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Let $X_i\sim U(0, 1)$, where $i = 1,\ldots,n$ and where all random variables are independent. Determine using the moment generating function (mgf) the distribution of the random variable $Y$

$$ Y = -2\ln\left(\prod_{i=1}^{n}X_i\right). $$

I know that in the case variables are independent and they're added together as to form a new variable. The new variable's moment generating function is just the product of the two old variables' mgfs but what confuse me here is that we're inside of the logarithm and I don't know how to deal with that.

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  • $\begingroup$ you have log of product which is just a sum of logs, then you can use what you know, since $\ln(X_k)$ are independent, too $\endgroup$ Nov 22, 2019 at 23:19
  • $\begingroup$ Spoiler alert. $\endgroup$
    – J.G.
    Nov 22, 2019 at 23:35
  • $\begingroup$ @DominikKutek : Since you're going to exponentiate in order to get the moment-generating function, I don't think that's the best way to look at it. $\endgroup$ Nov 23, 2019 at 0:16

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\begin{align} M_Y(t) & = \operatorname E(e^{tY}) \\[8pt] & = \operatorname E\left( e^{-2t\ln\prod_{i=1}^n X_i} \right) \\[8pt] & = \operatorname E\left(\left( e^{\ln\prod_{i=1}^n X_i} \right)^{-2t} \right) \\[8pt] & = \operatorname E\left( \left( \prod_{i=1}^n X_i \right)^{-2t} \right) \\[8pt] & = \operatorname E \left( \prod_{i=1}^n X_i^{-2t} \right) \\[8pt] & = \prod_{i=1}^n \operatorname E(X_i^{-2t}) \text{ by independence} \\[8pt] & = \left( \operatorname E\left(X_1^{-2t}\right) \right)^n \text{ by identical distribution}. \end{align}

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  • $\begingroup$ What happened to the $t$? $\endgroup$
    – Math1000
    Nov 23, 2019 at 0:16
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    $\begingroup$ Okay, so how do we recognize a distribution from this expression? I compute $$\mathbb E[X_1^{-2t}] = \int_0^1 x^{-2t} = 1-2t, $$ so $M_Y(t) = (1-2t)^n$. So this is just a $\chi^2$ distribution? $\endgroup$
    – Math1000
    Nov 23, 2019 at 0:25
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    $\begingroup$ For $x\ge0$ we have $$ \begin{align} & \Pr(-2\ln X \le x) \\ {} \\ = {} & \Pr( X \ge e^{-x/2}) \\ {} \\ = {} & 1 - e^{-x/2}, \end{align} $$ so that is an exponential distribution with expected value $2,$ and therefore a chi-square distribution with $2$ degrees of freedom. $\qquad$ $\endgroup$ Nov 23, 2019 at 0:36
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    $\begingroup$ When the number of degrees of freedom is anything besides $2,$ the chi-square distribution is not an exponential distribution. It is a gamma distribution (regardless of whether $\text{d.f.}=2$ or not). $\qquad$ $\endgroup$ Nov 23, 2019 at 0:38
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    $\begingroup$ A subtler point here is whether the chi-square distribution is determined by its moments. Some distributions are not. For example, there is a distribution that has the same moments, and thus the same moment-generating function, as the lognormal distribution, but is nonetheless a different distribution, not a lognormal. But some distributions are determined by their moments, and so do not share the same moment-generating function with any other distribution. $\endgroup$ Nov 23, 2019 at 0:47

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