# Product of a shifted Log-Normal and a Log-Normal distribution

Let $X$ and $Y$ follow Log-Normal distributions, with $\ln X \sim \mathcal{N}(\mu_x, \sigma_x^2)$ and $\ln Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$. $X$ and $Y$ are independent. Let $W = X (Y + c)$, where $c$ is a constant. Is $W$ still Log-Normally distributed? If not, can $W$ be approximated by a Log-Normal distribution?

I know that the product of Log-Normal distributions is Log-Normal. Unfortunately, $Y+c$ is no longer a Log-Normal, since its support is $[c, \infty)$. On the contrary, the support of $W$ is again $[0, \infty)$, which does not exclude the possibility that $W$ follows a Log-Normal distribution.

Moreover, the numerical approximation of $W$ seems indeed to be Log-Normal - see the superposition of the histograms below, with $\mu_x = 2$, $\mu_y = 4$, $\sigma^2_x = \sigma^2_y = 1$, and $c = 5$.

• How are $X$ and $Y$ related? Are they jointly normal? perfectly correlated? independent? – Glen_b Sep 8 '17 at 3:15
• I indeed meant independent, sorry for the confusion. – Louis Seg Sep 8 '17 at 8:20
• "On the contrary, the support of WW is again [0,∞)[0,∞), which does not exclude the possibility that WW follows a Log-Normal distribution." - This doesn't matter. Because, yes, it doesn't exclude the possibility, but not every function that has that support is a log-normal distribution. So it shouldn't cause one to think this IS log-normal either. – Mathemagical Sep 8 '17 at 8:30
• Of course Mathemagical ! It was simply to point out that the support of the distribution cannot be used to disprove that $W$ is a Log-Normal. – Louis Seg Sep 8 '17 at 8:50

## 2 Answers

Assuming the intent is that $X$ and $Y$ are independent, then $X(Y+c)$ is not lognormal.

(edit: consider that by Cramer's theorem $\log(W)$ will only be normal if both $\log(X)$ and $\log(Y+c)$ are, but the second term is not normal; Mathemagical mentions this argument below)

However, sums of bivariate lognormals are often very close to lognormal and $XY$ and $cX$ are jointly lognormal; for a fair range of values of the parameters, the approximation by a lognormal will probably be fairly decent.

With your particular example, a large simulation (say $10^6$ points) clearly shows that the lower tail of $\log(W)$ is a bit lighter and the upper tail a bit heavier than for a normal so you can tell without even worrying about the algebra. Repeated simulations show the same picture each time.

Nonetheless the lognormal approximation is excellent in that particular case, especially in the middle 98% or so of the distribution.

However, you can't always rely on it working without checking; it's not all that hard to come up with examples where the sum of lognormals is not close to lognormal. See some of the discussion and references in this post on CrossValidated -- The sum of independent lognormal random variables appears lognormal? for some additional information (though the question is about the independent case while $XY$ and $Xc$ are not independent, nonetheless the information is relevant)

Some papers for the correlated case:

Mehta, N.B., Wu, J., Molisch, A. F., and Zhang, J., (2007),
"Approximating a Sum of Random Variables with a Lognormal,"
IEEE Transactions on Wireless Communications, 6(7), 2690-2699.

Rook CJ and Kerman MC, (2015),
"Approximating the Sum of Correlated Lognormals: An Implementation"
arXiv:1508.07582 [q-fin.GN]
https://arxiv.org/ftp/arxiv/papers/1508/1508.07582.pdf

Lo, C.F. (2012),
The Sum and Difference of Two Lognormal Random Variables Journal of Applied Mathematics Volume 2012, Article ID 838397, 13 pages http://dx.doi.org/10.1155/2012/838397

• Thank you for your answer ! Do you have any thoughts on how to show theoretically that $X(Y + c)$ is not Log-Normally distributed ? – Louis Seg Sep 8 '17 at 8:24
• Louis, $\log W = \log X + \log (Y+c)$. $\log W$ is normal if and only if you know $\log (Y+c)$ is normal since the first term is normal. But, you know $\log (Y+c)$ is not normal. So $\log W$ is not normal and $W$ is not log-normal. Also, see my comment on your your original question above. – Mathemagical Sep 8 '17 at 8:36
• Thanks Mathemagical ! I did not know that for $W = X + Y$ to be normal, if $X$ is normal, $Y$ also had to be normal. Any reference on this? – Louis Seg Sep 8 '17 at 8:52
• For independent X and Y there's Cramer's theorem. Interestingly I had typed in the argument Mathemagical gave above but - expecting you would ask exactly the followup you did ask him -- I went off to find Cramer's theorem before pressing "add comment" (I'd forgotten the name of it which made it hard to locate), and when I got back Mathemagical had posted the same argument. But at least I have the name of that theorem now... so even though I no longer need to add the comment, it was worth looking -- you have the whole argument now. – Glen_b Sep 8 '17 at 9:13
• Thank you very much Mathemagical and Glen_b ! – Louis Seg Sep 8 '17 at 9:18

Of course the numerical simulation cannot tell you that much since lot of distribtuions have a similar shape. There are some trivial approaches for approximation for $c>0$: $$P(X(Y+c)\leq w)\leq \min\{P(XY\leq w),P(Xc\leq w\}\}.$$ So the CDF is upper bounded by log-normal distributions. And for the lower bound for all $\delta\in(0,1)$: $$P(XY\leq w(1-\delta),cX\leq \delta w)\leq P(X(Y+c)\leq w),$$ which has log-normal tail so no surprise that you see log-normal similarity in your numerical evaluations.

You can try to explicitly characterize the distribution, although not trivial:

$$P(W\leq w)=P(X(Y+c)\leq w)=P(\ln X+\ln(Y+c)\leq \ln w)\\ =\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}P(\ln(Y+c)\leq \ln w-x)\mathrm dx..$$ Using the following $$P(Y+c\leq e^{w-x})=P(Y\leq e^{w-x}-c)=P(\ln Y\leq \ln(e^{\ln w-x}-c))=F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c)),$$ we get $$P(W\leq w) =\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c))\mathrm dx\\ =\int_{-\infty}^{\ln\frac wc}\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c))\mathrm dx.$$ Even for the standard normal distribution, this will not be log-normal.