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For example, if we had random variables $X$ and $Y$ and we know that $corr(X,Y)=\rho$, how would you solve for $Cov(e^X,Y)$?

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  • $\begingroup$ Because the exponential function preserves order, if one of the correlations is $\pm 1,$ then so is the other. Apart from that, any ordered pair $(\rho,\rho^\prime)$ for which both values lie between $-1$ and $1$ can be the correlations of $(X,Y)$ and $(e^X,Y).$ $\endgroup$
    – whuber
    Commented Apr 6, 2022 at 18:15

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The following is given, $$ \int dxdy \ p(x, y)(x - \mu_x)(y - \mu_y) = \sigma_x\sigma_y\rho, $$ where $\mu_x(\mu_y)$ and $\sigma_x(\mu_y)$ are the mean and standard deviation of $X(Y)$, the requested covariance is given by $$ \begin{aligned} {\rm Cov}(e^X, Y) &= \int dxdy \ p(x, y)(e^x - \mu_{e^x})(y - \mu_y)\\ &=\int dxdy \ p(x, y)\left(1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots - \mu_{e^x}\right)(y - \mu_y) \end{aligned} $$ The mean $\mu_{e^x}$ $$ \begin{aligned} \mu_{e^x} &= \int dxdy \ p(x, y) e^x\\ &= M_X(t=1)\\ &= 1 + \mu_x + \frac{1}{2!}m_{2,x} + \frac{1}{3!}m_{3,x} + \cdots \end{aligned} $$ where $M_X$(t) is the moment generating function (see the definition here), and $m_{n,x}$ is the $n$-th moment of $p_X(x) = \int dy \ p(x, y)$.

We can further simplify the expression as $$ \begin{aligned} {\rm Cov}(e^X, Y) &= \int dxdy \ p(x, y)(y - \mu_y)\left(1 + x + \frac{1}{2!}x^2 + \cdots - 1 - \mu_x - \frac{1}{2!}m_{2,x} - \cdots\right)\\ &= \sigma_x\sigma_y\rho + \sum_{n=2}^{\infty}\frac{1}{n!}\int dxdy \ p(x, y)(y - \mu_y)(x^n - m_{n,x}). \end{aligned} $$ Therefore we would need a lot more information to obtain the desired correlation.

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    $\begingroup$ The conclusion does not necessarily follow, because it's logically possible that the covariance could be computed in some other way that doesn't require all moments to exist and be known. One logically correct way to respond would be to exhibit two variables $(X_1,Y_1)$ and $(X_2,Y_2)$ with the same correlation coefficient but with different values for the correlation of $(e^{X_i},Y_i).$ An even more complete response would exhibit a family of such variables $(X,Y)$ realizing the full locus of $\left(\operatorname{Cor}(X,Y), \operatorname{Cor}\left(e^X,Y\right)\right).$ $\endgroup$
    – whuber
    Commented Apr 6, 2022 at 17:29

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