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I need to compute the following covariance: \begin{equation} Cov(X, exp(-a X)) \end{equation} where X follows a normal distribution, $X = N(0.0, \sigma^2)$, and $a$ is a constant scalar.

My findings: From the definition of covariance I concluded that \begin{equation} Cov(X, exp(-a X)) = E[X\ exp(-ax)] \end{equation} as X is zero-mean. Hence it boils down to finding the first moment of the normal lognormal mixture.

Upon searching stackexchange and the internet I only found one result which treats this topic (the work by Yang): http://repec.org/esAUSM04/up.21034.1077779387.pdf

I gives the first moments of the mixture $u=e^{1/2 \eta} \epsilon$. The one I am interested in is stated as: \begin{equation} E(u) = \frac{1}{2} \rho \sigma e^{\frac{1}{8} \sigma^2} \end{equation}

I cannot follow the "derivation" of this equation (none is actually given in the paper), but I believe that it is readily applicable to my LNL mixture.

The expected value has a factor which contains the covariance of the random variables considered by Yang and the other contains the exponential of the process $\eta$. In my case $\epsilon$ does not have unit variance, but variance $\sigma^2$. Also my $\eta$ is defined as $-2 a X$ to apply the logic of Yang. Since these two processes are fully negatively correlated, I assume that the Expected value should be:

\begin{equation} E[X\ exp(-ax)] = - a \sigma^2 \exp(\frac{1}{2} a^2 \sigma^2) \end{equation}

In simulations, this expectation matches the monte-carlo derived moment very well, hence I guess that above reasoning is correct.

My questions:

1) Is above reasoning really correct?

2) How did Yang compute the expected value? Understanding the derivation would allow me to directly start from $X exp(-aX)$, instead of fitting my mixture to his shape.

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  • $\begingroup$ Can't speak for how someone else calculated the expected value, but a reasonable approach would seem to be using $xe^{-ax} = -\frac{\partial}{\partial a}e^{-ax}$ and then the MGF of the normal $\endgroup$
    – Nadiels
    Commented Dec 18, 2018 at 12:16

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So while adapting the result from the paper to the perfectly correlated case would work it isn't the approach I would suggest, instead I would go with the approach in my comment -- it doesn't make anything simpler by considering the bivariate case. That said if you are curious as to how the result you are interested in is derived then the following would be one way of going about it.

Let $$ \begin{bmatrix} X \\ Y \end{bmatrix} \sim \mathcal{N}\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 & \rho \sigma \\ \rho \sigma & \sigma^2 \end{bmatrix} \right), $$ then $$ Y \mid X=x \sim \mathcal{N}(\rho\sigma x, \sigma^2(1-\rho^2). $$ So the idea is just going to be to use the MGF, and properties of the conditional expectation, it is a little tedious, but it should go something like $$ \begin{align} \mathbb{E}\left[Xe^{\frac{Y}{2}}\right] &= \mathbb{E}\left[ X\mathbb{E}\left[ e^{\frac{Y}{2}} \; \big| \; X \right]\right] \\ &= e^{\frac{\sigma^2(1-\rho^2)}{2^3}}\mathbb{E}\left[ X e^{\frac{\rho \sigma X}{2}}\right] \\ &=\frac{2}{\rho}e^{\frac{\sigma^2(1-\rho)}{2^3}}\frac{\partial}{\partial \sigma}\mathbb{E}\left[e^{\frac{\rho\sigma X}{2}} \right] \\ &=\frac{2}{\rho}e^{\frac{\sigma^2(1-\rho^2)}{2^3}}\frac{\partial}{\partial \sigma} e^{\frac{\rho^2 \sigma^2}{2^3}} \\ &= \frac{2}{\rho}e^{\frac{\sigma^2(1-\rho^2)}{2^3}} \cdot \frac{2 \rho^2 \sigma}{2^3}e^{\frac{\rho^2 \sigma^2}{2^3}} \\ &=\frac{1}{2} \rho \sigma e^{\frac{\sigma^2}{2^3}} \end{align} $$

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  • $\begingroup$ This is great! After reading up on the law of total expectation and following your first comment I was able to understand the steps you carried out. In fact the good news is that they can be applied in the same sequence to solve my original problem using $\rho = -1$, $X \sim \mathcal{N}(0.0, \sigma^2)$ and $Y \sim \mathcal{N}(0.0, a^2 \sigma^2)$. The result obtained in this way is the one I assumed and postulated in the original post. Thanks!! $\endgroup$
    – Elarion
    Commented Dec 18, 2018 at 19:17

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