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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?

Thanks

alezok
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user135172
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    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. – alezok Nov 04 '15 at 13:16

1 Answers1

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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)$.

BCLC
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alezok
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  • I see. Thank you!! – user135172 Nov 04 '15 at 15:46
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    @Ale Shouldn't it be $f_Y(y)=2\lambda ye^{-\lambda y^2}$? I.e. $y$ instead of $y^2$? – Mick A Nov 04 '15 at 15:57
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    @Mick Correct, I edited my answer. – alezok Nov 04 '15 at 15:59
  • @alezok, could you please explain how you could derive the equality $\mathbb{P}(\sqrt{X} \leq y) = \mathbb{P}(X \leq y^2)$? So, it's true in general $\mathbb{P}(Z \leq z) = \mathbb{P}(Z^n \leq z^n)$ where $n$ is a real value, isn't it? Also, you said you would edit your answer to show $ f_Y(y) = 2 \lambda y e^{-\lambda y^2} $ as suggested by Mick, but it still shows $ f_Y(y) = 2 \lambda y e^{-\lambda y^2} \mathrm{1}{[0,\infty)}(y)$? What does $\mathrm{1}{[0,\infty)}(y)$ mean? Thanks in advance. – Nemo Feb 01 '20 at 00:11
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    @Nemo, $X$ is a non-negative random variable (by definition) and by writing the event $\sqrt{X} \leq y$ we tacitly assume that $y \geq 0$ and thus $\sqrt{X} \leq y$ is equivalent $X \leq y^2$ by "taking the square on both sides". The general statement is not true in general, it depends on the assumptions on the random variable $Z$ and on the scalar $z$. Mike was only correcting a typo in the exponent of my expression, the correct expression should still have the indicator function $\mathrm{1}_{[0,\infty)}(y)$ which is simply equal to $1$ if $y\in [0,\infty)$ (i.e. $y \geq 0$) and $0$ otherwise – alezok Feb 03 '20 at 18:18
  • Thanks for your clarification, @alezok. From my understanding, user135172 did manage to derive the correct distribution ($f(y)=2ye^{-y^2}$) because $\lambda$ = 1 in this case. If so, why didn't he think his derivation was expected because it was not any known distribution, but then he accepted your answer (which is essentially the same)? Also, why did we need to have the indicator function? Without it, the expression $f_Y(y)=2\lambda ye^{-\lambda y^2}$ is still a distribution function wouldn't it? Do we need to prove integration of $f_Y(y)=2\lambda ye^{-\lambda y^2}$ converges to 1? – Nemo Feb 04 '20 at 04:00