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Let Z1 and Z2 be independent standard normal random variables. Find the following The probability density function of $e^{3Z_1+2Z_2}$

The given solutions is as follows:

enter image description here

But what doesn't make sense to me is; why are we integrating, when there is a formula

$\frac{e^{\frac{-(x-u)^2}{2a^2}}}{\sqrt{2\pi a}}$

for the standard normal p.d.f

If $3z_1 + 3z_2 = 5Z$ and $5 Z$ is an RV on its own

then according to the following ( which is the way I solved it at first)

enter image description here

the $\frac{1}{5z} $ factor shouldn't be there so the final solution would be

$\frac{e^{\frac{-(\frac{1}{5}ln(z))^2}{2}}}{\sqrt{2\pi }}$

Are the solutions incorrect or am I thinking about this wrong? This is a student written solution so it is not as reliable, hence the skepticism.

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  • $\begingroup$ If a RV $Y = e^X$ for $X$ normal, then $Y$ is said to be distributed log-normal. en.wikipedia.org/wiki/Log-normal_distribution $\endgroup$ Apr 20, 2019 at 20:20
  • $\begingroup$ @pbn990 that does not clarify anything. $\endgroup$
    – Kal
    Apr 20, 2019 at 21:05
  • $\begingroup$ I've explained this in an answer $\endgroup$ Apr 20, 2019 at 21:24

1 Answer 1

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For $e^{5Z}$, there does not appear to be a ratio of two normals, so I am not sure if the theorem you cited will be relevant to the problem. You can update your answer and include your work if you want specific comments on that.

With that being said, the student's answer is correct. As noted, $3Z_1 + 2Z_2$ is distribute normal with mean 0 and variance 25. By the link I attached in the comments, this means that $e^{5Z} \sim LogNormal(0, 25)$. In the article the pdf for this distribution is given, and if you plug in the parameters you will get the exact same answer as the student. The article also gives a way to derive the pdf, so this will give another way to solve the problem.

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