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If $X$ and $Y$ are two correlated Gaussian Random variables with parameters $(\mu_X, \sigma_X)$ and $(\mu_Y, \sigma_Y)$. If $Z=\exp(X - Y)$ then what is the distribution of $Z$.

Math world seems to suggest $X-Y$ should have a Normal Difference Distribution, based on this would that mean $Z$ has a log normal difference distribution ? I am unsure if there is any distribution named that. Also would the mean and standard deviation of $Z$ be simply $\exp(\mu(X - Y))$ and $\exp(\sigma(X - Y))$ where $\mu$ and $\sigma$ imply the mean and standard deviations one gets from Normal Difference distribution of $X-Y$.

Any help would be much appreciated.

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up vote 1 down vote accepted

If $X$ and $Y$ are joint normal, then $X-Y$ being a linear transformation of normal variables is again normal, implying that $\exp(X-Y)$ is log-normal. If no joint normality holds, log-normality may no longer follow. In general, you need to specify the joint distribution of $(X,Y)$ in order to derive that of $\exp(X-Y)$.

In case of joint normality, $X-Y$ is normal with mean, say, $\mu_d=\mu_X-\mu_Y$ and variance $\sigma_d^2=\sigma_X^2 + \sigma_Y^2-2 \text{Cov}(X,Y)$. In that case $\exp(X-Y)$ will a log-normal with those same parameters (that is, $\mu_d$ and $\sigma_d^2$). See this on how to specify the parameters of a log-normal distribution, and other information on that distribution.

Additionally, one point to be careful about here is that $\mu_d$ and $\sigma_d^2$ are the two parameters of log-normal distribution and NOT its mean and variance. For formulas for its two moments, you could look here.

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Even if they are joint normal. What about a scenerio if both $X$ and $Y$ shared the same mean but difference standard deviations, let's say standard deviation of $Y$ is abotu $1/3$rd that of $X$. Then $X-Y$ would have a bimodal distribution ? would this still be a log-normal distrbution ? – Comic Book Guy Dec 7 '12 at 5:49
It doesn't matter. I will edit my answer to clarify that point. – Learner Dec 7 '12 at 5:52
Thanks i look forwrd to it. – Comic Book Guy Dec 7 '12 at 5:54

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