# Expected value of Normal Lognormal Mixture

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.

• 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 Dec 18, 2018 at 12:16

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}
• 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!! Dec 18, 2018 at 19:17