What is the range of attainable correlations for the pair of exponentially distributed random variables $X_1 \sim {\rm Exp}(\lambda_1)$ and $X_2 \sim {\rm Exp}(\lambda_2)$, where $\lambda_1, \lambda_2 > 0$ are the rate parameters?

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    $\begingroup$ This question is linked to a side comment here . $\endgroup$
    – QuantIbex
    Aug 7, 2013 at 17:48

1 Answer 1


Let $\rho_{\min}$ (resp. $\rho_{\max}$) denote the lower (resp. upper) bound of the attainable correlation between $X_1$ and $X_2$. The bounds $\rho_{\min}$ and $\rho_{\max}$ are reached when $X_1$ and $X_2$ are respectively countermonotonic and comonotonic (see here).

Lower bound
To determine of the lower bound $\rho_{\min}$ we construct a pair countermonotonic exponential variables and compute their correlation.

The necessary and sufficient condition mentioned here and the probability integral transform provide a convenient way to construct the random variables $X_1$ and $X_2$ such that they are countermonotonic.
Recall that the exponential distribution function is $F(x) = 1 - \exp(-\lambda x)$, so the quantile function is $F^{-1}(q) = -\lambda^{-1}\log (1-q)$.

Let $U\sim U(0, 1)$ be a uniformly distributed random variables, then $1-U$ is also uniformly distributed and the random variables $$ X_1 = -\lambda_1^{-1}\log (1-U), \quad \text{and } X_2 = -\lambda_2^{-1}\log (U) $$ have the exponential distribution with rate $\lambda_1$ and $\lambda_2$ respectively. In addition, they are countermonotonic since $X_1 = h_1(U)$ and $X_2 = h_2(U)$, and the functions $h_1(x)=-\lambda_1^{-1}\log (1-x)$ and $h_2(x)=-\lambda_1^{-1}\log (x)$ are respectively increasing and deacreasing.

Now, let's compute the correlation of $X_1$ and $X_2$. By the properties of the exponential distribution we have ${\rm E}(X_1) = \lambda_1^{-1}$, ${\rm E}(X_2) = \lambda_2^{-1}$, ${\rm var}(X_1) = \lambda_1^{-2}$, and ${\rm var}(X_2) = \lambda_2^{-2}$. Also, we have \begin{align} {\rm E}(X_1 X_2) &= \lambda_1^{-1} \lambda_2^{-1} {\rm E}\{\log (1-U) \log (U)\}\\ &= \lambda_1^{-1} \lambda_2^{-1} \int_0^1 \log (1-u) \log (u) f_U(u) {\rm d}u \\ &= \lambda_1^{-1} \lambda_2^{-1} \int_0^1 \log (1-u) \log (u) {\rm d}u \\ &= \lambda_1^{-1} \lambda_2^{-1} \left (2 - \frac{\pi^2}{6} \right ), \end{align} where $f_U(u) \equiv 1$ is the density function of the standard uniform distribution. For the last equality I relied on WolframAlpha.

Thus, \begin{align} \rho_{\min} &= {\rm corr}(X_1, X_2)\\ &= \frac{ \lambda_1^{-1} \lambda_2^{-1} (2 - \pi^2/6 ) - \lambda_1^{-1} \lambda_2^{-1}}{ \sqrt{\lambda_1^{-2} \lambda_2^{-2}} } \\ &= 1 - \pi^2/6 \approx −0.645. \end{align} Note that the lower bound doesn't depend on the rates $\lambda_1$ and $\lambda_2$, and that the correlation never reaches $-1$, even when both margins are equal (i.e., when $\lambda_1 = \lambda_2$).

Upper bound
To determine of the upper bound $\rho_{\max}$ we follow a similar approach with a pair of comonotonic exponential variables. Now, let $X_1 = g_1(U)$ and $X_2 = g_2(U)$ where $g_1(x)=-\lambda_1^{-1}\log (1-x)$ and $g_2(x)=-\lambda_2^{-1}\log (1-x)$, which are both increasing functions. So, these random variables are comonotonic and both exponentialy distributed with rates $\lambda_1$ and $\lambda_2$.

We have \begin{align} {\rm E}(X_1 X_2) &= \lambda_1^{-1} \lambda_2^{-1} {\rm E}\{\log (1-U) \log (1-U)\}\\ &= \lambda_1^{-1} \lambda_2^{-1} \int_0^1 \left\{\log (1-u) \right\}^2 {\rm d}u \\ &= 2 \lambda_1^{-1} \lambda_2^{-1} , \end{align} and thus, \begin{align} \rho_{\max} &= {\rm corr}(X_1, X_2)\\ &= \frac{ 2\lambda_1^{-1} \lambda_2^{-1} - \lambda_1^{-1} \lambda_2^{-1}}{ \sqrt{\lambda_1^{-2} \lambda_2^{-2}} } \\ &= 1 . \end{align} Similarly to the lower bound, the upper bound doesn't depend on the rates $\lambda_1$ and $\lambda_2$.

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    $\begingroup$ Thanks for your calculations. I just wanted to add that $\rho_{max}=1$ could have been found immediately, noticing that $X_1$ and $X_2$ are of the same type: $\frac{\lambda_1}{\lambda_2}X_1$ has distribution $Exp(\lambda_2)$, i.e. the same distribution of $X_2$. $\endgroup$
    – user48713
    Jun 20, 2014 at 11:03
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    $\begingroup$ (+1). Note that the upper bound is obvious upon observing two exponential variables differ only by a scale factor. It is equally obvious that the lower bound cannot attain $-1$ when $\lambda_1\ne\lambda_2$ (for otherwise the skewness would be zero). $\endgroup$
    – whuber
    May 14, 2015 at 18:54

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