# Show that $x < {n \over 1-P_x}$ where $P_x$ represents the CDF of gamma distribution.

It is given that $$P_x = \frac{\int_0^x w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}, \text{ for } x>0$$

My approach:

$$P_x = 1 - \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}$$

$$1 - P_x = \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}$$

$$\frac{1 - P_x}{n} = \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{(1 +{n \over 2})}\ 2^{1 + {n \over 2}}}$$

How should I proceed from here? I have to show that $$x < {n \over 1-P_x}$$.

This is a simple application of Markov's inequality which states that for a non-negative random variable $$X$$ and $$x>0$$, $$\dfrac{\mathbb{E}(X)}x\ge \mathbb{P}(X\ge x)$$
Choosing $$X\sim\Gamma\left(\frac{n}2,\frac12\right)$$, where the parameters are shape and rate parameters respectively, $$\mathbb{E}(X)=n, \mathbb{P}(X\ge x)=1-P_x$$ since $$X$$ is a continuous random variable, we get the inequality you desire.