Find the distribution of $Y=\sqrt X$ when $X$~$N(0,1)$

Question

PDF of $X$ is $f_x(t)=\frac{1}{\sqrt{2pi}}e^{\frac{-x^2}{2}}$.

I tried to found the CDF of $Y$: $F_Y(t)=P(Y \leq t)=P(\sqrt X \leq t)=P(X \leq t^2)=F_X(t^2)$

How do i get distrbution of Y from this? If I were to find the $P(X\leq t^2)$ I would get error function and i don't have the CDF of X to find $F_X(t^2)$

I tried to find the PDF of $Y$: $f_Y(t)=F'_Y(t)=F'_X(t^2)=2tf_X(t^2)$
Does PDF define the distribution of random variable? If not how to i find it?

@Surb what if we only consider $X>0$? How do I proceed then? — HiThere, Mar 08 '21 at 17:35
$X<0$ has non-zero chance, so you can only study the distribution of $Y$ conditional on the event $X>0$. If $X$ is, unconditionally, normally distributed, then you first need to compute the conditional distribution of $X$ conditional on the event $X>0$, which is no longer normal. Do you know how to compute conditional distributions that depend on events? — Nap D. Lover, Mar 08 '21 at 17:46
hint: In general: $P(X\leq x|X \in A)=P(X\leq x, X\in A)/P(X\in A)$ for all events with non-zero chance $P(X\in A)>0$. Here then $A=(0, \infty)$. — Nap D. Lover, Mar 08 '21 at 18:19

Nap D. Lover · Accepted Answer · 2021-03-08T19:00:54.253

1

Here is a solution to finding the distribution of $Y=\sqrt{X}$ conditional on $X>0$ where unconditionally $X\sim \mathcal{N}(0,1)$. In this case, $Y=\sqrt{X}$, given $X>0$ is properly defined, because the square-root is undefined for negative reals. The conditional distribution of $X$ is given by the basic definition of conditional probability conditional on events: $$P(X\leq x| X>0)=P(0<X\leq x)/P(X>0)=\frac{1}{1-\Phi(0)}[\Phi(x)-\Phi(0)],$$ so that then $$P(Y\leq y|X>0)=P(\sqrt{X} \leq y |X >0)$$ $$=P(X\leq y^2 |X>0)=\frac{1}{1-\Phi(0)}[\Phi(y^2)-\Phi(0)],$$ now differentiate with respect to $y$ to find that the conditional probability-density-function of $Y$, conditional on $X>0$, is given by $$f_Y(y|x>0)=\frac{2y \cdot \phi(y^2)}{1-\Phi(0)}$$

Here is a visual/numerical verification in R (black is the empirical density, blue is the exact density derived above):

Please comment for further clarifications/questions/corrections.

edited Mar 08 '21 at 19:00

answered Mar 08 '21 at 18:30

Nap D. Lover

1,207

Does PDF defines the distribution of random variable? I only heard that CDF defines the distribution of random variable. Everything except this one is clear for me right now. = – HiThere Mar 08 '21 at 18:51
The "distribution of a RV" can be used generically (as I am doing here) to refer to either or both of the PDF or the CDF. It is technically an abuse of terminology and some authors are more strict about avoiding such ambiguous terminology as it can be confusing to beginners. But the ambiguity is hardly a problem, since if you have the PDF, you can compute the CDF by integrating it, and if you have the CDF you can potentially differentiate it to get the PDF, at least for continuous RVs. So by "distribution of the RV" I simply mean, either the PDF or the CDF or both. – Nap D. Lover Mar 08 '21 at 18:59
@HiThere "Everything except this one is clear for me right now" Then it would be nice, if you accepted the answer. – callculus42 Mar 08 '21 at 19:09

score 0 · Answer 2 · answered Mar 08 '21 at 17:51

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If you consider the absolute value of $X$ instead (half-normal distribution), you should be able to derive the pdf of $Y:=\sqrt{|X|}$ via the density transformation formula.

answered Mar 08 '21 at 17:51

huhu27

3

If I do this $$F_Y(t)=P(Y\leq t)=P(\sqrt{|X|}\leq t$=P(|X|\leq t^2)=P(x\in [0,t^2])=F_X(t^2)-F_X(0)$$ then the PDF is going to be exactly the same as the one I got in my try $f_Y(t)=2tf_X(t^2)$ – HiThere Mar 08 '21 at 18:10
Yes, that's true. I think it is a generalized gamma distribution. – huhu27 Mar 08 '21 at 18:14
@HiThere $|X|\leq t^2 \iff -t^2 \leq X \leq t^2$ – Nap D. Lover Mar 08 '21 at 18:15
So I would get $f_Y(t)=2tf_X(t^2)-2tf_X(-t^2)=0$? – HiThere Mar 08 '21 at 18:23

Find the distribution of $Y=\sqrt X$ when $X$~$N(0,1)$

2 Answers2