# CDF of continuous random variable $Y=X^2$ [closed]

It is stated that X is exponentially distributed with parameter $$\lambda$$. How do I go about finding the CDF? My original intuition is to find the PDF, but I'm not sure how to do that if $$P(Y\leq y) = P(X^2 \leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y})$$. I am totally lost, plesae give me some guidance.

• X being exponential cannot be negative. So $P(X \le \sqrt{y}) = 1 - e^{-\lambda\sqrt{y}}$
– sku
Mar 20, 2022 at 2:33

You are good to go so far.$$\mathsf P(Y\leqslant y) = \mathsf P(X\leqslant\surd y)-\mathsf P(X\leqslant-\surd y)$$
Now you have been promised that $$X$$ is exponentially distributed with parameter $$\lambda$$, so therefore you should know the CDF for $$X$$ is: $$\mathsf P(X\leqslant x) ~=~ (1-\mathrm e^{-\lambda x})\,\mathbf 1_{0\leqslant x}$$ and its pdf is $$f_X(x)= \lambda\mathrm e^{-\lambda x}\,\mathbf 1_{0\leqslant x}$$ .
So to find the pdf for $$Y$$, either substitute the CDF for $$X$$ at those values and take the derivative with respect for $$y$$, or take the derivative first then substitute the pdf for $$X$$ at those values.