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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Jul 1, 2021 at 14:13
  • $\begingroup$ Thank for the advices @whuber! $\endgroup$
    – POC
    Commented Jul 2, 2021 at 12:06

1 Answer 1


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 $Y$.

  • 1
    $\begingroup$ 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 $\endgroup$
    – ir7
    Commented Jun 17, 2021 at 14:58
  • $\begingroup$ @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, $\endgroup$
    – POC
    Commented Jul 6, 2021 at 13:36
  • $\begingroup$ 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? $\endgroup$
    – jcken
    Commented Jul 6, 2021 at 18:13
  • $\begingroup$ Yes I have. Thank you for the informations! $\endgroup$
    – POC
    Commented Jul 6, 2021 at 19:38
  • $\begingroup$ @jcken, last comment : if you don't mind checking this related question, I would really appreciate : stats.stackexchange.com/questions/533520/… $\endgroup$
    – POC
    Commented Jul 6, 2021 at 20:13

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