The construction in this section generalizes a model investigated by Nair and Sankaran. It has close and interesting connections to the semigroups isomorphic to \( ([0, \infty), +) \) that were studied in Section 2 and Section 3. Our starting point is an interval \([a, b)\) where \(-\infty \lt a \lt b \le \infty\). As usual, the collection \(\ms B\) Borel measurable sets it the reference \(\sigma\)-algebra and Lebesgue measure \(\lambda\) is the reference measure. For \(n \in \N_+\) we give \([a, b)^n\) the product \(\sigma\)-algebra \(\ms B^n\) and produce measure \(\lambda^n\). We begin with an extension of the general definition of multivariate exponential distribution from Section 4.
Suppose that we have a class of special parametric distributions on \([a, b)\). If \(n \in \N_+\) and the distribution of \(X_i\) is in this class for \(i \in \{1, 2, \ldots, n\}\) then in the most general sense, \(\bs{X} = (X_1, X_2, \ldots, X_n)\) has a multivariate distribution in the class.
Suppose now that \(U\) and \(V\) have continuous distributions on \([a, b)\) with respective reliability functions \(F\) and \(G\) for the graph \(([a, b), \le)\). We assume that \(F\) and \(G\) have piecewise-continuous second derivatives, so that in particular, \(V\) has probability density function \(g = -G^\prime\). Also, we have our usual support assumptions in place, so \(F\) is strictly decreasing (and continuous), with \(F(a) = 1\) and \(F(x) \to 0\) as \(x \to b\). The same properties hold for \(G\). As noted in Section 3, \(R = -\ln F\) is the cumulative rate function of \(U\) for the graph \(([a, b), \le)\) and is a homeomorphism from \([a, b)\) to \([0, \infty)\). Define \[ x \oplus y = R^{-1}[R(x) + R(y)], \quad x, \, y \in [a, b) \] Then \( ([a, b), \oplus) \) is a semigroup isomorphic to the standard semigroup \( ([0, \infty), +) \), and the associated order is the ordinary order \(\le\). Moreover, \(U\) has an exponential distribution relative to \(([a, b), \oplus)\). Here is the main definition:
Suppose that \[G(x_1 \oplus y_1) + G(x_2 \oplus y_2) \ge G(x_1 \oplus y_2) + G(x_2 \oplus y_1), \quad a \le x_1 \le x_2 \lt b, \, a \le y_1\le y_2 \lt b\] Then \(H\) defined by \(H(x, y) = G(x \oplus y)\) for \(x, \, y \in [a, b)\) is a reliability function on the product graph \(([a, b)^2, \le)\). The corresponding distribution is the bivariate distribution associated with \(U\) and \(V\) or equivalently the bivariate distribution associated with \(\oplus\) and \(V\).
Recall that \(a\) is the identity of \(\oplus\) so \(H(z, a) = H(a, z) = G(z)\) for \(z \in [a, b)\). Also, \(H\) is positive and decreasing in \(x\) and \(y\) with \(H(a, a) = 1\), \(H(x, y) \to 0\) as \(x \to b\) for fixed \(y \in [a, b)\) and \(H(x, y) \to 0\) as \(y \to b\) for fixed \(x \in [a, b)\). Finally, the displayed equation is precisely the condition needed so that the measure associated with \(H\) assigns nonnegative measure to the rectangle \([x_1, x_2) \times [y_1, y_2)\) where \(a \le x_1 \le x_2 \lt b\) and \(a \le y_1 \le y_2 \lt b\): \[H(x_1, y_1) - H(x_1, y_2) - H(x_2, y_1) + H(x_2, y_2) \ge 0, \quad \]
Of course, only the distributions of \(U\) and \(V\) are relevant so we don't need random variables defined on a common probability space. But as usual, the language of random variables is more natural. Recall also that if \(R\) is a cumulative rate function on \([a, b)\) then so is \(c R\) for \(c \in (0, \infty)\) and moreover, \(R\) and \(c R\) generate the same operator \(\oplus\). So we can ignore such parameters and use a standard
distribution of \(U\).
Suppose that \((X, Y)\) has the bivariate distribution associated with \(U\) and \(V\). Then \(X\) and \(Y\) have the same distribution as \(V\), with reliability function \(G\).
This follows immediately from the definition since \(X\) has reliability function \(x \mapsto H(x, a) = G(x)\) and similarly \(Y\) has reliability function \(y \mapsto H(a, y) = G(y)\).
So proposition provides a method of constructing a bivariate distribution on \([a, b)^2\) with identically distributed maginals. If \(V\) is a member of a special parameter class of distributions on \([a, b)\) then \((X, Y)\) has a bivariate distribution in the class in the sense of definition . Since the marginal distributions are known, our interest is in the special properties of the bivariate distribution.
Let \begin{align*} h(x, y) = \frac{\partial^2}{\partial x \, \partial y} G(x \oplus y) &= G^{\prime \prime}(x \oplus y) \frac{\partial} {\partial x} (x \oplus y) \frac{\partial} {\partial y} (x \oplus y) + G^\prime(x \oplus y) \frac{\partial^2} {\partial x \, \partial y} (x \oplus y) \\ &= -g^\prime(x \oplus y) \frac{\partial} {\partial x} (x \oplus y) \frac{\partial} {\partial y} (x \oplus y) - g(x \oplus y) \frac{\partial^2} {\partial x \, \partial y} (x \oplus y) \end{align*} If \(h\) is nonnegative then the bivariate distribution associated with \(U\) and \(V\) is well defined, with probability density function \(h\).
Let \(H(x, y) = G(x \oplus y)\) for \(x, \, y \in [a, b)\). Then as noted before, \(H\) is decreasing in each variable with \(H(a, a) = 1\) and \(H(x, y) \to 0\) as \(y \to b\) for \(x \in [a, b)\) and \(H(x, y) \to 0\) as \(x \to b\) for \(y \in [a, b)\). These properties, together with the condition that \(h(x, y) = \partial^2 H(x, y) / \partial x \, \partial y \ge 0\) ensures that \(h\) is a probability density function on \([a, b)^2\) with \(H\) as the corresponding reliability function.
When the two distributions are the same, the construction generates independent, identically distributed variables:
Suppose that \(U\) and \(V\) have the same distribution. Then the bivariate distribution associated with \(U\) and \(V\) is well defined, and corresponds to independent random variables with the common distribution.
In the notation that we have established, \(F = G\) and \(F = e^{-R}\). Hence \[H(x, y) = F(x \oplus y) = \exp\left[-R\left(R^{-1}(R(x) + R(y))\right)\right] = \exp[-(R(x) + R(y))] = e^{-R(x)} e^{-R(y)} = F(x) F(y), \quad (x, y) \in [a, b)^2 \]
Suppose that \((X, Y)\) has the bivariate distribution associated with \(U\) and \(V\). Given \(X = x \in [a, b)\), the conditional reliability function of \(Y\) is \[G(y \mid x) = \frac{g(x \oplus y)}{g(x)} \frac{\partial}{\partial x}(x \oplus y), \quad y \in [a, b)\]
In this subsection, suppose that \([a, b) = [0, \infty)\). The most important special case (and a slight generalization of the one considered by Nair and Sankaran) is when \(V\) has an exponential distribution.
Suppose that \(V\) has the exponential distribution on \(([0, \infty), +)\) with rate \(\alpha \in (0, \infty)\). If \[\alpha \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) \ge \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \] then the bivariate distribution associated with \(U\) and \(V\) is well defined and is the bivariate exponential distribution corresponding to \(U\) and \(\alpha\). The reliabiltiy function \(H\) and density function \(h\) are given by \begin{align*} H(x, y) &= \exp[-\alpha (x \oplus y)], \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= \alpha \exp[-\alpha (x \oplus y)] \left[ \alpha \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) - \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \right] \end{align*}
If \((X, Y)\) has the bivariate exponential distribution corresponding to \(U\) and \(\alpha\) as in then \(X\) and \(Y\) have exponential distributions on \(([0, \infty), +)\) with rate \(\alpha\), so \((X, Y)\) really does have a bivariate exponential distribution in the general sense of definition . The rate function of \((X, Y)\) is given by \[ (x, y) \mapsto \alpha^2 \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) - \alpha \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \] The special case \(\alpha = 1\) is the setting of the paper by Nair and Sankaran (but of course, not in our semigroup notation).
Suppose that \(V\) has the exponential distribution on \(([0, \infty), +)\) with rate 1. If \[ \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) \ge \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \] then the bivariate distribution associated with \(U\) and \(V\) is well defined and is the bivariate exponential distribution corresponding to \(U\). The reliabiltiy function \(H\) and density function \(h\) are given by \begin{align*} H(x, y) &= \exp[-(x \oplus y)], \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= \exp[-(x \oplus y)] \left[ \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) - \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \right] \end{align*}
The rate function is given by \[ (x, y) \mapsto \frac{\partial}{\partial x} (x \oplus y) \frac{\partial}{\partial y} (x \oplus y) - \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) \]
Suppose that \((X, Y)\) has the bivariate exponential distribution associated with \(U\) and \(\alpha \in (0, \infty)\). Given \(X = x \in [0, \infty)\), the conditional reliability function of \(Y\) is \[ G(y \mid x) = \exp[-\alpha(x \oplus y - x)] \frac{\partial}{\partial x} (x \oplus y), \quad y \in [0, \infty) \]
When both of the defining distributions are exponential, the bivariate distribution corresponds to independent, identically distributed exponential variables by :
Suppose that \(U\) has an exponential distributions on \(([0, \infty), +)\). The bivariate exponential distribution corresponding to \(U\) and \(\alpha \in (0, \infty)\) is well defined, and if \((X, Y)\) has this distribution then \(X\) and \(Y\) are independent, each with the exponential distribution with rate \(\alpha\). In particular, \((X, Y)\) has an exponential distribution for the product semigroup \(([0, \infty), +)\).
Suppose that \(U\) has a Weibull distribution on \([0, \infty)\) with shape parameter \(k \in (0, \infty)\). The operator associated with \(U\) is given by \[ x \oplus y = (x^k + y^k)^{1 / k}, \quad (x, y) \in [0, \infty)^2\] If \(k \ge 1\) then the bivariate exponential distribution assocated with \(U\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-\left(x^k + y^k\right)^{1 / k}\right], \quad (x, y) \in [0, \infty)^2\\ h(x, y) &= \exp\left[-\left(x^k + y^k\right)^{1 / k}\right] x^{k-1} y^{k-1} \left(x^k + y^k\right)^{1 / k - 2} \left[(k - 1) + \left(x^k + y^k\right)^{1 / k}\right], \quad (x, y) \in [0, \infty)^2 \end{align*}
The reliability function \(F\) of \(U\) with scale parameter \(\alpha \in (0, \infty)\) is given by \(F(x) = \exp\left[-(x / \alpha)^k\right]\) for \(x \in [0, \infty)\). Hence the cumulative rate function \(R\) is given by \(R(x) = x^k / \alpha^k\) for \(x \in [0, \infty)\) and so we can take \(\alpha = 1\). The operator is \[ x \oplus y = (x^k + y^k)^{1 / k}, \quad x, \, y \in [0, \infty)\] For the partial derivatives, \begin{align*} \frac{\partial}{\partial x} (x \oplus y) & = x^{k - 1} (x^k + y^k)^{1 / k - 1} \\ \frac{\partial}{\partial y} (x \oplus y) & = y^{k - 1} (x^k + y^k)^{1 / k - 1} \\ \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) & = (1 - k) x^{k - 1} y^{k - 1} (x^k + y^k)^{1 / k - 2} \end{align*} The condition in is satisfied if \(k \ge 1\).
The distribution in is the Type 3 bivariate exponential distribution studied by Gumbel, but we could refer to it as the bivariate exponential distribution associated with the Weibull operator with parameter \(k \in [1, \infty)\). The rate function is given by \[ (x, y) \mapsto x^{k-1} y^{k-1} \left(x^k + y^k\right)^{1 / k - 2} \left[(k - 1) + \left(x^k + y^k\right)^{1 / k}\right], \quad (x, y) \in [0, \infty)^2 \]
Suppose that \(U\) has a Lomax distribution on \([0, \infty)\) with scale parameter \(\beta \in (0, \infty)\). The operator associated with \(U\) is given by \[ x \oplus y = x + y + \frac{x y}{\beta}, \quad (x, y) \in [0, \infty)^2\] If \(\beta \ge 1\) then the bivariate exponential distribution assocated with \(U\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-\left(x + y + \frac{x y}{\beta}\right)\right], \quad (x, y) \in [0, \infty)^2\\ h(x, y) &= \exp\left[-\left(x + y + \frac{x y}{\beta}\right)\right]\left[\left(1 - \frac{1}{\beta}\right) + \frac{x}{\beta} + \frac{y}{\beta} + \frac{x y}{\beta^2}\right], \quad (x, y) \in [0, \infty)^2 \end{align*}
The reliability function \(F\) of \(U\) with shape parameter \(\alpha \in (0, \infty)\) is given by \(F(x) = [\beta / (x + \beta)]^\alpha\) for \(x \in [0, \infty)\). Hence the cumulative rate function \(R\) is given by \(R(x) = \alpha[\ln(x + \beta) - \ln(\beta)]\) for \(x \in [0, \infty)\) and so we can take \(\alpha = 1\). The operator is \[ x \oplus y = x + y + \frac{x y}{\beta}, \quad x, \, y \in [0, \infty)\] For the partial derivatives, \(\partial (x \oplus y) / \partial x = 1 + y / \beta \), \(\partial (x \oplus y) / \partial y = 1 + x / \beta \), and \(\partial^2(x \oplus y) / \partial x \, \partial y = 1 / \beta\). The condition in is satisfied if \(\beta \ge 1\).
The distribution in is the Type 1 bivariate exponential distribution studied by Gumbel, but we could refer to it as the bivariate exponential distribution associated with the Lomax operator with parameter \(\beta \in (0, \infty)\) . The rate function is given by \[ (x, y) \mapsto \left[\left(1 - \frac{1}{\beta}\right) + \frac{x}{\beta} + \frac{y}{\beta} + \frac{x y}{\beta^2}\right], \quad (x, y) \in [0, \infty)^2 \] and hence the bivariate distribution has increasing rate.
Suppose that \(U\) has a Gompertz distribution on \([0, \infty)\) with scale parameter \(1 / \beta \in (0, \infty)\). The operator associated with \(U\) is given by \[ x \oplus y = \frac{1}{\beta} \ln\left(e^{\beta x} + e^{\beta y} - 1\right), \quad (x, y) \in [0, \infty)^2\] The bivariate exponential distribution assocated with \(U\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \left(e^{\beta x} + e^{\beta y} - 1\right)^{-1 / \beta}, \quad (x, y) \in [0, \infty)^2\\ h(x, y) &= (1 + \beta) e^{\beta(x + y)} \left(e^{\beta x} + e^{\beta y} - 1\right)^{-(1 / \beta + 2)}, \quad (x, y) \in [0, \infty)^2 \end{align*}
The reliability function \(F\) of \(U\) with shape parameter \(\alpha \in (0, \infty)\) is given by \(F(x) = \exp\left[-\alpha \left(e^{\beta x} - 1\right)\right]\) for \(x \in [0, \infty)\). Hence the cumulative rate function \(R\) is given by \(R(x) = \alpha \left(e^{\beta x} - 1\right)\) for \(x \in [0, \infty)\) and so we can take \(\alpha = 1\). The operator is \[ x \oplus y = \frac{1}{\beta} \ln\left(e^{\beta x} + e^{\beta y} - 1\right)\] For the partial derivatives, \begin{align*} \frac{\partial}{\partial x} (x \oplus y) & = \frac{e^{\beta x}}{e^{\beta x} + e^{\beta y} - 1} \\ \frac{\partial}{\partial y} (x \oplus y) & = \frac{e^{\beta y}}{e^{\beta x} + e^{\beta y} - 1} \\ \frac{\partial^2}{\partial x \, \partial y} (x \oplus y) & = -\frac{\beta e^{\beta x} e^{\beta y}}{\left(e^{\beta x} + e^{\beta y} - 1\right)^2} \end{align*} The condition in is trivially satisfied for \(\beta \gt 0\).
We can refer to this as the bivariate exponential distribution associated with the Gompertz operator with parameter \(\beta \in [1, \infty)\). The rate function is given by \[(x, y) \mapsto (1 + \beta) e^{\beta (x + y)}\left(e^{\beta x} + e^{\beta y} - 1\right)^{-2}, \quad (x, y) \in [0, \infty)^2\]
In this subsection we assume again \([a, b) = [0, \infty)\) but now that \(U\) rather than \(V\) has an exponential distribution. Just as in the details of , the operator \(\oplus\) reduces to ordinary addition \(+\) regardless of the rate parameter of \(U\). Here is the main result
If \(g\) is decreasing (equivalently \(G\) is convex) then the bivariate distribution associated with \(U\) and \(V\) is well defined. The reliability function \(H\) and density function \(h\) of this distribution are given by \begin{align*} H(x, y) &= G(x + y), \quad (x, y) \in [0, \infty)^2\\ h(x, y) &= G^{\prime \prime}(x + y) = -g^\prime(x + y), \quad (x, y) \in [0, \infty)^2 \end{align*}
If \((X, Y)\) has the bivariate distribution associated with \(U\) and \(V\) then of course \(X\) and \(Y\) both have the distribution of \(V\). The rate function of \((X, Y)\) is \[ (x, y) \mapsto \frac{G^{\prime \prime}(x + y)}{G(x + y)} \] Note also that proposition gives another proof of proposition , since \(G\) given by \(G(x) = e^{-\alpha x}\) for \(x \in [0, \infty)\) is convex for \(\alpha \in (0, \infty)\). Our next result concerns the conditional distributions.
Suppose that \((X, Y)\) has the bivariate distribution described in . Then given \(X = x \in [0, \infty)\),
From parts (a) and (b) of , the conditional rate function of \(Y\) given \(X = x \in [0, \infty)\) is \[y \mapsto -\frac{G^{\prime \prime}(x + y)}{G^\prime(x + y)} = -\frac{g^\prime(x + y)}{g(x)} \] From part (c) we have the interesting result that \(\E(Y \mid x)\) is the reciprocal of the rate function of \(Y\) at \(x\). Of course by symmetry, the conditional distribution of \(X\) given \(Y = y \in [0, \infty)\) has the same properties. There is also a simple expression for the correlation of the bivariate distribution in terms of moments of \(V\).
Suppose that \((X, Y)\) has the bivariate distribution described in . Then \(\E(X Y) = \frac{1}{2} \E(V^2)\) and hence \[\cor(X, Y) = \frac{1}{2}\left[1 - \frac{\E^2(V)}{\var(V)}\right]\]
From the basic moment result in Section 1.3, \[\E(X Y) = \int_0^\infty \int_0^\infty H(x, y) \, dx \, dy = \int_0^\infty \int_0^\infty G(x + y) \, dx \, dy \] For the inside integral \begin{align*} \int_0^\infty G(x + y) \, dx & = \int_y^\infty G(u) \, du = \int_u^\infty \P(V \ge u) \, du \\ & = \int_y^\infty \E[ \bs{1}(V \ge u)] du = \E\left[\int_y^\infty \bs{1}(V \ge u) \, du\right] = \E\left[\int_0^\infty \bs{1}(y \le u \le V) du \right] = \E[(V - y), \, V \ge y] \end{align*} Hence \begin{align*} \E(X Y) & = \int_0^\infty \E[(V - y), \, V \ge y] dy = \E\left[\int_0^\infty(V - y) \bs{1}(V \ge y) \, dy\right] \\ & = \E\left[\int_0^V (V - y) \, dy\right] = \E\left(V^2 - \frac{1}{2} V^2\right) = \frac{1}{2} \E(V^2) \end{align*} Since \(X\) and \(Y\) both have the same distribution as \(V\) we have \[\cov(X, Y) = \frac{1}{2} \E(V^2) - [\E(V)]^2 = \frac{1}{2}[\var(V) - \E^2(V)] \] Finally, dividing by \( \sd(X) \sd(Y) = \var(V)\) gives the result.
We can also give a proof of the first result by conditioning and using : \[ \E(X Y) = \E[\E(X Y \mid X)] = \E[X \E(Y \mid X) = \left[-X \frac{G(X)} {G^\prime(X)}\right] \] But \(x \mapsto -G^{\prime}(x)\) is a density of \(X\) so by our basic moment result, \[\E(X Y) = \int_0^\infty x G(x) \, dx = \frac{1}{2} \E(V^2) \]
As a corollary of , if \(V\) has a convex reliability function \(G\) then \(\E^2(V) \le 3 \var(V)\) or equivalently, \[\E(V^2) \ge \frac{4}{3} \E^2(V)\] an improvement over the standard result \(\E(V^2) \ge \E^2(V)\) from Jensen's inequality. We will consider a few concrete examples that parallel the ones in the previous subsection. Note that if \(G\) is a reliability function, then so is \(G_c\) for \(c \in (0, \infty)\) where \(G_c(x) = G(c x)\) for \(x \in [0, \infty)\). The transformation corresponds to adding a scale parameter \(1 / c\) to the distribution of \(G\). Moreover, \(G\) is convex if and only if \(G_c\) is convex. Of course correlation is unchanged by a scale parameter so the formula in Proposition is still valid. So in the examples that follow we will eliminate scale parameters when appropriate.
Suppose that \(V\) has the standard Weibull distribution with shape parameter \(k \in (0, \infty)\). If \(k \le 1\) then the bivariate distribution associated with \(U\) and \(V\) is well defined, with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-(x + y)^k\right], \quad (x, y) \in [0, \infty)^2]\\ h(x, y) & = \exp\left[-(x + y)^k\right] k (x + y)^{k - 2} \left[(1 - k) + k (x + y)^k \right], \quad (x, y) \in (0, \infty)^2 \end{align*}
The reliability function \(G\) of \(V\) is given by \(G(x) = \exp(-x^k)\) for \(x \in [0, \infty)\). Hence \[G^{\prime \prime}(x) = k x^{k - 2} \exp(-x^k)\left[(1 - k) + k x^k\right], \quad x \in (0, \infty) \] so \(G\) is convex when \(k \in (0, 1]\).
So the rate function is given by \[(x, y) \mapsto k (1 - k) (x + y)^{k - 2} + k^2 (x + y)^{2 k - 2}, \quad (x, y) \in (0, \infty)^2 \] and hence the bivariate Weibull distribution has decreasing rate.
Suppose that \((X, Y)\) has the bivariate Weibull distribution with parameter \(k \in (0, 1]\) as defined in . Then \(\E(Y \mid X = x) = x^{1 - k} / k\) for \(x \in [0, \infty]\). Moreover, given \(X = x \in (0, \infty)\), the conditional distribution of \(Y\) is the same as the distribution of \(\min\{W, Z\}\) where
From Proposition , the conditional reliability function of \(Y\) given \(X = x \in (0, 1)\) reduces (after some simplification) to \[ G(y \mid x) = \frac{G^\prime(x + y)}{G^\prime(x)} = \left(1 + \frac{y}{x}\right)^{k - 1} \exp\left[-(x + y)^k + x^k\right], \quad y \in [0, \infty) \]
Suppose again that \((X, Y)\) has the bivariate Weibull distribution with parameter \(k \in (0, 1]\) as defined in . Then \[\cor(X, Y) = \frac{1}{2} \left[1 - \frac{\Gamma^2(1 + 1 / k)}{\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)}\right]\] So \(X\) and \(Y\) are uncorrelated when \(k = 1\), positively correlated for \(k \in (0, 1)\) with \(\cor(X, Y) \uparrow \frac{1}{2}\) as \(k \downarrow 0\).
Of course when \(k = 1\), \(V\) has an exponential distribution so we are in the setting of and hence \(X\) and \(Y\) are independent.
Suppose that \(V\) has the standard Lomax distribution with shape parameter \(\alpha \in (0, \infty)\). Then the bivariate distribution associated with \(U\) and \(V\) is well defined, with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= (x + y + 1)^{-\alpha}, \quad (x, y) \in [0, \infty)^2\\ h(x, y) &= \alpha (\alpha + 1) (x + y + 1)^{-(\alpha + 2)}, \quad (x, y) \in [0, \infty)^2 \end{align*}
The reliability function \(G\) of \(V\) is given by \(G(x) = (x + 1)^{-\alpha}\) for \(x \in [0, \infty)\). Hence \[G^{\prime \prime}(x) = \alpha (\alpha + 1)(x + 1)^{-\alpha - 2}, \quad x \in (0, \infty) \] so \(G\) is convex.
So the rate function is given by \[(x, y) \mapsto \alpha (\alpha + 1)(x + y + 1)^{-2}, \quad (x, y) \in [0, \infty)^2\] and hence the bivariate Lomax distribution has decreasing rate.
Suppose that \((X, Y)\) has the bivariate Lomax distribution with parameter \(\alpha \in (0, \infty)\) as defined in , and let \(x \in [0, \infty)\). Then given \(X = x\), random variable \(Y\) has a Lomax distribution with shape parameter \(\alpha + 1\) and scale parameter \(x + 1\) and in particular, \(\E(Y \mid X = x) = (x + 1) / \alpha\)
From Proposition , the conditional reliabity function of \(Y\) given \(X = x \in [0, \infty)\) is \[ G(y \mid x) = \frac{(x + y + 1)^{-(\alpha + 1)}}{(x + 1)^{-(\alpha + 1)}} = \left(1 + \frac{y}{x + 1}\right)^{-(\alpha + 1)}, \quad y \in [0, \infty) \]
Suppose again that \((X, Y)\) has the bivariate Lomax distribution with parameter \(\alpha \in (0, \infty)\) as defined in . If \(\alpha \gt 2\) then \(\cor(X, Y) = 1 / \alpha\).
Note that \(\cor(X, Y) \uparrow \frac{1}{2}\) as \(\alpha \downarrow 2\) and \(\cor(X, Y) \downarrow 0\) as \(\alpha \uparrow \infty\).
Suppose that \(V\) has the Gompertz distribution with scale parameter 1 and shape parameter \(\alpha \in (0, \infty)\). If \(\alpha \ge 1\) then the bivariate distribution associated with \(U\) and \(V\) is well defined, with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-\alpha \left(e^{x + y} - 1\right)\right], \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= \alpha e^{x + y} \left(\alpha e^{x + y} - 1\right) \exp\left[-\alpha \left(e^{x + y} - 1\right)\right], \quad (x, y) \in [0, \infty)^2 \end{align*}
The reliability function \(G\) of \(V\) is given by \(G(x) = \exp\left[-\alpha \left(e^x - 1\right)\right]\) for \(x \in [0, \infty)\). Hence \[G^{\prime \prime}(x) = \alpha e^x \left(\alpha e^x - 1\right) \exp\left[-\alpha \left(e^x - 1\right)\right], \quad x \in (0, \infty) \] so \(G\) is convex.
So the rate function is given by \[ (x, y) \mapsto \alpha e^{x + y} \left(\alpha e^{x + y} - 1\right), \quad (x, y) \in [0, \infty)^2 \] and hence the bivariate Gompertz distribution has increasing rate.
Suppose that \((X, Y)\) has the bivariate Gompertz distribution with parameter \(\alpha \in [1, \infty)\) as described in , and let \(x \in [0, \infty)\). Then \(\E(Y \mid X = x) = e^{-x} / \alpha\). Moroever given \(X = x\), the conditional distribution of \(Y\) is the same as the distribution of \(\min\{W, Z\}\) where
Another interesting case is when the distributions of \(U\) and \(V\) belong to the same special parametric class. In a sense, the bivariate distribution associated with \(U\) and \(V\) is a pure bivariate distribution associated with the class.
Suppose that the base interval is \([0, \infty)\) and that \(U\) and \(V\) have Weibull distributions with shape parameters \(j, \, k \in (0, \infty)\). So as in , the operator associated with \(U\) is given by \[x \oplus y = (x^j + y^j)^{1 / j}, \quad (x, y) \in [0, \infty)^2\] while as in the reliability function \(G\) of \(V\) is given by \[G(x) = \exp\left(-x^k\right), \quad x \in [0, \infty)\]
If \(j \ge k\) then the bivariate distribution associated with \(U\) and \(V\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-\left(x^j + y^j\right)^{k / j}\right], \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= k x^{j - 1} y^{j - 1} \left(x^j + y^j\right)^{k / j - 2} \left[(j - k) + k \left(x^j + y^j\right)^{k / j}\right] \exp\left[-\left(x^j + y^j\right)^{k / j}\right], \quad (x, y) \in (0, \infty)^2 \end{align*}
From the general theory, \(H\) is given by \(H(x, y) = G(x \oplus y)\) for \((x, y) \in [0, \infty)^2\). The formula for \(h = \partial^2 H(x, y) / \partial x \, \partial y\) then follows from standard calculus. Note that \(h \ge 0\) if \(j \ge k\).
Of course if \((X, Y)\) has the bivariate Weibull distribution with parameters \(j, \, k \in (0, \infty)\) with \(j \ge k\) then \(X\) and \(Y\) have the same distribution as \(V\), namely Weibull with shape parameter \(k\). The rate function of \((X, Y)\) is given by \[ (x, y) \mapsto k x^{j - 1} y^{j - 1} \left(x^j + y^j\right)^{k / j - 2} \left[(j - k) + k \left(x^j + y^j\right)^{k / j}\right] \] In the special case \(k = 1\) and \(j \ge 1\), we are in the setting of , a bivariate exponential distribution corresponding to the Weibull operator. In the special \(j = 1\) and \(k \le 1\) we are in the setting of , a bivariate Weibull distribution associated with standard operator. Finally if \(j = k\) then \(X\) and \(Y\) are independent by
Suppose that \((X, Y)\) has the pure bivariate Weibull distribution with parameter \(j \in (0, \infty)\) and \(k \in (0, j)\), as defined in . The conditional reliability function of \(Y\) given \(X = x \in [0, \infty)\) is \[G(y \mid x) = x^{j - k} \left(x^j + y^j\right)^{k / j - 1} \exp\left[-\left(x^j + y^j\right)^{k / j} + x^k\right], \quad y \in [0, \infty) \]
The base interval is once again \([0, \infty)\). Suppose that \(U\) has the Lomax distribution with shape parameter 1 and scale parameter \(\alpha \in (0, \infty)\) and that \(V\) has the Lomax distribution with shape parameter \(\beta \in (0,\infty)\) and scale parameter 1. So as in , the operator associated with \(U\) is given by \[x \oplus y = x + y + \frac{x y}{\alpha}, \quad (x, y) \in [0, \infty)^2\] while as in , the reliability function \(G\) of \(V\) is given by \[G(x) = (1 + x)^{-\beta}, \quad x \in [0, \infty)\]
The bivariate distribution associated with \(U\) and \(V\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \left(1 + x + y + \frac{x y}{\alpha}\right)^{-\beta}, \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= \beta (\beta + 1) \left(1 + x + y + \frac{x y}{\alpha}\right)^{-\beta - 2} + \frac{\beta}{\alpha} \left(1 + x + y + \frac{x y}{\alpha}\right)^{-\beta - 1}, \quad (x, y) \in (0, \infty)^2 \end{align*}
From the general theory, \(H\) is given by \(H(x, y) = G(x \oplus y)\) for \((x, y) \in [0, \infty)^2\). The formula for \(h = \partial^2 H(x, y) / \partial x \, \partial y\) then follows from standard calculus. Note that \(h \ge 0\).
Of course if \((X, Y)\) has the bivariate Lomas distribution with parameters \(\alpha, \, \beta \in (0, \infty)\) then \(X\) and \(Y\) have the same distribution as \(V\), namely Lomax with shape parameter \(\beta\). The rate function of \((X, Y)\) is given by \[ (x, y) \mapsto \frac{\beta (1 + \beta) + (1 + x + y + x y / \alpha) \beta / \alpha} {(1 + x + y + x y / \alpha)^2} \] so \((X, Y)\) has decreasing rate. If \(\alpha = 1\) then \(X\) and \(Y\) are independent by .
Suppose that \((X, Y)\) has the bivariate Lomax distribution with parameters \(\alpha, \, \beta \in (0, \infty)\) as defined in . The conditional reliability function of \(Y\) given \(X = x \in [0, \infty)\) is \[G(y \mid x) = \left(\frac{1 + x + y + x y / \alpha}{1 + x}\right)^{-(\beta + 1)}(1 + y / \alpha), \quad y \in [0, \infty)\]
The base interval is once again \([0, \infty)\). Suppose that \(U\) has the Gompertz distribution with shape parameter 1 and scale parameter \(1 / \beta \in (0, \infty)\), and that \(V\) has the Gompertz distribution with shape parameter \(\alpha \in (0,\infty)\) and scale parameter 1. So as in , the operator associated with \(U\) is given by \[x \oplus y = \frac{1}{\beta} \ln \left(e^{\beta x} + e^{\beta y} - 1\right), \quad (x, y) \in [0, \infty)^2\] while as in , the reliability function \(G\) of \(V\) is given by \[G(x) = \exp\left[-\alpha \left(e^x - 1\right)\right], \quad x \in [0, \infty)\]
If \(\beta \ge 1\), the bivariate distribution associated with \(U\) and \(V\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= \exp\left[-\alpha \left(\left[e^{\beta x} + e^{\beta y} - 1\right]^{1 / \beta} - 1\right)\right], \quad (x, y) \in [0, \infty)^2 \\ h(x, y) &= H(x, y) \alpha e^{\beta x} e^{\beta y}\left(e^{\beta x} + e^{\beta y} -1\right)^{1 / \beta - 2} \left[\alpha \left(e^{\beta x} + e^{\beta y} - 1\right)^{1 / \beta} + (\beta - 1)\right], \quad (x, y) \in (0, \infty)^2 \end{align*}
From the general theory, \(H\) is given by \(H(x, y) = G(x \oplus y)\) for \((x, y) \in [0, \infty)^2\). The formula for \(h = \partial^2 H(x, y) / \partial x \, \partial y\) then follows from standard calculus. Note that \(h \ge 0\) if \(\beta \ge 1\).
Of course if \((X, Y)\) has the bivariate Gompertz distribution with parameters \(\alpha, \, \beta \in (0, \infty)\) then \(X\) and \(Y\) have the same distribution as \(V\), namely Gompertz with shape parameter \(\alpha\). The rate function of \((X, Y)\) is given by \[ (x, y) \mapsto \alpha e^{\beta x} e^{\beta y}\left(e^{\beta x} + e^{\beta y} -1\right)^{1 / \beta - 2} \left[\alpha \left(e^{\beta x} + e^{\beta y} - 1\right)^{1 / \beta} + (\beta - 1)\right] \] If \(\beta = 1\) then \(X\) and \(Y\) are independent by .
Suppose that the base interval is \([0, 1)\) and that \(U\) and \(V\) have beta distributions with left shape parameters \(\alpha, \, \beta \in (0, \infty)\) and and right shape parameters 1. So the reliability functions \(F\) and \(G\) of \(U\) and \(V\) are given by \(F(x) = 1 - x^\alpha\) and \(G(x) = 1 - x^\beta\) for \(x \in [0, 1)\).
If \(\alpha \ge \beta\) then the bivariate distribution associated with \(U\) and \(V\) is well defined with reliability function \(H\) and density function \(h\) given by \begin{align*} H(x, y) &= 1 - \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{\beta / \alpha}, \quad (x, y) \in [0, 1)^2 \\ h(x, y) &= \beta x^{\alpha - 1} y^{\alpha - 1} \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{\beta / \alpha -2} \left[(\alpha - \beta) + \beta \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)\right], \quad (x, y) \in (0, 1)^2 \end{align*}
The cumulative rate function \(R\) of \(U\) is given by \(R(x) = - \ln[F(x)] = -\ln(1 - x^\alpha)\) for \(x \in [0, 1)\) and hence the associated operator is \[x \oplus y = \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{1 / \alpha}, \quad (x, y) \in [0, 1)^2 \] So then \[ H(x, y) = G(x \oplus y) = 1 - (x \oplus y)^\beta = 1 - \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{\beta / \alpha}, \quad (x, y) \in [0, 1)^2 \] After some calculus and algebra, \[ h(x, y) = \frac{\partial^2}{\partial x \, \partial y} H(x, y) = \beta x^{\alpha - 1} y^{\alpha - 1} \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{\beta / \alpha -2} \left[(\alpha - \beta) + \beta \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)\right], \quad (x, y) \in (0, 1)^2 \] So if \(\alpha \ge \beta\) the \(h\) is nonnegative and so \(H\) is the reliability function and \(h\) the density function of a bivariate distribution on \([0, 1)^2\).
Of course if \((X, Y)\) has the bivariate beta distribution with parameters \(\alpha, \, \beta \in (0, \infty)\) then \(X\) and \(Y\) have the same distribution as \(V\), namely beta with left shape parameter \(\beta\) and right shape parameter 1. Also, if \(\alpha = \beta\) then \(X\) and \(Y\) are independent by . When \(\alpha = 1\) we can compute the correlation in a reasonbly concise way using the harmonic number function \(\hn\): \[ \hn(x) = \int_0^1 \frac{1 - t^x}{1 - t} dt = -\sum_{k=1}^\infty \binom{x}{k} \frac{(-1)^k}{k}, \quad x \in [0, \infty)\]
Suppose that \((X, Y)\) has the bivariate beta distribution with parameters \(\alpha = 1\) and \(\beta \in (0, 1]\) as defined in . Then \(\E(X Y) = 1 - \hn(1 + \beta) / (1 + \beta)\) and hence \[ \cor(X, Y) = \frac{2 + \beta}{\beta}[1 + 2 \beta - (1 + \beta) \hn(1 + \beta)] \]
So \(X\) and \(Y\) are positively correlated and the correlation decreases from about 0.7 to 0 as \(\beta\) increases from \(0\) to \(1\).
Suppose that \((X, Y)\) has the bivariate beta distribution with parameters \(\alpha \in (0, \infty)\) and \(\beta \in (0, \alpha]\) as defined in . The conditional reliability function of \(Y\) given \(X = x \in [0, 1)\) is \[ G(y \mid x) = x^{\alpha - \beta} \left(1 - y^\alpha\right) \left(x^\alpha + y^\alpha - x^\alpha y^\alpha\right)^{\beta / \alpha - 1}, \quad y \in [0, 1) \]