Quadratic replacement for the distribution function

23 Views Asked by At

How to prove that for any random variable it is true $F_{\xi^2}(x^{2}) = F_{\xi}(x) - F_{\xi}(-x) $ ? Where $F$ is the distribution function

I know that $\lim_{x\to \infty} F_{\xi}(x) = 1$ and $\lim_{x\to -\infty} F_{\xi}(x) = 0$

Therefore, I decided to go to the limit in both parts $$1 = \lim_{x\to \infty} F_{\xi^2}(x^{2}) = \lim_{x\to \infty}F_{\xi}(x) - \lim_{x\to \infty}F_{\xi}(-x) = 1 - 0 = 1 $$

It is correct? I don't t understand what to do next. Help me, please.

1

There are 1 best solutions below

2
On BEST ANSWER

Simply using the definition

$$F_{\xi^2}(x^2)=\mathbb{P}[\xi^2 \leq x^2]=\mathbb{P}[\xi \leq |x|]=\mathbb{P}[-x\leq \xi\leq x]=F_{\xi}(x)-F_{\xi}(-x)$$

thus you are all set!