# Prove Chi-squared distribution by induction

$\forall n \in \mathbb{N}, {X_n\sim N(0,1)}$ independent. I wanna prove the pdf of $X=\sum_{k=1}^{n}X_k^2$ by induction :

Define $$m_n=\int_0^\infty t^{n/2-1}exp(-t/2)dt$$ we have $$\forall t > 0, f_n(t)=\dfrac 1 {m_n} t^{n/2-1}exp(-t/2)$$

When $n=1$, it's easy to prove that $f_1(t)=\dfrac{exp(-\frac t 2)}{\sqrt{2\pi t}}$.

Suppose the argument holds for $n$, when for $n+1$, I'm stuck at this step:

$$f_{n+1}(t)=\int_0^tf_n(u)f_1(t-u)du=...=\dfrac{exp(-t/2)}{m_1\cdot m_n}t^{\frac{n-1}2}\int_0^1 \dfrac {v^{n/2-1}} {\sqrt{1-v}}dv$$

How could I simplify further this expression?

Thanks a lot~

Your notation is a bit odd, but here is the induction step. Define for each positive integer $n$, $$S_n = \sum_{i=1}^n X_i^2$$ where $X_i \sim \operatorname{Normal}(0,1)$. Now suppose there exists a positive integer $\nu$ such that $S_\nu$ follows a chi-square distribution with $\nu$ degrees of freedom, with density $$f_{S_\nu} (x) = \frac{x^{\nu/2 - 1} e^{-x/2}}{2^{\nu/2} \Gamma(\nu/2)}, \quad x > 0.$$ Then \begin{align*} f_{S_{\nu + 1}}(x) &= \int_{s=0}^x f_{S_\nu}(s) f_{S_1}(x-s) \, ds \\ &= \int_{s=0}^x \frac{s^{\nu/2-1} e^{-s/2}}{2^{\nu/2} \Gamma(\nu/2)} \cdot \frac{(x-s)^{-1/2} e^{-(x-s)/2}}{2^{1/2} \Gamma(1/2)} \, ds \\ &= \frac{e^{-x/2}}{2^{(\nu+1)/2}} \int_{s=0}^x \frac{s^{\nu/2-1} (x-s)^{-1/2}}{\Gamma(\nu/2)\Gamma(1/2)} \, ds, \quad s = xu, \quad ds = x \, du \\ &= \frac{e^{-x/2}}{2^{(\nu+1)/2}} \cdot \frac{x^{(\nu+1)/2-1}}{\Gamma(\frac{\nu+1}{2})} \int_{u=0}^1 \frac{\Gamma(\frac{\nu+1}{2}) }{\Gamma(\nu/2)\Gamma(1/2)} u^{\nu/2-1} (1-u)^{-1/2}\, du, \end{align*} and this last integral is simply the integral of a $\operatorname{Beta}(\nu/2, 1/2)$ density, and therefore equals $1$. Hence $$f_{S_{\nu+1}}(x) = \frac{x^{(\nu+1)/2 - 1} e^{-x/2}}{2^{(\nu+1)/2} \Gamma(\frac{\nu+1}{2})}$$ which establishes $S_{\nu+1}$ is also chi-square but with $\nu+1$ degrees of freedom, completing the induction step.

This is not the easiest way to show this result.

• you made it look pretty easy. whats an easier way? Commented Nov 1, 2018 at 7:35
• @Timkinsella An easier way is to avoid the convolution calculation by using the properties of the moment-generating function of $X^2$. The MGF is calculated as $$M_{X^2}(t) = \operatorname{E}[e^{tX^2}] = \int_{x=-\infty}^\infty e^{tx^2} \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx = 2 \int_{x=0}^\infty \frac{e^{-(1-2t) x^2/2}}{\sqrt{2\pi}} \, dx = \frac{2}{\sqrt{1-2t}} \int_{u=0}^\infty \frac{e^{-u^2/2}}{\sqrt{2\pi}} \, du = \frac{1}{\sqrt{1-2t}},$$ hence $M_{S_n}(t) = M_{X^2}(t)^n = (1-2t)^{-n/2}$; calculate the MGF of a chi-square distribution with $n$ degrees of freedom, and the result matches. Commented Nov 1, 2018 at 19:04
• ah, ok. to be fair though that proof uses the fairly deep fact that MGF determines distribution. Commented Nov 2, 2018 at 2:46

You don't need to simplify the expression any further. The key is that the integral factor $$\int_0^1 \frac{v^{n/2-1}}{\sqrt{1-v}}dv$$ is just a constant, and thus the it's just part of the normalization of the density function. The fact that you've shown it's correct up to a constant is enough.

So you really just have a recursive definition for the normalization constant $m_{n+1}$ as $$\frac{1}{m_{n+1}} = \frac{1}{m_1m_n}\int_0^1 \frac{v^{n/2-1}}{\sqrt{1-v}}dv.$$

That being said, the integral can be 'done' in a sense. I believe it's a Beta function, so has an expression in terms of Gamma functions.