0
$\begingroup$

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 $x < {n \over 1-P_x}$.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.