2
$\begingroup$

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 $n$)?

My annotations

$\log{Y/n}=\frac{\log(X_1)+\log(X_2)+...+\log(X_n)}{n}$

Let's call $\log(X_i)$ from $Z_i$, so we have $$\frac{Z_1+Z_2+Z_3+...+Z_n}{n}=\overline{Z}n$$

by Central limit theorem: $\frac{\overline{Z}n-\mu_z}{\sigma_z/\sqrt{n}}\rightarrow_dNormal(0,1)$

But I'm stuck here, I can not find the distribution of $\log{Y/n}$

$\endgroup$
2
  • $\begingroup$ Please beware that $Y=\frac{\log(X_1)+\log(X_2)+...+\log(X_n)}{n}$ is false with this notation. You should take logarithms in both sides: $\log(Y)=\log(X_1)+\log(X_2)+...+\log(X_n)$ $\endgroup$
    – Pere
    Commented Mar 20, 2018 at 13:53
  • $\begingroup$ yes, my bad sorry $\endgroup$
    – beginner
    Commented Mar 20, 2018 at 15:36

1 Answer 1

2
$\begingroup$

There aren't enough assumptions stated yet. We need to know something about the mean, variance of the $X$s. At least you must know they're finite. If so, you're on the right track with CLT. Another way of expressing the CLT is this: $\bar{X} \approx_d \mathcal{N}\left( \mu, \sigma^2/n\right)$. That is, the sample mean is approximately normally distributed with mean equal to the population mean of $X$ and variance equal to the $1/n$ scaled population variance of $X$. The last bit is accounting for the change in scale. Log $X$ has an approximate variance as well, given by the $\delta$-method.

$\endgroup$
4
  • $\begingroup$ These are the information I have. I have neither the mean nor the variance of $Xs$ ... $\endgroup$
    – beginner
    Commented Mar 20, 2018 at 15:38
  • $\begingroup$ So $\mathcal{N}\left(E(\log X), \frac{\operatorname{Var}(\log X)}{n}\right)$? $\endgroup$
    – beginner
    Commented Mar 20, 2018 at 15:42
  • $\begingroup$ @beginner You must state the assumptions you must make to give the most general answer possible. Always remember the Cauchy dist'n as a counterexample. Your expression of the mean is correct: we can say nothing of the mean of the log-X, although bounds can be given by Jensen's inequality. The variance can be expressed as a function of the variance of X (and not log-X) if you use the $\delta$method. $\endgroup$
    – AdamO
    Commented Mar 20, 2018 at 16:25
  • 1
    $\begingroup$ To be honest, I don't know the $\delta$ method $\endgroup$
    – beginner
    Commented Mar 20, 2018 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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