12
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ This question is linked to a side comment here . $\endgroup$ – QuantIbex Aug 7 '13 at 17:48
9
$\begingroup$

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 $\lambda_2$.

$\endgroup$
2
  • 1
    $\begingroup$ 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 '14 at 11:03
  • 2
    $\begingroup$ (+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 '15 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.