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Suppose that $X_1\sim \operatorname{Gamma}(p,1)$ and independently, $X_2\sim \operatorname{Gamma}(p+1/2,1)$. Show that $Y=2\sqrt{X_1X_2}\sim\operatorname{Gamma}(2p,1)$.

This problem followed a section on bivariate transformations, so I made attempts to solve this problem using that method. However, I was getting an integral that does not properly integrate, as was my professor.

I also attempted to use CDFs and mgfs, but I was not able to make any progress. How would you solve this problem?

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    $\begingroup$ If bivariate transformations are in the next chapter, I'd suggest looking at other methods for this. Maybe start by exploring distributions of $\sqrt{}$ of gamma and product of gammas. // Simulated this for $p=2$ and it works fine, so I don't doubt it's true. I'll continue comment in Answer format to show simulation. $\endgroup$ – BruceET Oct 23 '19 at 7:42
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    $\begingroup$ Possible duplicate of product of two gamma random variables $\endgroup$ – BruceET Oct 23 '19 at 7:52
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    $\begingroup$ @Xi'an: Hence your awesome answer (+1). $\endgroup$ – BruceET Oct 23 '19 at 10:01
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    $\begingroup$ The relationship between $p$ and $q=p+1/2$ is special: I believe that when $q\ne p\pm1/2,$ $\sqrt{X_1X_2}$ is not a multiple of any Gamma distribution. Following the analysis in @Xi'an's answer, it looks like the result has a density proportional to a power of $z$ times a Bessel $K$ function, reducing to a multiple of $e^{-z}$ only when $|q-p|=1/2.$ $\endgroup$ – whuber Oct 23 '19 at 15:29
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    $\begingroup$ @Xi'an I remember reading about this property years ago and have been hunting for the reference, because I cannot readily formulate an intuitive explanation. There are some very closely related results in the exercises to Chapter 16 of Kendall & Stuart Volume 1 (the chapter on distributions related to the Normal). $\endgroup$ – whuber Oct 23 '19 at 18:02
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If $X$ is a Gamma $\mathcal G(p,1)$ variate, its density is $$f_X(x)= \frac{x^{p-1}}{(p-1)!}e^{-x}\mathbb I_{\mathbb R^*_+}(x)$$ Therefore, the probability density of $Y_1=\sqrt{X_1}$ is $$f_{Y_1}(y)=\frac{y^{2(p-1)}}{(p-1)!}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y)\times\overbrace{\left|\frac{\text{d}x}{\text{d}y}\right|}^{\text{Jacobian}} =2y\times\frac{y^{2(p-1)}}{(p-1)!}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y) =\frac{2y^{2p-1}}{(p-1)!}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y)$$ And the probability density of $Y_2=\sqrt{X_2}$ is $$f_{Y_2}(y)=\frac{2y^{\overbrace{2p+1-1}^{2p}}}{\Gamma(p+1/2)}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y)$$ Hence the joint density of $(Y_1,Y_2)$ is $$g(y_1,y_2)=\frac{4y_1^{2p-1}y_2^{2p}}{(p-1)!\Gamma(p+1/2)}e^{-y_1^2-y_2^2}\mathbb I_{\mathbb R^*_+}(y_1)\mathbb I_{\mathbb R^*_+}(y_1)$$ Considering the change of variables from $(y_1,y_2)$ to $(z=y_1y_2,y_2)$, the joint density of $(Z,Y_2)$ is $$h(z,y_2)=g(z/y_2,y_2)\overbrace{\left|\frac{\text{d}(y_1,y_2)}{\text{d}(z,y_2)}\right|}^{\text{Jacobian}}=g(z/y_2,y_2)\left|\frac{\text{d}y_1}{\text{d}z}\right|=g(z/y_2,y_2)y_2^{-1}$$ and the density of $Z$ is the marginal \begin{align*} f_Z(z) &= \int_0^\infty g(z/y_2,y_2)y_2^{-1}\,\text{d}y_2\\ &=\int_0^\infty \frac{4z^{2p-1}y_2^{2p-(2p-1)}}{(p-1)!\Gamma(p+1/2)}e^{-z^2y_2^{-2}-y_2^2}y_2^{-1}\,\text{d}y_2\\ &= \frac{4z^{2p-1}}{(p-1)!\Gamma(p+1/2)}\int_0^\infty e^{-z^2y_2^{-2}-y_2^2}\,\text{d}y_2\\ &= \frac{4z^{2p-1}}{(p-1)!\Gamma(p+1/2)} \frac{\sqrt{\pi}}{2}e^{-2z} \end{align*} [the last integral is formula 3.325 in Gradshteyn & Ryzhik, 2007]

Hence the density of $S=2Z$ is $$f_S(s)=\frac{\sqrt{\pi}2^{1-2p}s^{2p-1}}{(p-1)!\Gamma(p+1/2)}e^{-s}=\frac{s^{2p-1}}{\Gamma(2p)}e^{-s}$$ [where the constant simplifies by the Legendre duplication formula]

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    $\begingroup$ Wow. This is a beautiful solution, very well explained. Thank you so much for your help. $\endgroup$ – Ron Snow Oct 23 '19 at 23:33
  • $\begingroup$ @Edison: this is very standard, just playing by the book, apply a change of variables and integrate out the extra component ($y_2$ in this case). $\endgroup$ – Xi'an Oct 24 '19 at 7:34
  • $\begingroup$ Thank you! Would you know how to solve this problem with mgfs? $\endgroup$ – Ron Snow Oct 26 '19 at 22:25
  • $\begingroup$ With mgfs, you need first find the mgf of $\sqrt{X_i}$ which is not straightforward. $\endgroup$ – Xi'an Oct 27 '19 at 6:57
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... Comment continued: R code for the simple simulation is as shown below. Unfortunately, the simulation gives no clue how to work the problem. (See @Xi'an's Answer.)

set.seed(1023)
p = 2;  m = 10^6
x1 = rgamma(m,p,1);  x2 = rgamma(m,p+.5,1)
y = 2*sqrt(x1*x2)
hist(y, br=60, prob=T, col="skyblue2")
 curve(dgamma(x,2*p,1), add=T, col="red", n=10001)

enter image description here

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    $\begingroup$ Thank you for providing the visualization. This is extremely helpful! $\endgroup$ – Ron Snow Oct 23 '19 at 23:33
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I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$.

Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $$f(x)=\frac{e^{-x}x^{p-1}}{\Gamma(p)}\mathbf1_{x>0}$$ with $p>0$ as in @Xi'an's answer.

We have

\begin{align} E\left[e^{t\ln Y}\right]&=E\left[Y^t\right] \\&=2^tE\left[X_1^{t/2}\right]E\left[X_2^{t/2}\right] \end{align}

For $t>-2p$ where $p>0$, clearly

$$E\left[X_1^{t/2}\right]=\frac{\Gamma\left(p+\frac t2\right)}{\Gamma(p)}$$

And $$E\left[X_2^{t/2}\right]=\frac{\Gamma\left(p+\frac t2+\frac12\right)}{\Gamma\left(p+\frac12\right)}$$

Using Legendre's duplication formula,

\begin{align} E\left[e^{t\ln Y}\right]&=2^t\cdot \frac{\Gamma(2p+t)\sqrt\pi/2^{2p+t-1}}{\Gamma(2p)\sqrt\pi/2^{2p-1}} \\&=\frac{\Gamma(2p+t)}{\Gamma(2p)} \end{align}

This is the MGF of the logarithm of a $\mathsf{Gamma}(2p,1)$ distribution evaluated at $t$ (or simply the $t$th order raw moment about $0$) where $t\in (-2p,\infty)$. As the MGF exists in an open interval containing $0$, we can conclude that $Y\sim \mathsf{Gamma}(2p,1)$ by uniqueness theorem of MGF. In effect we see that this Gamma distribution is uniquely determined by its moments.

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