I am looking for an analytical proof, that the sum of two log normal random variables is not log-normal. Couldn't find it anywhere, does somebody know where to find it or know how to do it?
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1$\begingroup$ Do you agree that $\ln (A+B) \ne \ln(A) + \ln(B)$ ? Where A and B are lognormal variables. $\endgroup$– nbbo2Feb 28, 2017 at 18:16
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3$\begingroup$ try to find the characteristic function and see whether it agrees with that of a lognormal. $\endgroup$– GordonFeb 28, 2017 at 18:17
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$\begingroup$ @Gordon I had to check myself but thge characteristic function is very difficult (no closed form en.wikipedia.org/wiki/Log-normal_distribution). So the question is valid and I think noob2 s answer is the simpliest. $\endgroup$– Richi WaMar 1, 2017 at 7:57
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$\begingroup$ @Richard: Is this article helpful? I had not carefully gone through it. $\endgroup$– GordonMar 1, 2017 at 15:31
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$\begingroup$ @Gordon thank you ! the article is very interesting but I don't have the time right now to try the chf approach using the formula therein ... $\endgroup$– Richi WaMar 1, 2017 at 15:57
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2 Answers
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Let lnA be N(0,1) and lnB be N(0,k) where we will let k tend to zero. Then B has all of its density at 1, so A+B>1 in the limit. Hence A+B is not lognormal.
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$\begingroup$ You try to find one example for the contradiction. Can't we find any easier one? And the density of B accumulates around 0 - not 1, right? $\endgroup$– Richi WaMar 1, 2017 at 7:52
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$\begingroup$ @Richard: The statement that $B$ has its mass around 1 looks correct to me. $\endgroup$ Mar 1, 2017 at 11:16
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$\begingroup$ @LocalVolatility right! .. I forgot the log .. sorry! $\endgroup$– Richi WaMar 1, 2017 at 11:51
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$\begingroup$ Ok, so could i write it down like that: An intuition why the sum of log-normal distributed random variables is not log-normal, could be given like this: Let $X_1\sim\mathcal{N}(0,1)$ and $X_2\sim\mathcal{N}(0,k)$ and let k tend to zero. Then it follows, that $\exp(X_2)$ has all of its probability density concentrated at 1, which implies that $\exp(X_1)+\exp(X_2)\geq 1$ a.s. .This however implies $ln(\exp(X_1)+\exp(X_2)\geq 0$ a.s., which is a contradiction to being normal, as the domain of the normal distribution is $\mathbb{R}$. $\endgroup$– Mh AztecMar 1, 2017 at 11:58
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$\begingroup$ You need to be careful with how you write this argument- saying things like "in the limit" and then referring to a situation where $k>0$ is sloppy. $\endgroup$ Mar 1, 2017 at 13:17
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Let $Z_1$ and $Z_2$ be normal random variables. Therefore, $e^{Z_1}$ and $e^{Z_2}$ are log-normal random variables.
The central limit theorem says that the sum of random variables tends toward a normal distribution even if their sampling distributions are not normally distributed. Therefore $Z_1$ + $Z_2$ will tend toward normally distributed. However, $e^{Z_1} + e^{Z_2} \ne e^{Z_1+Z_2}.$
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$\begingroup$ I don't think this is a valid answer, though you reach the correct conclusion. 1) The CLT applies in the limit a the number of random variables becomes large. Here we have a sum of two. 2) In its common form, it requires i.i.d. random variables. It is not clear from the question that the OP is only interested in two identically distributed log-normals. $\endgroup$ Feb 28, 2017 at 23:17
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$\begingroup$ @LocalVolatility, I'm not sure I agree. Assume for a contradiction that the sum of two lognormals is always lognormal. Then it certainly follows that the sum of two iid lognormals is lognormal , and also that the sum of multiple iid lognormals is lognormal. This then leads to the CLT contradiction. $\endgroup$– dm63Mar 1, 2017 at 4:21
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2$\begingroup$ @dm63, not sure I agree with you. You could take any distribution closed under convolution and apply your reasoning to show that this is not really a "contradiction" per se. Let me take an example, the sum of two Poisson variables is Poisson distributed, this is a fact. Then it certainly follows that the sum of two iid Poisson is Poisson, and also that the sum of multiple iid Poisson is Poisson. Then it leads to the CLT contradiction. Well no, because CLT is a convergence result. $\endgroup$ Mar 1, 2017 at 8:59
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1$\begingroup$ @quantuple that is pretty convincing. The Poisson looks more and more like a Normal distribution as the parameter tends to infinity, but it still a Poisson not a Normal. I stand corrected $\endgroup$– dm63Mar 2, 2017 at 3:12