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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 $\log{Y/n}$

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  • $\begingroup$ 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
    Mar 20, 2018 at 13:53
  • $\begingroup$ yes, my bad sorry $\endgroup$
    – beginner
    Mar 20, 2018 at 15:36

1 Answer 1

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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.

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  • $\begingroup$ These are the information I have. I have neither the mean nor the variance of $Xs$ ... $\endgroup$
    – beginner
    Mar 20, 2018 at 15:38
  • $\begingroup$ So $\mathcal{N}\left(E(\log X), \frac{\operatorname{Var}(\log X)}{n}\right)$? $\endgroup$
    – beginner
    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
    Mar 20, 2018 at 16:25
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    $\begingroup$ To be honest, I don't know the $\delta$ method $\endgroup$
    – beginner
    Mar 20, 2018 at 22:50

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