Attainable correlations for exponential random variables 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?
 A: 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 $\lambda_2$.
