0
$\begingroup$

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.

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

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .