I am trying to define a bivariate exponential distribution $(S, T)$ with marginals $S\sim\mathrm{Exp}(\lambda_S)$ and $T\sim\mathrm{Exp}(\lambda_T)$ for $\lambda_S > \lambda_T$. I would like the joint distribution of $S$ and $T$ to have two properties:

  1. $\rho(S, T)$ can be controlled by some parameter of the joint distribution
  2. $S\leq T$

The literature has many examples of distributions that fulfill one of these two properties. For instance, the BVE of Marshall and Olkin (1967) can be constructed from independent random variables $\tilde S, \tilde T$, and $C$ and selected constant $\lambda_C < \lambda_T$ as

\begin{align*} \tilde S &\sim \mathrm{Exp}(\lambda_S-\lambda_C) \\ \tilde T &\sim \mathrm{Exp}(\lambda_T-\lambda_C) \\ C &\sim \mathrm{Exp}(\lambda_C) \\ S &= \min(\tilde S, C) \\ T &= \min(\tilde T, C) \end{align*}

Here, $S$ and $T$ have the desired marginal distributions, and $\lambda_C$ controls the degree to which they are correlated (requirement 1). However, we cannot guarantee that $S\leq T$ (requirement 2).

Alternately, we could use independent random variables $\tilde S$ and $T$ to construct

\begin{align*} \tilde S &\sim \mathrm{Exp}(\lambda_S-\lambda_T) \\ T &\sim \mathrm{Exp}(\lambda_T) \\ S &= \min(\tilde S, T) \end{align*}

Again, $S$ and $T$ have the desired marginal distributions. This time $S\leq T$ (requirement 2), but we have no way to control the correlation (requirement 1).

Is there a bivariate exponential distribution that meets both of my requirements?


1 Answer 1


Consider $E \sim \mathcal{E}( \lambda_S - \lambda_C)$ and $C \sim \mathcal{E}(\lambda_C)$. Now define \begin{align*} S & = \min \{ E , C \} \\ T & = \min \{ a E , C\} \end{align*} where $$ a = \frac{ \lambda_S - \lambda_C}{\lambda_T - \lambda_C} > 1 . $$

You get $ S \sim \mathcal {E}(\lambda_S)$ and $ T \sim \mathcal{E} (\lambda_T)$, by construction $S \leq T$ and you have some ability to control correlation between two variables. Analytically, $$\rho(S,T) = \frac{\lambda_S+\lambda_C(1-\lambda_T/\lambda_S-\lambda_S/\lambda_T)}{\lambda_S-\lambda_C} . $$ If $ \lambda_C \approx 0$ you get $S \approx E$ and $S\approx a E$ hence $\rho \approx 1$, but if $\lambda_C \approx \lambda_T$ then $T \approx C$ and $S \approx \min\{E, C\}$, meaning $\rho\approx\lambda_T/\lambda_S$.

With the following simulation I obtained an estimation of $0.625$ for the correlation with $\lambda_S = 2$, $\lambda_T = 1.1$ and $\lambda_C = 1$ (analytically we know $\rho(S,T)=\frac{139}{220}\approx 0.632$).

lambda_S = 2; lambda_T = 1.1; lambda_C = 1
a = (lambda_S -lambda_C)/(lambda_T -lambda_C)

E = rexp(10000,  lambda_S - lambda_C)
C = rexp(10000,  lambda_C)

S =  apply(rbind(E,C), 2, min)
TT = apply(rbind( a*E,C), 2, min)

cor(S, TT)
# 0.6254737

Changing $\lambda_S = 10$ gives $\rho(S,T)=\frac{1979}{9900}\approx 0.200$.


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