# Finding the PDF of $\ T=\frac{\pi}{n}\sum_{i=1}^{n} X^2_i$

Consider a random samples $X_1,X_2,..,X_n$ from a variables with density function $$f_X(x)=2\lambda\pi xe^{-\lambda\pi x^2} \ \ \ \ \ \ \ x>0$$

I have shows that for $i=1,..,n$, $X^2_i\sim \text{Gamma}(1,\frac{1}{\pi\lambda})$

Now show that the density function of $T=\frac{\pi}{n}\sum_{i=1}^{n} X^2_i$ is $$f_T(t)=\frac{n^n\lambda^n t^{n-1}e^{-n\lambda t}}{\Gamma(n)} \ \ \ \ \ \ \ \ t>0$$

I tried to solve this in a similar way to showing $X^2_i\sim \text{Gamma}(1,\frac{1}{\pi\lambda})$. I let $$Z=\sum_{i=1}^{n} X^2_i \ \ \ \ \ \text{such that} \ \ Z\sim \text{Gamma}(1,\frac{1}{\pi\lambda})\Rightarrow f_Z(z)=\lambda\pi e^{-\lambda\pi z}$$ Therefore $$T=\frac{\pi}{n} Z$$ which is monotonic over $z>0$. Hence \begin{align*} f_T(t)&=f_Z(t)\Big|\frac{dx}{dt}\Big| \\ &=\lambda\pi e^{-\lambda\pi t}\Big|\frac{n}{\pi}\Big| \\ &=\lambda n e^{-\lambda\pi t} \ \ \ \ t>0 \\ \end{align*} But this is clearly not the density required. Where have I made a mistake?

• $Z$ is the sum of the $X_i$'s squared, it has a gamma distribution but a scaled version of that of the $X_i^2$'s. That is, not the same distribution as the $X_i^2$'s. Regarding the proof, are you familiar with moment generating functions? – StubbornAtom Jul 1 '18 at 5:59
• A well known result – Tony Hellmuth Jul 1 '18 at 6:09
• @stubbornAtom: Yep. Can this be calculated somehow using the mgf? – user557493 Jul 1 '18 at 6:19
• Yes. One can show that the MGF of $Z$ is also the MGF of a Gamma variate with some parameter $(a,b)$, and by using the uniqueness property of MGF, one can conclude that $Z$ is also a Gamma variate with parameters $(a,b)$. – StubbornAtom Jul 1 '18 at 6:23
• @StubbornAtom well almost, one can conclude $Z$ is also a Gamma variate with parameters $(n \cdot a,b)$ – Tony Hellmuth Jul 1 '18 at 6:38

## 1 Answer

Some Hints:

Let $X_i$ be a Gamma distribution r.v. with parameters $(a_i,b)$: $$X_i\sim \text{Gamma}(a_i,b)$$

One can show (by moment generating functions or direct derivation of pdf) that: $$\sum_i^nX_i \sim Gamma(\sum_i^n a_i,b)$$

One may also show that by the scaling property of a Gamma distribution: $$c \ X_i \sim Gamma(a,c \cdot b)$$

Can you take it from here?

• I believe I can, but I have a question. Does it matter that in this case, $X^2_i\sim\text{Gamma}(a,b)$? How does the property $\sum_i^n X_i\sim\text{Gamma}\Big(\sum_i^n a_i,b\Big)$ change if $X_i$ is $X^2_i$? – user557493 Jul 2 '18 at 0:06
• It is just a generalisation. Any random variable, be it $X_i$ or $\sqrt{ \sin(e^ \frac { X_i}{ln (a)})}$, if they are both distributed gamma with known parameters, then the statements hold. Easier yet, assume $Y_i = X_i^2$. – Tony Hellmuth Jul 2 '18 at 1:43
• I have solved it :) thanks for your help mate! – user557493 Jul 2 '18 at 1:51