# If $X_1,X_2$ are independent beta then showing $\sqrt{X_1X_2}$ is beta

Here is a problem that came in a semester exam in our university few years back which I am struggling to solve.

If $X_1,X_2$ are independent $\beta$ random variables with densities $\beta(n_1,n_2)$ and $\beta(n_1+\dfrac{1}{2},n_2)$ respectively then show that $\sqrt{X_1X_2}$ follows $\beta(2n_1,2n_2)$.

I used the Jacobian method to obtain that the density of $Y=\sqrt{X_1X_2}$ is as follows: $$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_y^1\dfrac{1}{x^2}(1-x^2)^{n_2-1}(1-\dfrac{y^2}{x^2})^{n_2-1}dx$$

I am lost at this point actually. Now, in the main paper, I found a hint had been supplied. I tried to use the hint but could not obtain the desired expressions. The hint is verbatim as follows:

Hint: Derive a formula for the density of $Y=\sqrt{X_1X_2}$ in terms of the given densities of $X_1$ and $X_2$ and try to use a change of variable with $z=\dfrac{y^2}{x}$.

So at this point, I try to make use of this hint by considering this change of variable. Hence I get, $$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_{y^2}^y\dfrac{z^2}{y^4}(1-\dfrac{y^4}{z^2})^{n_2-1}(1-y^2.\dfrac{z^2}{y^4})^{n_2-1}\dfrac{y^2}{z^2}dz$$which after simplification turns out to be (writing $x$ for $z$)$$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_{y^2}^y\dfrac{1}{y^2}(1-\dfrac{y^4}{x^2})^{n_2-1}(1-\dfrac{x^2}{y^2})^{n_2-1}dx$$

I do not really know how to proceed. I am not even sure that I am interpreting the hint properly. Anyway, here goes the rest of the hint:

Observe that by using the change of variable $z=\dfrac{y^2}{x}$, the required density can be expressed in two ways to get by averaging $$f_Y(y)=constant.y^{2n_1-1}\int_{y^2}^1(1-\dfrac{y^2}{x})^{n_2-1}(1-x)^{n_2-1}(1+\dfrac{y}{x})\dfrac{1}{\sqrt{x}}dx$$Now divide the range of integration into $(y^2,y)$ and $(y,1)$ and write $(1-\dfrac{y^2}{x})(1-x)=(1-y)^2-(\dfrac{y}{\sqrt{x}}-\sqrt{x})^2$ and proceed with $u=\dfrac{y}{\sqrt{x}}-\sqrt{x}$.

Well, honestly, I cannot understand how one can use these hints: it seems I am getting nowhere. Help is appreciated. Thanks in advance.

You have probably discovered the answer to this by now, but just so that this is answered...

We can find the distribution of $Y$ by using the 'moment matching method'.

The $r$th order raw moment of $Y$ is given by

$\mathbb{E}(Y^r)=\mathbb{E}(X_1^{r/2})\mathbb{E}(X_2^{r/2})=\dfrac{B(n_1+\frac{r}{2},n_2)}{B(n_1,n_2)}.\dfrac{B(n_1+\frac{r}{2}+\frac{1}{2},n_2)}{B(n_1+\frac{1}{2},n_2)}$

$=\dfrac{\Gamma(n_1+\frac{r}{2})\Gamma(n_1+\frac{r}{2}+\frac{1}{2})}{\Gamma(n_1)\Gamma(n_1+\frac{1}{2})}.\dfrac{\Gamma(n_1+n_2)\Gamma(n_1+n_2+\frac{1}{2})}{\Gamma(n_1+n_2+\frac{r}{2})\Gamma(n_1+n_2+\frac{r}{2}+\frac{1}{2})}$

Now using Legendre's Duplication formula we simplify the above to get

$\mathbb{E}(Y^r)=\dfrac{\Gamma(2n_1+r)\Gamma(2n_2)}{\Gamma(2n_1+2n_2+r)}.\dfrac{\Gamma(2n_1+2n_2)}{\Gamma(2n_1)\Gamma(2n_2)}$

$\qquad\quad=\dfrac{B(2n_1+r,2n_2)}{B(2n_1,2n_2)}$,

which is the $r$th order raw moment of a $\beta(2n_1,2n_2)$ distribution. As $Y$ is a bounded random variable with $\mathbb{P}(0<Y<1)=1$, its distribution is uniquely determined from its moments and hence the result.

The problem can be solved more generally, and then the OP's problem is just a special case. This also nests problems presently unanswered on mathSE such as: Product of two Beta distributed random variables

Let $X \sim \text{Beta}(a,b)$ with pdf $f(x)$ be independent of $Y \sim \text{Beta}(\alpha,\beta)$ with pdf $g(y)$:

Then, a general solution for the pdf of the product $Z = X Y$, say $h(z)$, is:

where I am using the TransformProduct function from the mathStatica package for Mathematica to automate the nitty-gritties (as disclosure: I am one of the authors), expressed as a function of Hypergeometric $_2F_1$ functions.

In the OP's special case where $\alpha = a + \frac12$ and $\beta = b$, the pdf of the product $Z$ simplifies to:

It is then easy to show that the pdf of $\sqrt{Z} \sim \text{Beta}(2a,2b)$.