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I am doing some independent study in asymptotic statistics and point estimation and am aware that you can get from log transformations of uniform random variables (exponential) all the way up to chi-squared distributions but I came across an intriguing question and I'm not sure how to handle it. I was thinking CLT or Delta Method but am unsure of myself. The thing I am trying to sort out is:

Given $X_1,\ldots X_n\overset{iid}{\sim}Uniform(0,\theta)$ and $Z_n=\left(\prod_{i=1}^nX_i\right)^{1/n}$ what is an example of a function of $Z_n$, i.e. $g(Z_n)$, so that $g(Z_n)\overset{d}{\rightarrow}\chi_{(1)}^2$, where $\overset{d}{\longrightarrow}$ denotes convergence in distribution and $\chi_{(1)}^2$ is a chi-squared random variable with 1 degree of freedom?

Maybe something like $g(h(Z_n))$ where $h(Z_n)\overset{d}{\rightarrow}Normal(0,1)$ and $g(t)=t^2$? Any help is greatly appreciated.

I have tried:

Let $Y_1=X_i/\theta$ I get $Y\sim Uniform(0,1)$ and so $-\log(Y)\sim Exponential(1)$. From here I know $-2\log(Y)$ is $\chi_{(2)}^2$ but I don't see how to relate $-\log(Y)$ back to a function of $Z_n$ that is $\chi_{(1)}^2$

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As you write,

$$Y_i = X_i/\theta \Rightarrow Y_i \sim U(0,1), \;\; -2\ln(Y_i) \sim \chi_{(2)}^2$$

Consider

$$Z_n=\left(\prod_{i=1}^nX_i\right)^{1/n} = \left(\prod_{i=1}^n(\theta Y_i)\right)^{1/n}$$

$$\ln Z_n = \frac 1n \sum_{i=1}^n \left(\ln \theta + \ln Y_i\right) = \ln \theta + \frac 1n \sum_{i=1}^n \ln Y_i$$

$$\Rightarrow 2\ln \theta - 2\ln Z_n = \frac 1n \sum_{i=1}^n (-2\ln Y_i)$$

The sum of independent ch-squares is a chi-square also, having as degrees of freedom the sum of the degrees of freedom of the components. So

$$\sum_{i=1}^n (-2\ln Y_i) = W \sim \chi_{(2n)}^2$$

Then also,

$$G=\frac 1n W \sim Gamma (k= n, m = 2/n)$$

(shape-scale parametrization). This Gamma distribution has mean $E(G) = km = 2$ and variance $\operatorname{Var}(G) = km^2 = 4/n$. Centered and scaled, this r.v. converges to a standard normal, as $k=n \rightarrow \infty$

$$\frac {G-2}{\sqrt{4/n}} \rightarrow_d N(0,1),\;\; n\rightarrow \infty$$

But we also have

$$\frac {G-2}{\sqrt{4/n}} = \sqrt{n}\frac {2\ln \theta - 2\ln Z_n-2}{2} = \sqrt{n}\left(\ln \theta - \ln Z_n-1\right)$$

So the random variable

$$Q_n = \sqrt{n}\left(\ln \theta - \ln Z_n-1\right) \rightarrow_d Q \sim N(0,1)$$

Then by the continuous mapping (Mann-Wald) theorem

$$ Q_n^2 \rightarrow_d Q^2 \sim \chi_{(1)}^2$$

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