I have the following problem. Given i.i.d. standard Gaussian random variables $x_i$ with mean $0$ and variance $1$. We compute $X=\sum_{i=1}^n x_i^2$. Find $c$ such that $\Pr(c\geq X)\leq 1/2$.
I know that $X$ is a chi squared random variable with $n$ degrees of freedom. The probability that $X$ lies in an interval $[0,c]$ is therefore given by the distribution function which is in this case $F_n(y)=P(n/2, y/2)$ where $P$ is the incomplete Gamma function. Hence I have to find $c$ such that $P(n/2, c/2) \leq 1/2$. Unfortunately the incomplete Gamma function is (for me) hard to handle and I am stuck on how to explicitly compute this $c$.
Best Answer
Hint
The density function of $\ X\ $ is bounded above by $\ \frac{x^{\frac{n}{2}-1}}{2^\frac{n}{2}\Gamma\big(\frac{n}{2}\big)}\ $, so $$ P\big(c\ge X\big)\le\frac{1}{2^\frac{n}{2}\Gamma\big(\frac{n}{2}\big)}\int_0^cx^{\frac{n}{2}-1}dx\ . $$