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


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

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

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 $N\left(E(\log X), \frac{\operatorname{Var}(\log X)}{n}\right).$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.