the probability density function of \((U, V, W) = (X + Y, Y + Z, X + Z)\).
\(g(u, v, w) = \frac{1}{2}\) for \((u, v, w)\) in the rectangular region \(T \subset \R^3\) with vertices \(\{(0,0,0), (1,0,1), (1,1,0), (0,1,1), (2,1,1), (1,1,2), (1,2,1), (2,2,2)\}\). So \((U, V, W)\) is uniformly distributed on \(T\).
Suppose that \((X_1, X_2, \ldots, X_n)\) is a sequence of independent random variables, each with the standard uniform distribution. Find the distribution function and probability density function of the following variables.
Both distributions in are beta distributions. More generally, all of the order statistics from a random sample of standard uniform variables have beta distributions, one of the reasons for the importance of this family of distributions.
In the order statistic experiment, select the uniform distribution.
Let \(f\) denote the probability density function of the standard uniform distribution.
In , you can see the behavior predicted by the central limit theorem beginning to emerge. Recall that if \((X_1, X_2, X_3)\) is a sequence of independent random variables, each with the standard uniform distribution, then \(f\), \(f^{*2}\), and \(f^{*3}\) are the probability density functions of \(X_1\), \(X_1 + X_2\), and \(X_1 + X_2 + X_3\), respectively. More generally, if \((X_1, X_2, \ldots, X_n)\) is a sequence of independent random variables, each with the standard uniform distribution, then the distribution of \(\sum_{i=1}^n X_i\) (which has probability density function \(f^{*n}\)) is known as the Irwin-Hall distribution with parameter \(n\).
Open the Special Distribution Simulator and select the Irwin-Hall distribution. Vary the parameter \(n\) from 1 to 3 and note the shape of the probability density function. (These are the density functions in ). For each value of \(n\), run the simulation 1000 times and compare the empricial density function and the probability density function.
A remarkable fact is that the standard uniform distribution can be transformed into almost any other distribution on \(\R\). This is particularly important for simulations, since many computer languages have an algorithm for generating random numbers, which are simulations of independent variables, each with the standard uniform distribution. Conversely, any continuous distribution supported on an interval of \(\R\) can be transformed into the standard uniform distribution.
Suppose first that \(F\) is a distribution function for a distribution on \(\R\) (which may be discrete, continuous, or mixed), and let \(F^{-1}\) denote the quantile function.
Suppose that \(U\) has the standard uniform distribution. Then \(X = F^{-1}(U)\) has distribution function \(F\).
The critical property satisfied by the quantile function (regardless of the type of distribution) is \( F^{-1}(p) \le x \) if and only if \( p \le F(x) \) for \( p \in (0, 1) \) and \( x \in \R \). Hence for \(x \in \R\), \(\P(X \le x) = \P\left[F^{-1}(U) \le x\right] = \P[U \le F(x)] = F(x)\).
Assuming that we can compute \(F^{-1}\), the previous exercise shows how we can simulate a distribution with distribution function \(F\). To rephrase the result, we can simulate a variable with distribution function \(F\) by simply computing a random quantile. Most of the apps in this project use this method of simulation. The first image below shows the graph of the distribution function of a rather complicated mixed distribution, represented in blue on the horizontal axis. In the second image, note how the uniform distribution on \([0, 1]\), represented by the thick red line, is transformed, via the quantile function, into the given distribution.
There is a partial converse , for continuous distributions.
Suppose that \(X\) has a continuous distribution on an interval \(S \subseteq \R\) Then \(U = F(X)\) has the standard uniform distribution.
For \( u \in (0, 1) \) recall that \( F^{-1}(u) \) is a quantile of order \( u \). Since \( X \) has a continuous distribution, \[ \P(U \ge u) = \P[F(X) \ge u] = \P[X \ge F^{-1}(u)] = 1 - F[F^{-1}(u)] = 1 - u \] Hence \( U \) is uniformly distributed on \( (0, 1) \).
Show how to simulate the uniform distribution on the interval \([a, b]\) with a random number. Using your calculator, simulate 5 values from the uniform distribution on the interval \([2, 10]\).
\(X = a + U(b - a)\) where \(U\) is a random number.
Suppose that \(X\) has the probability density function \(f\) given by \(f(x) = 3 x^2\) for \(0 \le x \le 1\). Find the probability density function of each of the following:
Random variables \(X\), \(U\), and \(V\) in have beta distributions, the same family of distributions that we saw in for the minimum and maximum of independent standard uniform variables. In general, beta distributions are widely used to model random proportions and probabilities, as well as physical quantities that take values in closed bounded intervals (which after a change of units can be taken to be \( [0, 1] \)). On the other hand, \(W\) has a Pareto distribution, named for Vilfredo Pareto.
Suppose that the radius \(R\) of a sphere has a beta distribution probability density function \(f\) given by \(f(r) = 12 r^2 (1 - r)\) for \(0 \le r \le 1\). Find the probability density function of each of the following:
Suppose that the grades on a test are described by the random variable \( Y = 100 X \) where \( X \) has the beta distribution with probability density function \( f \) given by \( f(x) = 12 x (1 - x)^2 \) for \( 0 \le x \le 1 \). The grades are generally low, so the teacher decides to curve
the grades using the transformation \( Z = 10 \sqrt{Y} = 100 \sqrt{X}\). Find the probability density function of
Recall that a Bernoulli trials sequence is a sequence \((X_1, X_2, \ldots)\) of independent, identically distributed indicator random variables. In the usual terminology of reliability theory, \(X_i = 0\) means failure on trial \(i\), while \(X_i = 1\) means success on trial \(i\). The basic parameter of the process is the probability of success \(p = \P(X_i = 1)\), so \(p \in [0, 1]\). The random process is named for Jacob Bernoulli.
For \(i \in \N_+\), the probability density function \(f\) of the trial variable \(X_i\) is \(f(x) = p^x (1 - p)^{1 - x}\) for \(x \in \{0, 1\}\).
By definition, \( f(0) = 1 - p \) and \( f(1) = p \). These can be combined succinctly with the formula \( f(x) = p^x (1 - p)^{1 - x} \) for \( x \in \{0, 1\} \).
Now let \(Y_n\) denote the number of successes in the first \(n\) trials, so that \(Y_n = \sum_{i=1}^n X_i\) for \(n \in \N\).
\(Y_n\) has the probability density function \(f_n\) given by \[ f_n(y) = \binom{n}{y} p^y (1 - p)^{n - y}, \quad y \in \{0, 1, \ldots, n\}\]
We have seen this derivation before. The number of bit strings of length \( n \) with 1 occurring exactly \( y \) times is \( \binom{n}{y} \) for \(y \in \{0, 1, \ldots, n\}\). By the Bernoulli trials assumptions, the probability of each such bit string is \( p^n (1 - p)^{n-y} \).
The distribution of \( Y_n \) is the binomial distribution with parameters \(n\) and \(p\).
For \( m, \, n \in \N \)
Part (a) can be proved directly from the definition of convolution, but the result also follows simply from the fact that \( Y_n = X_1 + X_2 + \cdots + X_n \).
From part (b) it follows that if \(Y\) and \(Z\) are independent variables, and that \(Y\) has the binomial distribution with parameters \(n \in \N\) and \(p \in [0, 1]\) while \(Z\) has the binomial distribution with parameter \(m \in \N\) and \(p\), then \(Y + Z\) has the binomial distribution with parameter \(m + n\) and \(p\).
Find the probability density function of the difference between the number of successes and the number of failures in \(n \in \N\) Bernoulli trials with success parameter \(p \in [0, 1]\)
\(f(k) = \binom{n}{(n+k)/2} p^{(n+k)/2} (1 - p)^{(n-k)/2}\) for \(k \in \{-n, 2 - n, \ldots, n - 2, n\}\)
Recall that the Poisson distribution with parameter \(t \in (0, \infty)\) has probability density function \(f\) given by \[ f_t(n) = e^{-t} \frac{t^n}{n!}, \quad n \in \N \] This distribution is named for Simeon Poisson and is widely used to model the number of random points in a region of time or space; the parameter \(t\) is proportional to the size of the region.
If \( a, \, b \in (0, \infty) \) then \(f_a * f_b = f_{a+b}\).
Let \( z \in \N \). Using the definition of convolution and the binomial theorem we have \begin{align} (f_a * f_b)(z) & = \sum_{x = 0}^z f_a(x) f_b(z - x) = \sum_{x = 0}^z e^{-a} \frac{a^x}{x!} e^{-b} \frac{b^{z - x}}{(z - x)!} = e^{-(a + b)} \frac{1}{z!} \sum_{x=0}^z \frac{z!}{x!(z - x)!} a^{x} b^{z - x} \\ & = e^{-(a+b)} \frac{1}{z!} \sum_{x=0}^z \binom{z}{x} a^x b^{n-x} = e^{-(a + b)} \frac{(a + b)^z}{z!} = f_{a+b}(z) \end{align}
The last result means that if \(X\) and \(Y\) are independent variables, and \(X\) has the Poisson distribution with parameter \(a \gt 0\) while \(Y\) has the Poisson distribution with parameter \(b \gt 0\), then \(X + Y\) has the Poisson distribution with parameter \(a + b\). In terms of the Poisson model, \( X \) could represent the number of points in a region \( A \) and \( Y \) the number of points in a region \( B \) (of the appropriate sizes so that the parameters are \( a \) and \( b \) respectively). The independence of \( X \) and \( Y \) corresponds to the regions \( A \) and \( B \) being disjoint. Then \( X + Y \) is the number of points in \( A \cup B \).
Recall that the exponential distribution with rate parameter \(r \in (0, \infty)\) has probability density function \(f\) given by \(f(t) = r e^{-r t}\) for \(t \in [0, \infty)\). This distribution is often used to model random times such as failure times and lifetimes. In particular, the times between arrivals in the Poisson model of random points in time have independent, identically distributed exponential distributions.
Show how to simulate, with a random number, the exponential distribution with rate parameter \(r\). Using your calculator, simulate 5 values from the exponential distribution with parameter \(r = 3\).
\(X = -\frac{1}{r} \ln(1 - U)\) where \(U\) is a random number. Since \(1 - U\) is also a random number, a simpler solution is \(X = -\frac{1}{r} \ln U\).
For next, recall that the floor and ceiling functions on \(\R\) are defined by \[ \lfloor x \rfloor = \max\{n \in \Z: n \le x\}, \; \lceil x \rceil = \min\{n \in \Z: n \ge x\}, \quad x \in \R\]
Suppose that \(T\) has the exponential distribution with rate parameter \(r \in (0, \infty)\). Find the probability density function of each of the following random variables:
Note that the distributions are geometric distributions on \(\N\) and on \(\N_+\), respectively. In many respects, the geometric distribution is a discrete version of the exponential distribution.
Suppose that \(T\) has the exponential distribution with rate parameter \(r \in (0, \infty)\). Find the probability density function of each of the following random variables:
In \(Y\) has a Pareto distribution while \(Z\) has an eextreme value distribution.
Suppose that \(X\) and \(Y\) are independent random variables, each having the exponential distribution with parameter 1. Let \(Z = \frac{Y}{X}\).
Suppose that \(X\) has the exponential distribution with rate parameter \(a \gt 0\), \(Y\) has the exponential distribution with rate parameter \(b \gt 0\), and that \(X\) and \(Y\) are independent. Find the probability density function of \(Z = X + Y\) in each of the following cases.
Suppose that \((T_1, T_2, \ldots, T_n)\) is a sequence of independent random variables, and that \(T_i\) has the exponential distribution with rate parameter \(r_i \gt 0\) for each \(i \in \{1, 2, \ldots, n\}\).
Note that the minimum \(U\) in part (a) of has the exponential distribution with parameter \(r_1 + r_2 + \cdots + r_n\). In particular, suppose that a series system has independent components, each with an exponentially distributed lifetime. Then the lifetime of the system is also exponentially distributed, and the failure rate of the system is the sum of the component failure rates.
In the order statistic experiment, select the exponential distribution.
Suppose again that \((T_1, T_2, \ldots, T_n)\) is a sequence of independent random variables, and that \(T_i\) has the exponential distribution with rate parameter \(r_i \gt 0\) for each \(i \in \{1, 2, \ldots, n\}\). Then \[ \P\left(T_i \lt T_j \text{ for all } j \ne i\right) = \frac{r_i}{\sum_{j=1}^n r_j} \]
When \(n = 2\), the result was shown in the section on joint distributions. Returning to the case of general \(n\), note that \(T_i \lt T_j\) for all \(j \ne i\) if and only if \(T_i \lt \min\left\{T_j: j \ne i\right\}\). Note that he minimum on the right is independent of \(T_i\) and by , has an exponential distribution with parameter \(\sum_{j \ne i} r_j\).
The result in the previous exercise is very important in the theory of continuous-time Markov chains. If we have a bunch of independent alarm clocks, with exponentially distributed alarm times, then the probability that clock \(i\) is the first one to sound is \(r_i \big/ \sum_{j = 1}^n r_j\).
Recall that the (standard) gamma distribution with shape parameter \(n \in \N_+\) has probability density function \[ g_n(t) = e^{-t} \frac{t^{n-1}}{(n - 1)!}, \quad t \in (0, \infty) \] With a positive integer shape parameter, as we have here, it is also referred to as the Erlang distribution, named for Agner Erlang. This distribution is widely used to model random times under certain basic assumptions. In particular, the \( n \)th arrival times in the Poisson model of random points in time has the gamma distribution with parameter \( n \).
Let \( g = g_1 \), and note that this is the probability density function of the exponential distribution with parameter 1, which was the topic of our last discussion.
If \( m, \, n \in \N_+ \) then
Part (a) hold trivially when \( n = 1 \). Also, for \( t \in [0, \infty) \), \[ g_n * g(t) = \int_0^t g_n(s) g(t - s) \, ds = \int_0^t e^{-s} \frac{s^{n-1}}{(n - 1)!} e^{t-s} \, ds = e^{-t} \int_0^t \frac{s^{n-1}}{(n - 1)!} \, ds = e^{-t} \frac{t^n}{n!} = g_{n+1}(t) \] Part (b) follows from (a).
Part (b) means that if \(X\) has the gamma distribution with shape parameter \(m\) and \(Y\) has the gamma distribution with shape parameter \(n\), and if \(X\) and \(Y\) are independent, then \(X + Y\) has the gamma distribution with shape parameter \(m + n\). In the context of the Poisson model, part (a) means that the \( n \)th arrival time is the sum of the \( n \) independent interarrival times, which have a common exponential distribution.
Suppose that \(T\) has the gamma distribution with shape parameter \(n \in \N_+\). Find the probability density function of \(X = \ln T\).
\(h(x) = \frac{1}{(n-1)!} \exp\left(-e^x\right) e^{n x}\) for \(x \in \R\)
Recall that the Pareto distribution with shape parameter \(a \in (0, \infty)\) has probability density function \(f\) given by \[ f(x) = \frac{a}{x^{a+1}}, \quad 1 \le x \lt \infty\] Members of this family have already come up in several of the previous exercises. The Pareto distribution, named for Vilfredo Pareto, is a heavy-tailed distribution often used for modeling income and other financial variables.
Suppose that \(X\) has the Pareto distribution with shape parameter \(a\). Find the probability density function of each of the following random variables:
In , \(V\) also has a Pareto distribution but with parameter \(\frac{a}{2}\); \(Y\) has the beta distribution with parameters \(a\) and \(b = 1\); and \(Z\) has theexponential distribution with rate parameter \(a\).
Show how to simulate, with a random number, the Pareto distribution with shape parameter \(a\). Using your calculator, simulate 5 values from the Pareto distribution with shape parameter \(a = 2\).
Recall that the standard normal distribution has probability density function \(\phi\) given by \[ \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^2}, \quad z \in \R\]
Suppose that \(Z\) has the standard normal distribution, and that \(\mu \in (-\infty, \infty)\) and \(\sigma \in (0, \infty)\).
Random variable \(X\) has the normal distribution with location parameter \(\mu\) and scale parameter \(\sigma\). The normal distribution is perhaps the most important distribution in probability and mathematical statistics, primarily because of the cemtral limit theorem, one of the fundamental theorems. It is widely used to model physical measurements of all types that are subject to small, random errors.
Suppose that \(Z\) has the standard normal distribution. Find the probability density function of \(Z^2\) and sketch the graph.
\(g(v) = \frac{1}{\sqrt{2 \pi v}} e^{-\frac{1}{2} v}\) for \( 0 \lt v \lt \infty\)
Random variable \(V\) has the chi-square distribution with 1 degree of freedom.
Suppose that \( X \) and \( Y \) are independent random variables, each with the standard normal distribution, and let \( (R, \Theta) \) be the standard polar coordinates \( (X, Y) \). Find the probability density function of
Note that the joint PDF of \( (X, Y) \) is \[ f(x, y) = \phi(x) \phi(y) = \frac{1}{2 \pi} e^{-\frac{1}{2}\left(x^2 + y^2\right)}, \quad (x, y) \in \R^2 \] From , the PDF of \( (R, \Theta) \) is \[ g(r, \theta) = f(r \cos \theta , r \sin \theta) r = \frac{1}{2 \pi} r e^{-\frac{1}{2} r^2}, \quad (r, \theta) \in [0, \infty) \times [0, 2 \pi) \] From the factorization theorem for joint PDFs, it follows that \( R \) has probability density function \( h(r) = r e^{-\frac{1}{2} r^2} \) for \( 0 \le r \lt \infty \), \( \Theta \) is uniformly distributed on \( [0, 2 \pi) \), and that \( R \) and \( \Theta \) are independent.
The distribution of \( R \) is the (standard) Rayleigh distribution, and is named for John William Strutt, Lord Rayleigh.
The standard normal distribution does not have a simple, closed form quantile function, so the random quantile method of simulation does not work well. However, the last exercise points the way to an alternative method of simulation.
Show how to simulate a pair of independent, standard normal variables with a pair of random numbers. Using your calculator, simulate 6 values from the standard normal distribution.
The Rayleigh distribution in the last exercise has CDF \( H(r) = 1 - e^{-\frac{1}{2} r^2} \) for \( 0 \le r \lt \infty \), and hence quantle function \( H^{-1}(p) = \sqrt{-2 \ln(1 - p)} \) for \( 0 \le p \lt 1 \). Thus we can simulate the polar radius \( R \) with a random number \( U \) by \( R = \sqrt{-2 \ln(1 - U)} \), or a bit more simply by \(R = \sqrt{-2 \ln U}\), since \(1 - U\) is also a random number. We can simulate the polar angle \( \Theta \) with a random number \( V \) by \( \Theta = 2 \pi V \). Then, a pair of independent, standard normal variables can be simulated by \( X = R \cos \Theta \), \( Y = R \sin \Theta \).
Suppose that \(X\) and \(Y\) are independent random variables, each with the standard normal distribution. Find the probability density function of \(T = X / Y\).
As usual, let \( \phi \) denote the standard normal PDF, so that \( \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-z^2/2}\) for \( z \in \R \). Using the theorem on quotient in above, the PDF \( f \) of \( T \) is given by \[f(t) = \int_{-\infty}^\infty \phi(x) \phi(t x) |x| dx = \frac{1}{2 \pi} \int_{-\infty}^\infty e^{-(1 + t^2) x^2/2} |x| dx, \quad t \in \R\] Using symmetry and a simple substitution, \[ f(t) = \frac{1}{\pi} \int_0^\infty x e^{-(1 + t^2) x^2/2} dx = \frac{1}{\pi (1 + t^2)}, \quad t \in \R \]
Random variable \(T\) has the (standard) Cauchy distribution, named after Augustin Cauchy.
Suppose that a light source is 1 unit away from position 0 on an infinite straight wall. We shine the light at the wall an angle \( \Theta \) to the perpendicular, where \( \Theta \) is uniformly distributed on \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Find the probability density function of the position of the light beam \( X = \tan \Theta \) on the wall.
The PDF of \( \Theta \) is \( f(\theta) = \frac{1}{\pi} \) for \( -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} \). The transformation is \( x = \tan \theta \) so the inverse transformation is \( \theta = \arctan x \). Recall that \( \frac{d\theta}{dx} = \frac{1}{1 + x^2} \), so by the change of variables formula , \( X \) has PDF \(g\) given by \[ g(x) = \frac{1}{\pi \left(1 + x^2\right)}, \quad x \in \R \]
Thus, \( X \) also has the standard Cauchy distribution. Clearly we can simulate a value of the Cauchy distribution by \( X = \tan\left(-\frac{\pi}{2} + \pi U\right) \) where \( U \) is a random number. This is the random quantile method.
Open the Cauchy experiment, which is a simulation of the light problem in the previous exercise. Keep the default parameter values and run the experiment in single step mode a few times. Then run the experiment 1000 times and compare the empirical density function and the probability density function.