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I am struggling with a proof, and I am wondering if anyone can help or point me to the right direction. Suppose that we have two variables, $X$ and $Y$, and they follow a multivariate normal distribution with covariance matrix $Q$. Now, suppose that we have two more variables, $Z_1$ and $Z_2$ that follow a normal distribution with covariance $\Omega$, whose elements are,

$\Omega_{11} = exp(X)$, $\Omega_{22} = exp(Y)$, $\Omega_{21} = \Omega_{12}=\exp (X/2)\exp(Y/2)$,

And the question is, what is the expectation of $\Omega$?

For $\Omega_{11}$ and $\Omega_{22}$ it is clear that the expectation is $\exp(0.5Q_{11})$ and $\exp(0.5Q_{22})$ using the properties of log normal variables, but I am not sure how to compute it for the covariance term.

Thanks!

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    $\begingroup$ So the Variance of, say, $Z_1$, which is $\Omega_{11}$ is $e^{X}$? I.e., the variance is a random variable here? $\endgroup$ – Alecos Papadopoulos Jul 19 '14 at 17:29
  • $\begingroup$ Yes, exactly! This is related to stochastic volatility models, I just tried to simplify it as much as I could $\endgroup$ – Chucky Jul 20 '14 at 8:48
  • $\begingroup$ Then it should be conditional variance, ${\rm Var}[Z_1|X_1,X_2]$, etc. $\endgroup$ – StasK Oct 26 '14 at 15:52
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    $\begingroup$ Are you claiming that $Z_1$ and $Z_2$ are normal random variables but do not enjoy a multivariate normal distribution? Note that if you are saying that conditioned on the values of $X$ and $Y$, $Z_1$ and $Z_2$ have covariance matrix $\Omega$, then $\Omega$ has determinant $0$ and $Z_1, Z_2$ are conditionally perfectly correlated and thus just scalar multiples of each other. $\endgroup$ – Dilip Sarwate Dec 31 '14 at 15:17
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If $Z_1$ and $Z_2$ are random variables with covariance matrix $$\Omega = \left[\begin{matrix}\exp(x)&\exp((x+y)/2)\\\exp((x+y)/2) & \exp(y)\end{matrix}\right]\tag{1}$$ then $Z_1$ and $Z_2$ are perfectly posiitively correlated random variables: their (Pearson) correlation coefficient is $1$ and $$\frac{Z_2-\mu_2}{\exp(y/2)} = \frac{Z_1-\mu_1}{\exp(x/2)}.\tag{2}$$ This holds regardless of the specific distributions of $Z_1$ and $Z_2$ as long as the said distributions have the given means and variances. In particular, it is not necessary to assume that $Z_1$ and $Z_2$ are normal random variables, but doing so is fine.

If $(x,y)$ is a realization of a pair $(X,Y)$ of random variables, then $\Omega$ in $(1)$ can be thought of as the conditional covariance matrix of $Z_1$ and $Z_2$ given that $(X,Y)$ has value $(x,y)$. Thus, the expected value of $\Omega$ regarded as a function of $(X,Y)$ is $$E[\Omega(X,Y)] = \left[\begin{matrix}E[\exp(X)]&E[\exp((X+Y)/2)]\\E[\exp((X+Y)/2)] & E[\exp(Y)]\end{matrix}\right].\tag{3}$$ The diagonal terms of the right side of $(3)$ are just the moment-generating functions $M_X(t)$ and $M_Y(t)$ of $X$ and $Y$ respectively (evaluated at $1$) while the off-diagonal terms are also easily determined in the special case when $(X,Y)$ enjoy a bivariate normal distribution. Note that $X+Y = W$ is a normal random variable with mean $$E[W] = E[X]+E[Y], \quad \operatorname{var}(W) = \operatorname{var}(X) + \operatorname{var}(Y) + 2\operatorname{cov}(X,Y)\tag{4}$$ and so the off-diagonal terms are $M_W(\frac 12)$ where $W$ is a normal random variable with mean and variance given by $(4)$.

Note that the distributions of $Z_1$ and $Z_2$ are irrelevant to these calculations as long as their covariance matrix is given by $(1)$.

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Ok, I think I figured it out, but could anyone check if I am doing something wrong?

So, first define $\tilde{X} = X/2$ and the same for $\tilde{Y}$. Then if $R$ is a diagonal matrix with $0.5$ in each element of the diagonal, the covariance matrix of $\tilde{X}$ and $\tilde{Y}$ is $\tilde{Q}=RQR$.

Then, we can rewrite,

$\mathbb{E}[\exp(\tilde{X})\exp(\tilde{Y})] = Cov[\exp(\tilde{X}),\exp(\tilde{Y})] + \mathbb{E}[\exp(\tilde{X})]\mathbb{E}[\exp(\tilde{Y})]$

The product of the expectations is trivial (again, using log normal distribution properties), and for the covariance, we use the properties of multivariate log-normal distribution, so that,

$Cov[\exp(\tilde{X}),\exp(\tilde{Y})] = e^{0.5(\tilde{Q}_{11}+\tilde{Q}_{22})}(e^{\tilde{Q}_{12}}-1))$

What do you think?

PS: Also note that I am assuming the the process for $X$ and $Y$ is zero mean.

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