# what is this probability distribution?1

if $X_1,X_2,X_3,...,X_n$ is a sample size $n$ and $i.i.d.$ of a random variable with distribution $f(x)$, and

$$Y=X_1.X_2.X_3.....X_n$$

what is the approximate distribution of $\log{Y/n}$ for large sample (high $n$)?

Is not information missing from this question? Because $n$ is high I think I can assume that $f(x)$ is normal distribution, but I'm stuck on it

• Not quite following here. Did you mean for $Y$ to be the product of the $X$'s? – spaceisdarkgreen Mar 17 '18 at 22:05

## 1 Answer

If you mean to write $Y_n = X_1X_2\ldots X_n$ and are asking about the distribution of $\frac{1}{n}\log(Y_n),$ then this is the sample mean of independent samples of $\log(X).$ Provided that $E(|\log(X)|)<\infty,$ the law of large numbers says that this converges almost surely to $E(\log(X)).$ So the limiting distribution is a degenerate point at that value.

If we're looking to make a scale transformation so that the limiting distribution is non-degenerate, then provided we have $\operatorname{Var}(\log(X)) < \infty$ we can use the central limit theorem to conclude $$\frac{\sqrt{n}(\frac{1}{n}\log Y_n-E(\log X))}{\sqrt{\operatorname{Var}(\log X) }}\to_D N(0,1),$$ so for $n$ large the distribution of $\frac{1}{n}\log Y_n$ looks like a $N\left(E(\log X), \frac{\operatorname{Var}(\log X)}{n}\right).$

• $E(|\log(X)|)<\infty\to_{a.s.} E(\log(X))$? – Muradin Bronzebeard Mar 18 '18 at 16:07