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Let $X_1$ and $X_2$ be two normal random variables. Write $X_1\sim N(\mu_1, \sigma^2_1)$ and $X_2\sim N(\mu_2, \sigma^2_2)$, to fix ideas.

Consider the corresponding log-normal random variables: $Z_1 = \exp(X_1)$, $Z_2 = \exp(X_2)$.

Question: what is the distribution of the product of the two random variables, i.e., the distribution of $Z_1Z_2$?

If the normal random variables $X_1, X_2$ are independent, or they have a bivariate normal distribution, the answer is simple: we have $Z_1Z_2 = \exp(X_1+X_2)$ with the sum $X_1+X_2$ normal, hence the product $Z_1Z_2$ is still lognormal.

But suppose that $X_1, X_2$ are generally $not$ independent, say with correlation $\rho$. What can we say about the distribution of $Z_1Z_2$?

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  • $\begingroup$ This might be useful: stats.stackexchange.com/questions/19948/… $\endgroup$
    – Greenparker
    May 15, 2016 at 17:20
  • $\begingroup$ I doubt it though.. basically this question asks "if the marginals are normally distributed, can we say anything about their joint distribution?" And I don't think we can say much in general $\endgroup$
    – Ant
    May 15, 2016 at 19:27
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    $\begingroup$ $Z_1 Z_2 = \exp(X_1 + X_2)$ in general, so your real question is whether $X_1+X_2$ is normal (which it will be if $X_1, X_2$ are bivariate normal with correlation $\rho$) $\endgroup$
    – Henry
    May 15, 2016 at 22:12
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    $\begingroup$ If you don't have bivariate normality, merely specifying the correlation and the margins is not sufficient to pin down the bivariate distribution. $\endgroup$
    – Glen_b
    May 16, 2016 at 1:31

1 Answer 1

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Using Dilips answer here, if $X$ and $Y$ are bi-variate normal and $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ and the correlation between $X$ and $Y$ is $\rho$. Then

$$ Cov(X,Y) = \rho \sigma_1 \sigma_2,$$

$$X + Y \sim N(\mu_1 + \mu_2, \sigma^2_1 + \sigma^2_2 + 2\rho\sigma_1 \sigma_2). $$

Thus $Z_1Z_2$ will also be a lognormal distribution with parameters $\mu_1 + \mu_2$ and $\sigma^2_1 + \sigma^2_2 + 2\rho\sigma_1 \sigma_2$.

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    $\begingroup$ Add some qualification? This is true if the distribution of (X,Y) is bi-variate normal, but that the marginal distribution of X is normal and the marginal distribution of Y is normal does not imply the joint distribution is bi-variate normal... stats.stackexchange.com/questions/30159/… $\endgroup$
    – Matthew Gunn
    May 15, 2016 at 17:41
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    $\begingroup$ @MatthewGunn Interesting. I was blissfully unaware of this. As it stands, my answer does not address the question completely. I will wait a couple of hours before deleting it. Thanks. $\endgroup$
    – Greenparker
    May 15, 2016 at 17:44
  • $\begingroup$ Do you have alot of data from this distribution? In such case you can plot it, or otherwise test if bivariate normal is a good distribution approximation. Otherwise, maybe you could estimate a copula and go from there! $\endgroup$
    – kjetil b halvorsen ♦
    May 15, 2016 at 18:36
  • $\begingroup$ This is quite obvious, but you're missing the point: what happens if you don't know whether the two marginals have a bivariate distribution? $\endgroup$
    – RandomGuy
    May 16, 2016 at 10:53
  • $\begingroup$ Thanks! This amazing answer saved my day! $\endgroup$
    – Jinhua Wang
    May 26, 2019 at 15:45

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