Parameters of the log-normal from CDF of a composition of $n$ i.i.d

Let $$X_1,\ldots,X_n$$ be i.i.d. log-normal random variables such that $$\log(X_i)\sim N(\mu,\sigma^2)\ \ \forall i=1,\ldots,n$$

Now let $$Y$$ be equal to the $$\min(X_1,\ldots,X_n)$$. It is quite easy to obtain the relation between the corresponding CDFs: $$P(Y

Is there a closed form to calculate parameters $$\mu_y,\sigma_y$$ of a log-normal variable $$Y$$ given parameters $$\mu, \sigma$$ of a log-normal variable $$X$$? And vice versa (find $$\mu,\sigma$$ given $$\mu_y,\sigma_y$$)?

• Welcome to CV, Roman. Your question is predicated on a falsehood: when $n\gt 1,$ $Y$ does not have a lognormal distribution. One can compute the geometric mean and geometric sd of $Y$ using numerical methods, if that's what you're trying to ask.
– whuber
Commented Apr 13, 2023 at 16:36
• Many questions on site address the approximation of a sum of independent lognormals by a lognormal (which often works fairly well, even just by matching the first two moments). However, counterexamples where it's clearly not approximately lognormal are not so hard to identify. Commented Apr 14, 2023 at 0:34
• @whuber, Glen_b, thanks for your answers. So, in this example, $F_Y$ is not log-normal? Maybe there is a way to approximate it by one? The ultimate goal is to be able to draw randomly from $Y$ knowing the distribution of $X$, and/or be able to draw randomly from $X$ knowing the distribution of $Y$ Commented Apr 14, 2023 at 10:16
• I doubt a lognormal approximation would be very good, especially for large $n.$
– whuber
Commented Apr 14, 2023 at 12:07

Here $$Y$$ is the minimum of $$n$$ independent samples from a lognormal distribution $$X$$ with $$\ln(X)\sim N(\mu,\sigma)$$. Then $$Y$$ does not have a lognormal distribution, but one can still ask:
What are the mean and standard deviation of $$Y$$?
For large $$n$$, the distribution of $$Y$$ will be close to a Gumbel distribution with mean $$d_n + \gamma c_n$$, or $$e^\mu\left(1+\frac{\gamma\sigma}{\sqrt{2\ln n}}\right)\left( \ln\frac{n^4\ln n}{4\pi}\right)^{\sigma/\sqrt{8 \ln n}}$$ where $$\gamma$$ is Euler's gamma.