central limit theorem for a product

Question

Given $-1\leq x_i\leq 1$ identically distributed random variables for $i=1,2,\dots n$. What is the distribution function of their product? Is there a central limit theorem for products if $n$ is large?

Answer 1

The extension of the CLT to products would involve the $n^\text{th}$ root of $n$ variables. This raises problems when we consider random variables that might be negative. Therefore, let's consider random variables $x_k\in[0,1]$ where $P(x_k\lt a)=F(a)$.

Let $u_k=\log(x_k)$, then $P(u_k\lt a)=F(e^a)$.

The mean of $u_k$ is $$ \begin{align} \mu &=-\int_{-\infty}^0F(e^a)\,\mathrm{d}a\\ &=-\int_0^1F(t)\frac{\mathrm{d}t}{t} \end{align} $$ and the variance of $u_k$ is $$ \begin{align} \sigma^2 &=-2\int_{-\infty}^0aF(e^a)\,\mathrm{d}a-\left(\int_{-\infty}^0F(e^a)\,\mathrm{d}a\right)^2\\ &=-2\int_0^1F(t)\log(t)\frac{\mathrm{d}t}{t}-\left(\int_0^1F(t)\frac{\mathrm{d}t}{t}\right)^2 \end{align} $$ So, if $-\int_0^1F(t)\log(t)\frac{\mathrm{d}t}{t}\lt\infty$, the standard CLT applies to $\log(x_k)$ and the $n^\text{th}$ root of the product of $n$ variables tends to $$ e^{-\int_0^1F(t)\frac{\mathrm{d}t}{t}} $$ Thus, the product of $x_k$ approximates a log-normal distribution where the log of the product has mean $n\mu$ and variance $n\sigma^2$. That is, the distribution of the $n^\text{th}$ root of the product of $n$ variables approximates $$ \frac{\sqrt{n}}{x\sigma\sqrt{2\pi}}e^{-\frac{n}{2}\left(\frac{\log(x)-\mu}{\sigma}\right)^2} $$ which tends to a Dirac delta at $x=e^\mu$.

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The distribution of the logarithm of the product of $n$ of the $x_k$ will be the $n$-fold convolution of of the distribution of $\log(x_k)$, which is $e^aF'(e^a)$. The cumulative distribution of the logarithm of the product of $n$ of the $x_k$ is then $$ F_n(e^a)=\overbrace{e^aF'(e^a)\ast e^aF'(e^a)\ast\dots\ast e^aF'(e^a)}^{n-1\text{ terms}}\ast F(e^a) $$ The distribution of the product of $n$ of the $x_k$ is then $F_n'$

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Example 1: For a uniform distribution on $[0,1]$, we have $F(t)=t$ and the $n^\text{th}$ root of the product of $n$ variables tends to $e^{-1}$.

The distribution of the $n^\text{th}$ root of the product of $n$ uniform $[0,1]$ variables approximates $$ \frac{\sqrt{n}}{x\sqrt{2\pi}}e^{-\frac{n}{2}(\log(x)+1)^2} $$ which tends to a Diract Delta at $x=e^{-1}$.

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Example 2: To compute the distribution of the product of two variables, we need to consider $F$ on a wider domain: $$ F(t)=\left\{\begin{array}{l} 0&\text{if }t\lt0\\ t&\text{if }0\le t\le1\\ 1&\text{if }t\gt1 \end{array}\right. $$

we compute the convolution $$ F_2(e^a)=e^aF'(e^a)\ast F(e^a)=(1-a)e^a $$ Therefore, the cumulative distribution is $$ F_2(t)=(1-\log(t))t $$ and the distribution of the product of two uniformly distributed reals in $[0,1]$ is $$ F_2'(t)=-\log(t) $$

Answer 2

I wanted to give a one-line answer:

For $C_i$ i.i.d., $X=\prod^n_iC_i\implies log(X)=\sum^n_ilog(C_i)\implies \exists (\mu, \sigma)$ s.t. $log(X)\sim N(\mu, \sigma^2)$

Hence X follows a log-normal distribution.