# What is the pdf of a normal distribution divided by the square root of a log-normal over n?

We know that if $$Z \sim \mathcal{N}(0,1)$$, $$W \sim \chi^2(n)$$ and are independently distributed, then the variable $$Y = \frac{Z}{\sqrt{W/n}}$$ follows a $$t$$-distribution with degrees of freedom $$n$$. Now I am wondering, if $$X$$ is a log-normal like $$X \sim \log\mathcal{N}(a,b)$$ and is independent from $$Z$$, what is $$Y = \frac{Z}{\sqrt{X/n}}$$?

I know there is this answer, but it shows the pdf of $$Y = \frac{Z}{X}$$. Can anyone build on this to answer my question. Thank you!

• I indicated the answer in comments to your previous, now-deleted question. That's why we prefer you edit unanswered questions rather than post new ones.
– whuber
Commented Jul 1, 2021 at 14:13
• Thank for the advices @whuber!
– POC
Commented Jul 2, 2021 at 12:06

We can manipulate $$Y$$ to help us here. First note we can re-write $$Y$$ as

$$Y = \frac{n^{1/2}Z}{X^{1/2}}$$ and because $$Z\sim N(0,1)$$ then $$n^{1/2}Z \sim N(0, n)$$.

Now we know $$X \sim \log N(a, b)$$. By the definition of the log-normal we have $$\log(X) \sim N(a, b)$$. Using standard log laws we have $$\log(X^{1/2}) = \frac{1}{2} \log(X)$$ thus $$\log(X^{1/2}) \sim N(\frac{a}{2}, \frac{b}{4})$$.

So by the definition of the log-Normal we have $$\sqrt{X} \sim \log N(\frac{a}{2}, \frac{b}{4})$$.

So because $$Y$$ can actually be expressed as the ratio of a Normal RV and a log-Normal RV you can simply apply the result from the referenced question using the above representation of $$Y$$.

• Just an addendum to above answer: $\frac{1}{\sqrt{X}}$ is also log-normal. So $Y$ is a product of independent normal and log-normal variables. This [SE answer][2] points to this study of normal log-normal mixtures (author's term for $e^{1/2\eta}\epsilon$ with $\eta$ and $\epsilon$ correlated normal variables). [2]: stats.stackexchange.com/a/96623/40256
– ir7
Commented Jun 17, 2021 at 14:58
• @jken Are you sure $b$ is not divided $2$ rather than $4$? Simulations and other resources seems to be suggesting that. Also, will your answer works even if $Z$ is not standard? I would assume, but just want to be sure. Thank you,
– POC
Commented Jul 6, 2021 at 13:36
• Note $b/4$ is the variance not the standard deviation. A standard result is $var(aX) = a^2 var(X)$ so $b$ turns into $(1/2)^2b = b/4$. Have you taken $b$ to be the standard deviation? Commented Jul 6, 2021 at 18:13
• Yes I have. Thank you for the informations!
– POC
Commented Jul 6, 2021 at 19:38
• @jcken, last comment : if you don't mind checking this related question, I would really appreciate : stats.stackexchange.com/questions/533520/…
– POC
Commented Jul 6, 2021 at 20:13