Let $X$ be a random variable with an exponential distribution where: $f(x) = e^{-x}$ for all $x \geq 0$ (i.e. $X \sim \textrm{Exp}(\lambda = 1)$). Calculate >the distribution of $Y = X^{\frac{1}{2}}$.

I tried to do the following thing: $P(X=k)=\int_0^k e^{-x}dx=-e^{-x}|_0^k=1-e^{-k}$.

So, $P(X^{\frac{1}{2}}=k)=P(X=k^2)=\int_0^{k^2} e^{-x}dx=-e^{-x}|_0^{k^2}=1-e^{-k^2}$.

I think that if I differentiate this function by $k$, I should get the distribution (am I right?). So, I get:

$(1-e^{-k^2})'_k = 2ke^{-k^2}$. I get that for $Y=X^{\frac{1}{2}}$, $f(y)=2ye^{-y^2}$. But I don't think this is the expected answer since it is not any known distribution.

Where was I wrong?


  • 1
    $\begingroup$ The idea is good, the way you work it out is not. $\mathbb{P}(X=k)=0$ if $X$ is a continuous probability distribution, e.g. an exponential random variable. You need to word with CDF, see my answer below. $\endgroup$ – alezok Nov 4 '15 at 13:16

Assume that $X \sim \textrm{Exp}(\lambda)$ and let $Y = \sqrt{X}$. For any $y \geq 0$ we have

$\mathbb{P}(Y \leq y) = \mathbb{P}(\sqrt{X} \leq y) = \mathbb{P}(X \leq y^2) = 1- e^{-\lambda y^2}$.

Note that in the second equality we have used in a crucial way the fact that an exponential random variable is supported on $[0,+\infty)$. The pdf of $Y$ can then be calculated by differentiating the CDF obtained above:

$ f_Y(y) = 2 \lambda y e^{-\lambda y^2} \mathrm{1}_{[0,\infty)}(y)$.

  • $\begingroup$ I see. Thank you!! $\endgroup$ – user135172 Nov 4 '15 at 15:46
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    $\begingroup$ @Ale Shouldn't it be $f_Y(y)=2\lambda ye^{-\lambda y^2}$? I.e. $y$ instead of $y^2$? $\endgroup$ – Mick A Nov 4 '15 at 15:57
  • 1
    $\begingroup$ @Mick Correct, I edited my answer. $\endgroup$ – alezok Nov 4 '15 at 15:59

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