\(\newcommand{\P}{\mathbb{P}}\) \(\newcommand{\E}{\mathbb{E}}\) \(\newcommand{\R}{\mathbb{R}}\) \(\newcommand{\N}{\mathbb{N}}\) \(\newcommand{\var}{\text{var}}\) \(\newcommand{\sd}{\text{sd}}\) \(\newcommand{\cov}{\text{cov}}\) \(\newcommand{\skw}{\text{skew}}\) \(\newcommand{\kurt}{\text{kurt}}\) \(\newcommand{\sgn}{\text{sgn}}\) \(\newcommand{\erfc}{\text{erfc}}\) \(\newcommand{\erf}{\text{erf}}\) \(\newcommand{\ms}{\mathscr}\) \(\newcommand{\Rta}{\Rightarrow}\) \(\newcommand{\bs}{\boldsymbol}\)
  1. Reliability
  2. 3. Standard Continuous Spaces
  3. 1
  4. 2
  5. 3
  6. 4
  7. 5
  8. 6
  9. 7

6. Norm Graphs

Preliminaries

In this section we study an interesting collection of graphs on \(\R^n\) that are induced (in the sense of Section 1.7 ) by the standard graph \(([0, \infty), \le)\) and the standard norm functions. For \(n \in \N_+\), we give \(\R^n\) the usual Borel \(\sigma\)-algebra \(\ms R^n\) and \(n\)-dimensional Lebesgue measure \(\lambda^n\) as the reference measure. For \(m \in \N\) and \(m \le n\), we also use \(\lambda^m\) to denote Lebesgue measure (or more technically Hausdorff measure) on an \(m\)-dimensional Riemannian (smooth) manifold of \(\R^n\). In particular, recall that \(\lambda^0\) is counting measure.

Recall the standard norms on \(\R^n\) for \(n \in \N_+\):

  1. For \(k \in [1, \infty)\), the \(k\) norm is given by \[\|\bs{x}\|_k = \left(\sum_{i=1}^n |x_i|^k\right)^{1/k}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \R^n\]
  2. The \(\infty\) norm is defined by \[\|\bs x\|_\infty = \max\{|x_i| : i \in \{1, 2, \ldots, n\}\}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \R^n\]

When \(n = 1\), the norms are all the same: \(\|x\|_k = |x|\) for \(k \in [1, \infty]\) and \(x \in \R\). When \(n \in \{2, 3, \ldots\}\), the norms are all different, of course. For \(n \in \N_+\) and \(k \in [1, \infty]\), the mapping \(\|\cdot\|_k\) is a measurable function from \(\R^n\) onto \([0, \infty)\).

For \(n \in \N_+\) and \(k \in [1, \infty]\), let \((\R^n, \Rta_k)\) denote the graph induced by \(\|\cdot\|_k\) and the standard graph \(([0, \infty), \le)\), so that \(\bs x \Rta_k \bs y\) if and only if \(\|\bs x\|_k \le \|\bs y\|_k\) for \(\bs x, \, \bs y \in \R^n\).

Note that the relation \(\Rta_k\) is reflexive and transitive.

For \(n \in \N_+\) and \(k \in [1, \infty]\), define

  1. \(b_{n,k} = \lambda^n\{\bs{x} \in \R^n: \|\bs x\|_k \le 1\}\), the volume of the unit ball in \(\R^n\) under the \(k\) norm.
  2. \(c_{n,k} = \lambda^{n-1}\{\bs x \in \R^n: \|\bs x\|_k = 1\}\), the surface area of the unit ball in \(\R^n\) under the \(k\) norm.

Of course, the terms volume and area are used in a general sense. As usual, \(\Gamma\) denotes the ordinary gamma special function. The following result is well known (see for example the paper by Slavik).

Volumes of the unit ball.

  1. For \(n \in \N_+\) and \(k \in [1, \infty)\) \[b_{n,k} = 2^n \frac{\Gamma^n(1 + 1/k)}{\Gamma(1 + n / k)} \]
  2. \(b_{n,\infty} = 2^n\) for \(n \in \N_+\)

Note that \( \lim_{k \to \infty} b_{n,k} = b_{n, \infty}\), justifying the notation.

For \(n \in \N_+\), \(k \in [1, \infty]\), and \(t \in [0, \infty)\),

  1. \(b_{n,k} t^n = \lambda^n \{\bs x \in \R^n: \|\bs x\|_k \le t\}\), the volume of the ball in \(\R^n\) of radius \(t\) in the \(k\) norm.
  2. \(c_{n,k} t^{n-1} = \lambda^{n-1} \{\bs x \in \R^n: \|\bs x\|_k = t\}\), the surface area of the ball in \(\R^n\) of radius \(t\) in the \(k\) norm.

As functions of \(t \in [0, \infty)\), the surface area is a multiple of the derivative of the volume, or equivalently, for \(n \in \N_+\) and \(k \in [1, \infty]\) there exists \(a_{n,k} \in (0, \infty)\) such that \(c_{n,k} = n a_{n,k} b_{n,k}\)

For most values of \(n \in \N_+\) and \(k \in [1, \infty]\), there is no simple closed formula for \(a_{n,k}\), but there are a few exceptions:

Special cases:

  1. \(a_{n,2} = 1\) for \(n \in \N_+\) so \[c_{n,2} = n b_{n,2} = 2 \frac{\pi^{n/2}}{\Gamma(n/2)}, \quad n \in \N_+\]
  2. \(a_{n,\infty} = 1\) for \(n \in \N_+\) so \[c_{n,\infty} = n b_{n,\infty} = n 2^n, \quad n \in \N_+\]
  3. \(a_{n,1} = \sqrt n\) so \[c_{n,1} = n \sqrt n b_{n,1} = \frac{2^n \sqrt n}{(n - 1)!}, \quad n \in \N_+ \]

The basic assumption in Section 1.7 on induced graphs is satisfied:

For \(n \in \N_+\) and \(k \in [1, \infty]\), let \(\R^n_k(t) = \{\bs x \in \R^n: \|x\|_k = t\}\), the surface of the ball of radius \(t \in [0, \infty)\) relative to the \(k\) norm, and the inverse image at \(t\) of the mapping \(\bs x \mapsto \|\bs x\|_k\) from \(\R^n\) onto \([0, \infty)\). Then \[\lambda^n(A) = \int_0^\infty \frac{1}{a_{n, k}} \lambda^{n - 1}[A \cap \R^n_k(t)] dt, \quad A \in \ms R^n\]

This is essentially the co-area formula for computing \(\lambda^n(A)\). Again in the notation Section 1.7, note also that for \(n \in \N_+\) and \(k \in [1, \infty]\), the function \(\beta_{n,k}\) is given by \[\beta_{n,k}(t) = \frac{1}{a_{n,k}} \lambda^{n-1}[\R^n_k(t)] = \frac{c_{n,k}}{a_{n,k}} t^{n-1} = n b_{n,k} t^{n-1}, \quad t \in [0, \infty)\] Going forward, we will surpress the dependence on \(n\) and \(k\) except when necessary. Our first result is the walk function.

The left walk function \(u_m\) of order \(m \in \N_+\) for the graph \((\R^n, \Rta_k)\) is given by \[u_m(\bs x) = \frac{b^m_{n,k}}{m!} \|\bs x\|^{m n}_k, \quad \bs x \in \R^n\]

Details:

This follows from \(u_0 = 1\) and the recursion relation \[u_{m + 1}(\bs x) = \int_{\bs y \Rta \bs x} u_m(\bs y) \, d\lambda^n(\bs y), \quad \bs x \in \R^n\] The integral can be evaluated using the co-area formula in . But there is also a simple combinatorial argument. For \(\bs x \in \R^n\), the measure of the set \[\{(\bs y_1, \bs y_2, \ldots \bs y_m): \bs y_i \in \R^n, \, \|\bs y_i\|_k \le \|\bs x\|_k \text{ for each } i \in \{1, 2, \ldots, m\}\}\] is of course \(b^m_{n,k} \|\bs x\|_k^{m n}\). But another algorithm for constructing this set is to select a walk \((\bs x_1, \bs x_2, \ldots, \bs x_m)\) of length \(m\) ending in \(\bs x\) for the graph \((\R^n, \Rta_k)\) and then select an ordering \((\bs y_1, \bs y_2, \ldots, y_m)\) of \((\bs x_1, \bs x_2, \ldots \bs x_m)\). By definition, the measure of the set of walks is \(u_m(\bs x)\), and there are \(m!\) permutations of each path, so it follows that \[m! u_m(\bs x) = b^m_{n,k} \|\bs x\|_k^{m n}\]

Note that \(u_m(\bs x)\) is the left walk function \(t \mapsto t^m / m!\) of order \(m\) for the standard graph \(([0, \infty), \le)\), evaluated at \(b_{n,k} \|\bs x\|_k^n\) which is the volume of the ball of radius \(\|\bs x\|_k\) in the \(k\) norm. We will see this pattern repeatedly. In particular, it follows that the left generating function of \((\R^n, \Rta_k)\) is the left generating function of \(([0, \infty), \le)\) evaluated at \(b_{n,k} \|\bs x\|_k^n\):

The left generating function \(U\) of \((\R^n, \Rta_k)\) is given by \[U(\bs x, t) = \exp\left(b_{n,k} \|\bs x\|^n_k t\right), \quad \bs x \in \R^n, \, t \in \R\]

Details:

This follows easily from the definition: \[U(\bs x, t) = \sum_{m = 0}^\infty u_m(\bs x) t^m, \quad \bs x \in \R^n, \, t \in \R\]

Probability

Suppose now that \(\bs X = (X_1, X_2, \ldots, X_n)\) is a random variable in \(\R^n\), so that \(\|\bs X\|_k\) is the corresponding induced variable in \([0, \infty)\).

Basic functions:

  1. The reliability function \(\hat F\) of \(\|\bs X\|_k\) for \(([0, \infty), \le)\) is the usual reliability function: \(\hat F(t) = \P(\|\bs X\|_k \ge t)\) for \(t \in [0, \infty)\).
  2. The reliability function \(F\) of \(\bs X\) for \((\R^n, \Rta_k)\) \(F(\bs x) = \hat F(\|\bs x\|_k)\) for \(\bs x \in \R^n\).
  3. If \(\bs X\) has density function \(f\) then \(\|\bs X\|_k\) has density function \(\hat f\) given by \[\hat f(t) = \int_{\R^n_k(t)} \frac{1}{a_{n,k}} f(\bs x) \, d\lambda^{n-1}(\bs x), \quad t \in [0, \infty)\]

We assume that \(\hat f(t) \gt 0\) for almost all \(t \in [0, \infty)\), so that \(\hat F(t) \gt 0\) for almost all \(t \in [0, \infty)\) and \(F(\bs x) \gt 0\) for almost all \(\bs x \in \R^n\). For \(t \in [0, \infty)\), the conditional distribution of \(\bs X\) given \(\|\bs X\|_k = t\) has density function \(\bs x \mapsto f(\bs x \mid t) = f(\bs x) / \hat f(t)\) on \(\R^n_k(t)\). The general results of Section 1.7 apply, of course, but as usual we are most interested in constant rate distributions. Here are the main results:

For \(n \in \N_+\) and \(k \in [1, \infty]\), the graph \((\R^n, \Rta_k)\) has a unique distribution with constant rate \(\alpha\) for each \(\alpha \in (0, \infty)\). If random variable \(\bs X\) has this distribution then

  1. \(\|\bs X\|_k\) has the Weibull distribution with shape parameter \(n\) and scale parameter \(\left(\alpha b_{n,k}\right)^{-1 / n}\), with density \(\hat f\) defined by \[\hat f(t) = \alpha n b_{n, k} t^{n - 1} \exp\left(-\alpha b_{n,k} t^n\right), \quad t \in [0, \infty)\]
  2. \(\bs X\) has density function \(f\) defined by \[f(\bs x) = \alpha \exp\left(-\alpha b_{n,k} \|\bs x\|^n_k\right), \quad \bs x \in \R^n\]
  3. The conditional distribution of \(\bs X\) given \(\|\bs X\|_k = t\) is uniform on \(\R^n_k(t)\) for \(t \in [0, \infty)\).
Details:

The results follow from Section 1.7. Specifically, \(\bs X\) has constant rate \(\alpha \in (0, \infty)\) for \((\R^n, \Rta_k)\) if and only if \(\|\bs X\|_k\) has rate function \(\hat r\) for \(([0, \infty), \le)\) where \(\hat r(t) = \alpha \beta_{n,k}(t) = \alpha n b_{n,k} t^{n-1}\) for \(t \in [0, \infty)\). So \(\|\bs X\|_k\) has reliability function \(\hat F\) for \(([0, \infty), \le)\) given by \[\hat F(t) = \exp\left(-\int_0^t \hat r(s) \, ds\right) = \exp\left(-\alpha b_{n,k} t^n\right), \quad t \in [0, \infty)\] which we recognize as the Weibull distribution in (a). And then \(\bs X\) has reliability function \(F\) for \((\R^n, \Rta_k)\) given by \(F(\bs x) = \hat F(\|\bs x\|_k)\) for \(\bs x \in \R^n\), and has density function \(f\) given by \(f(\bs x) = \alpha \hat F(\|\bs x\|_k)\) for \(\bs x \in \R^n\).

Note that \(f(\bs x)\) is the ordinary exponential density function for the standard graph \(([0, \infty), +)\), with rate \(\alpha\), evaluated at \(b_{n,k} \|\bs x\|_k^n\), which again is the volume of the ball of radius \(\|\bs x\|_k\) in the \(k\) norm.

Let \(n \in \N_+\) and \(k \in [1, \infty]\), and suppose that \(\bs X = (X_1, X_2, \ldots, X_n)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^n, \Rta_k)\).

  1. \(\bs X\) is an exchangeable sequence of random variables.
  2. \(\bs X\) is a sequence of identically distributed variables.
  3. The distribution of \(\bs X\) is symmetric in each variable.
  4. \(\bs X\) is a sequence of mean 0 and pairwise-uncorrelated variables.
Details:

These results follow from the density function in part (b) of .

  1. Note that \(f(x_1, x_2, \ldots, x_n)\) is invariant under a permutation of \((x_1, x_2, \ldots, x_n)\).
  2. This follows from (a).
  3. Note that if \((e_1, e_2, \ldots, e_n) \in \{-1, 1\}^n\) then \(f(e_1 x_1, e_2 x_2, \ldots, e_n x_n) = f(x_1, x_2, \ldots, x_n)\). It follows that \((e_1 X_1, e_2 X_2, \ldots, e_n X_n)\) has the same distribution as \(\bs X\).
  4. This follows from (c): For distinct \(i, j \in \{1, 2, \ldots, n\}\), we have \(\E(X_i) = \E(-X_i) = -\E(X_i)\) and \(\E(X_i X_j) = \E(-X_i X_j) = - \E(X_i X_j)\). Hence \(\E(X_i) = 0\) and then \(\cov(X_i, X_j) = \E(X_i X_j) = 0\).

Let \(n \in \N_+\) and \(k \in [1, \infty]\) and consider the family of constant rate distributionn on \((\R^n, \Rta_k)\).

  1. The family of distributions is a scale family with scale parameter \(\alpha^{-1 / n}\). That is, if \(\bs{Z}\) has the standard distribution with rate 1 and if \(\alpha \in (0, \infty)\) then \(\bs X = \alpha^{-1 / n} \bs{Z}\) has the distribution with rate \(\alpha\)
  2. The family of distributions is a one-parameter exponential family with natural parameter \(-\alpha b_{n, k}\) and natural statistic \(\|x\|_k^n\).
Details:
  1. Suppose that \(\bs{Z}\) has the distribution with rate 1 and density function \(g\), so that \(g(\bs{z}) = \exp\left(-b_{n, k} \|z\|_k^n\right)\). Let \(c \in (0, \infty)\) and let \(\bs{X} = c \bs{Z}\). The transformation is \(\bs{x} = c \bs{z}\) and so the inverse tranformation is \(\bs{z} = \frac{1}{c} \bs{x}\) which has Jacobian \(\frac{1}{c^n}\). So \(\bs{X}\) has density function \(f\) given by \[ f(\bs{x}) = \frac{1}{c^n} g\left(\frac{\bs{x}}{c}\right) = \frac{1}{c^n} \exp\left(-b_{n, k} \left\|\frac{\bs{x}}{c}\right\|_k^n\right) = \frac{1}{c^n} \exp\left(-\frac{1}{c^n} b_{n, k} \|\bs{x}\|_k^n\right), \quad \bs{x} \in \R^n \] If we replace \(c\) with \(\alpha^{-1 / n}\) then we get the distribution with constant rate \(\alpha\).
  2. This follows from the definition of the general exponential family.

Suppose again that \(\bs X\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^n, \Rta_k)\). Then \(\bs X\) maximizes entropy over all random variables \(\bs Y\) in \(\R^n\) with \(\E(\|\bs Y\|_k^n) = 1 / (\alpha b_{n,k})\).

The Case \(n = k\)

The case where the norm index is the same as the dimension is particularly interesting. In this case, abbreviate \(b_{n,n}\) by \(b_n\) so that \[b_n = \left[\frac{2}{n} \Gamma\left(\frac{1}{n}\right)\right]^n = 2^n \Gamma^n\left(1 + \frac{1}{n}\right), \quad n \in \N_+\] where again \(\Gamma\) is the ordinary gamma special function.

Let \(n \in \N_+\) and suppose that \(\bs X = (X_1, X_2, \ldots, X_n)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^n, \Rta_n)\). Then \(\bs X\) is a sequence of independent, identically distributed variables with common density function \(g\) on \(\R\) define by \[g(x) = \alpha^{1/n} \exp(-\alpha b_n |x|^n), \quad x \in \R\]

Details:

These results follow immediately from part (b) of and the factorization theorem for independence.

The distributions on \(\R\) defined by the density function in (b) are generalized normal distributions (see the papers by Goodman and by Sinz et al.) The distribution actually makes sense for any \(n \in (0, \infty)\), not just positive integers, and forms an interesting class of special distributions. We will use terminology that is appropriate for our setting:

For \(n, \, \alpha \in (0, \infty)\), the distribution defined in is the generalized normal distribution with index \(n\) and rate \(\alpha\).

For \(n = 1, \, 2\) the generalized normal distributions reduce to standard special distributions:

Suppose that \(X\) has the generalized normal distribution on \(\R\) with index \(n \in (0, \infty)\) and rate \(\alpha \in (0, \infty)\).

  1. If \(n = 1\), \(X\) has the Laplace (double exponential) distribution with parameter \(2 \alpha\).
  2. If \(n = 2\), \(X\) has the normal distribution with mean 0 and variance \(1 / 2 \alpha \pi\).
Details:

Let \(g\) denote the PDF as in .

  1. When \(n = 1\), \[g(x) = \alpha e^{-2 \alpha |x|}, \quad x \in \R\]
  2. When \(n = 2\), \[g(x) = \sqrt{\alpha} e^{-\alpha \pi x^2}, \quad x \in \R\]

As in , \(\alpha^{-1 / n}\) is a scale parameter of the family of distributions. That is, if \(Z\) has the standard distribution with index \(n\) and rate \(1\), and with density function \(z \mapsto e^{-b_n |z|^n}\) then \(X = \alpha^{-1 / n} Z\) has the distribution with index \(n\) and rate \(\alpha\) and with density function \(g\) in . The following theorems gives some basic properties:

Let \(g\) denote the density function of the generalized normal distribution on \(\R\) with index \(n \in (0, \infty)\) and rate \(\alpha \in (0, \infty)\), given in .

  1. \(g\) is symmetric about 0.
  2. \(g\) increases on \((-\infty, 0)\) and decreases on \((0, \infty)\).
  3. If \(n \ge 1\), the maximum of \(g\) (the mode of the distribution) is at \(x = 0\). If \(0 \lt n \lt 1\), \(g\) has am asymptote at \(x = 0\).
  4. If \(n \gt 1\), \(g\) is concave upward on \((-\infty, -x_n)\) and \((x_n, \infty)\), and concave downward on \((-x_n, 0)\) and \((0, x_n)\) where \[x_n = \left(\frac{n - 1}{n \, \alpha \, b_n}\right)^{1 / n}\]
Details:

Part (a) is clear. Parts (b) and (c) follow from standard calculus.

Note that the inflection points in (c) converge to \(\pm 1 / 2\) as \(n \to \infty\).

Suppose again that \(X\) has the generalized normal distribution on \(\R\) with index \(n \in (0, \infty)\) and rate \(\alpha \in (0, \infty)\), with density function \(g\).

  1. If \(m \in \N_+\) is odd then \(\E(X^m) = 0\). If \(m \in \N_+\) is even then \[\E(X^m) = \alpha^{-m/n} \left(\frac{n}{2}\right)^m \frac{\Gamma\left(\frac{m+1}{n}\right)}{\Gamma^{m+1}\left(\frac{1}{n}\right)}\]
  2. The distribution of \(X\) converges to the uniform distribution on \([-\frac 1 2, \frac 1 2]\) as \(n \to \infty\).
  3. \(X\) has entropy \(H(X) = (1 + \ln \alpha) / n\).
Details:

The results follow from known results for the generalized normal distribution, but we will give separate proofs for completeness. Suppose that \(Z\) has the standard distribution with index \(n\) and rate \(1\) so that \(X = \alpha^{-1 / n}\) has the generalized normal distribution with index \(n\) are rate \(\alpha\).

  1. If \(m \in \N_+\), then by a well-known integration formula, \[\E(|Z|^m) = 2 \int_0^\infty z^m e^{-b_n z^n} \, dz = \left(\frac{n}{2}\right)^m \frac{\Gamma\left(\frac{m+1}{n}\right)}{\Gamma^{m+1}\left(\frac{1}{n}\right)}\] If \(m\) is odd, then \(\E(Z^m) = 0\) since the distribution is symmetric about 0. If \(m\) is even then of course \(\E(|Z|^m) = \E(Z^m)\). Finally, \(E(X^m) = \alpha^{-m / n} \E(Z^m)\).
  2. Again, it suffices to consider the standard variable \(Z\), since \(\alpha^{-1 / n} \to 1\) as \(n \to \infty\). But \(e^{-b_n |z|^n} \to 1\) as \(n \to \infty\) if \(|z| \lt \frac 1 2\) and \(e^{-b_n |z|^n} \to 0\) as \(n \to \infty\) if \(|z| \ge \frac 1 2\).

In particular, the variance and kurtosis are given by \[\var(X) = \alpha^{-2 / n} \left(\frac{n}{2}\right)^2 \frac{\Gamma(3 / n)}{\Gamma^3(1 / n)}, \; \kurt(X) = \frac{\Gamma(5 / n) \Gamma(1 / n)}{\Gamma^2(3 / n)}, \quad n \in \N_+\] This class of distributions on \(\R\) is interesting because it generalizes the Laplace, the 0 mean normal, and because of the surprising convergence result in part (d). Since the distribution of \(X\) is symmetric about \(0\), it also follows that \(|X|\) has density function \(2 g\) on \([0, \infty)\), \(\sgn(X)\) is uniformly distributed on \(\{-1, 1\}\), and that \(|X|\) and \(\sgn(X)\) are independent.

The app below is a simulation of the generalized normal distribution. The parameters \(n\) and \(\alpha\) can be varied with the scrollbars.

The (ordinary) distribution function can be expressed in terms of the gamma and (upper) incomplete gamma functions, but is not particularly helpful:

Suppose again that \(X\) has the generalized normal distribution on \(\R\) with index \(n \in (0, \infty)\) and rate \(\alpha \in (0, \infty)\). The distribution function \(G\) is given by \[G(y) = 1 - \frac{\Gamma\left[1 / n, 2^n y^n \Gamma^n(1 + 1 / n)\right]}{2 \Gamma(1 / n)}, \quad y \ge 0\] and by symmetry, \(G(y) = 1 - G(-y)\) for \(y \lt 0\).

We can rephrase parts (a) and (b) of :

Constant rate distributions for \(n = 1, \, 2\).

  1. If \(X\) has the Laplace distribution on \(\R\) with scale parameter \(a \in (0, \infty)\), then \(X\) has constat rate \(\alpha = 1 / (2 a)\) for the graph \((\R, \Rta)\).
  2. If \(X\) and \(Y\) are independent, and each has the normal distribution with mean 0 and standard deviation \(\sigma \in (0, \infty)\), then \((X, Y)\) has constant rate \(\alpha = 1 / (2 \pi \sigma^2)\) for the graph \((\R^2, \Rta_2)\).

Return to the simulationn of the generalized normal distribution in . Set \(n = 1\) and \(n = 2\) and note the shape of the graph. Run the simulation in both cases and compare the empirical density function to the probability density function.

Part (d) of suggests that the uniform distribution on \([-\frac 1 2, \frac 1 2]\) has constant rate for the graph corresponding to \(n = \infty\). That is indeed true, properly understood. Let \(I = [-\frac 1 2, \frac 1 2]\) and consder \(I^\infty\) with norm \[\|\bs x\|_\infty = \sup\{|x_i|: i \in \N_+\}, \quad \bs x = (x_1, x_2, \ldots) \in I^\infty\] The graph \((I^\infty, \Rta_\infty)\) induced by the function \(\bs x \mapsto \|\bs x\|_\infty\) and the standard graph \(([0, \infty), \le)\) is defined in the usual way: \(\bs x \Rta_\infty \bs y\) if and only if \(\|\bs x\|_\infty \le \|\bs y\|_\infty\) for \(\bs x, \, \bs y \in I^\infty\). In terms of the measure structure, we give \(I\) the usual Borel \(\sigma\)-algebra \(\ms I\) and \(I^\infty\) the corresponding product \(\sigma\)-algebra \(\ms I^\infty \), namely the \(\sigma\)-algebra generated by sets of the form \(\prod_{i = 1}^\infty A_i\) where \(A_i \in \ms I\) for each \(i \in \N_+\) and \(A_i = I\) for all but finitely many \(i \in \N_+ \). Finally, for the reference measures, let \(\lambda\) denote Lebesgue measure on \((I, \ms I)\) (which is simply the uniform probability distribution) and let \(\lambda^\infty\) the corresponding product measure on \((I^\infty, \ms I^\infty)\).

Suppose that \(\bs X = (X_1, X_2, \ldots)\) is a sequence of independent variables, each uniformly distributed on \([-\frac 1 2, \frac 1 2]\). Then \(\bs X\) has constant rate 1 for the graph \((I^\infty, \Rta_\infty)\).

Details:

Random variable \(X_i\) has constant density function \(1\) on \(I\) and similarly \(\bs X\) also has constant density function \(1\) on \(I^\infty\). On the other hand, the reliability function \(F\) of \(\bs X\) for \((I^\infty, \Rta_\infty)\) is \[F(\bs x) = \P(\bs x \Rta_\infty \bs X) = \P(\|\bs x\|_\infty \le \|\bs X\|_\infty), \quad \bs x \in I^\infty\] But \(\P(\|\bs X\|_\infty = \frac 1 2) = 1\). To see this, note that if \(\epsilon \gt 0\) then with probability \(1\), \(|X_i| \gt \frac 1 2 - \epsilon\) for some (and in fact infinitely many) \(i \in \N_+\). Hence, \(F\) is also the constant function 1 on \(I^\infty\) so \(F = f\).

Return again to the simulation of the generalized normal distribution in . Starting with \(n = 1\), increase \(n\) to the maximum value and note the change in shape of the graph.

Error Functions

There are two interesting and tractible cases in the bivariate setting (\(n = 2\)) corresponding to the extreme values of the norm parameter (\(k = 1\) and \(k = \infty\)). In both cases, various distribution functions are best expressed in terms of the error function and its complement.

The error function \(\erf\) is defined by \[ \erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt, \quad x \in \R \] and the complementary error function is \(\erfc = 1 - \erf\).

The error function is closely related to the standard normal distribution function \(\Phi\): \[ \Phi(x) = \frac 1 2 + \frac 1 2 \erf\left(\frac{x}{\sqrt{2}}\right), \quad x \in \R \]

The Case \(n = 2\) and \(k = 1\).

Suppose that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_1)\). Then \((X, Y)\) has density function \(f\) given by \[f(x, y) = \alpha \exp[-2 \alpha (|x| + |y|)^2], \quad (x, y) \in \R^2\]

Details:

This follows immediately from since \(b_{2, 1} = 2\).

Of course, the density is constant on the the squares \(|x| + |y| = t \) for \(t \in (0, \infty)\) in contrast to a bivariate nomral density, for example, which is constant on ellipses. From the general results in , \(X\) and \(Y\) are uncorrelated and identically distributed, and the distribution is symmetric about 0. So in particular, the variables have mean 0 and skewness 0. The family of distributions is a scale family with scale parameter \(1 / \sqrt{\alpha}\).

Suppose again that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_1)\). The common density \(g\) of \(X\) and \(Y\) is given by \[g(x) = \sqrt{\frac{\pi \alpha}{2}} \erfc\left(\sqrt{2 \alpha} |x|\right), \quad x \in \R \] The variance is \(1 / 6 \alpha\) and the kurtosis is \(18 / 5\).

The density function \(g\) in has the following properties:

  1. \(g\) increases on \((-\infty, 0]\) and decreases on \([0, \infty)\), so \(x = 0\) is the mode of the distribution.
  2. \(g\) is concave upward on \((-\infty, 0)\) and on \((0, \infty)\), so \(g\) has a cusp at \(x = 0\).
Details:

This follows from standard calculus:

  1. \(g^\prime(x) = - 2 \sgn(x) \alpha \exp\left(-2 \alpha x^2\right)\) for \(x \in \R\).
  2. \(g^{\prime \prime}(x) = 8 \alpha^2 |x| \exp\left(-2 \alpha x^2\right)\) for \(x \in \R\).

So this distribution might be considered a second-order version of the Laplace distribution.

The app below is a simulation of the norm distribution with \(n = 2\) and \(k = 1\). The rate parameter \(\alpha\) can be varied with the scrollbar.

Suppose again that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_1)\). The common distribution function \(G\) of \(X\) and \(Y\) is given by \[G(y) = 1 - \frac{1}{2} \exp\left(-2 \alpha y^2\right) + \sqrt{\frac{\alpha \pi}{2}} y \, \erfc\left(\sqrt{2 \alpha} \, y\right), \quad y \in [0, \infty)\] and \(G(y) = 1 - G(-y)\) for \(y \in (-\infty, 0)\).

The Case \(n = 2\) and \(k = \infty\)

Suppose that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_\infty)\). Then \((X, Y)\) has joint density function \(f\) given by \[f(x, y) = \alpha \exp\left[-4 \alpha \left(|x| \vee |y|\right)^2\right], \quad(x, y) \in \R^2\]

Details:

This follows immediately from since \(b_{2, \infty} = 4\).

This density is constant on the squares \(|x| \vee |y| = t\) for \(t \in (0, \infty)\). As in the general case, \(X\) and \(Y\) are uncorrelated (but dependent) with identical distributions that are symmetric about 0. So in particular, the variables have mean 0 and skewness 0.

Suppose again that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_\infty)\). Then \(X\) and \(Y\) have common density function \(g\) given by \[g(x) = 2 \alpha |x| \exp\left(-4 \alpha x^2\right) + \frac{\sqrt{\pi \alpha}}{2} \erfc\left(2 \sqrt{\alpha} |x|\right), \quad x \in \R \] The common variance is \(1 / 6 \alpha\) and the common kurtosis is \(27 / 10\).

The density function \(g\) in has the following properties:

  1. \(g\) increases on \((-\infty, 0]\) and decreases on \([0, \infty)\), so \(x = 0\) is the mode of the distribution.
  2. \(g\) is concave upward on \((-\infty, -z)\) and on \((z, \infty)\), and concave downward on \((-z, z)\) where \(z = 1 / 2 \alpha\).
Details:

This follows from standard calculus:

  1. \(g^\prime(x) = - \sgn(x) \alpha^2 16 x^2 \exp\left(-4 \alpha x^2\right)\) for \(x \in \R\).
  2. \(g^{\prime \prime}(x) = 32 \alpha^2 x (4 \alpha^2 x^2 - 1) \exp\left(-4 \alpha x^2\right) \) for \(x \in \R\).

It's interesting that the variance is the same as in the case \(n = 2\), \(k = 1\).

The app below is a simulation of the norm distribution with \(n = 2\) and \(k = \infty\). The rate parameter \(\alpha\) can be varied with the scrollbar.

Suppose again that \((X, Y)\) has the distribution with constant rate \(\alpha \in (0, \infty)\) for the graph \((\R^2, \Rta_\infty)\). The common distribution function \(G\) of \(X\) and \(Y\) is given by \[G(y) = 1 - \frac{1}{2} \exp\left(-4 \alpha y^2\right) + \frac{\sqrt{\alpha \pi}}{2} y \, \erfc\left(2 \sqrt{\alpha} \, y\right), \quad y \in [0, \infty)\] and \(G(y) = 1 - G(-y)\) for \(y \in (-\infty, 0)\).

Random Walks

Returning to the general case of \(n \in \N_+\) and \(k \in [1, \infty]\), consider the random walk \((\bs Y_1, \bs Y_2, \ldots)\) on \((\R^n, \Rta_k)\) associated with \(\bs X\), where \(\bs X\) has constant rate \(\alpha \in (0, \infty)\) for \((\R^n, \Rta_k)\). From our general results and the left walk function in we have

For \(m \in \N_+\), random variable \(\bs Y_m\) has density function \(f_m\) given by \[f_m(\bs x) = \alpha^m \frac{1}{(m - 1)!} (b_{n,k} \|\bs x\|^n_k)^{m - 1} \exp\left(-\alpha b_{n,k} \|x\|^n_k\right), \quad \bs x \in \R^n\]

Note that \(f_m(\bs x)\) is the ordinary gamma density function for the standard graph \(([0, \infty), \le)\), (with order \(m\) and rate \(\alpha\)), evaluated at \(b_{n,k} \|\bs x\|_k^n\).