0
$\begingroup$

The continuous random variable $X$ has a $\text{Uniform}[-1, 3]$ distribution and let $Y = X^2$ .

Find the probability density function of $ Y$.


My attempt.

I know that it has to be between $0 < y < 1$. So $F(Y = y) = P(Y \leq y) = P(X^2 \leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y}) = \frac{2\sqrt{y}}{3+1} = \frac{\sqrt{y}}{2}$

I'm not sure how to go forward from this

$\endgroup$
  • $\begingroup$ which is the typo? the $X\leq 3$ or $y\leq 1$? $\endgroup$ – LinAlg Jul 5 '18 at 23:05
  • $\begingroup$ In fact $0 \le Y \le 9$ though when $1 \lt y \le 9$, you have $P(Y \le y)=P(X \le \sqrt y)$ $\endgroup$ – Henry Jul 5 '18 at 23:06
  • $\begingroup$ Similar question:math.stackexchange.com/questions/1531572/…. $\endgroup$ – StubbornAtom Jul 19 '18 at 20:08
0
$\begingroup$

Let $f_X$ and $f_Y$ denote the densities of $X$ and $Y$ respectively.

$$f_X(x)=\frac{1}{4}\mathbf1_{-1<x<3}$$

We have $y=g(x)$ where $g(x)=x^2$. So the function $g$ must be such that

$$g:(-1,3)\mapsto(0,9)$$

We define $g_i(x)=x^2$ for $i=1,2$ such that $g_1:(-1,0)\mapsto(0,1)$ and $g_2:(0,3)\mapsto (0,9)$.

(I have excluded the end-points in the supports of the random variables as it does not make a difference for continuous distributions)

So, $y=g_1(x)\implies x=g_1^{-1}(y)=-\sqrt y$ and $y=g_2(x)\implies x=g_2^{-1}(y)=\sqrt y$

To directly apply the transformation formula, we have

\begin{align}f_Y(y)&=f_X(-\sqrt y)\left|\frac{d}{dy}(-\sqrt y)\right|\mathbf1_{0<y<1}+f_X(\sqrt y)\left|\frac{d}{dy}(\sqrt y)\right|\mathbf1_{0<y<9} \\&=\frac{1}{4}\cdot\frac{1}{2\sqrt y}\mathbf1_{0<y<1}+\frac{1}{4}\cdot\frac{1}{2\sqrt y}\mathbf1_{0<y<9} \\&=\frac{1}{4\sqrt y}\mathbf1_{0<y<1}+\frac{1}{8\sqrt y}\mathbf1_{1<y<9} \end{align}

$\endgroup$
  • $\begingroup$ @herbsteinberg It is pretty clear and straightforward to me at least, as it is. $\endgroup$ – StubbornAtom Jul 6 '18 at 17:58
  • $\begingroup$ A clearer approach to the solution can be made using $U=|X|$ as an intermediate, since $U$ and $Y$ are one-to-one. The density function for $X.\ f_X(x)=\frac{1}{4}$. Therefore the density function for $U,\ f_U(u)=\frac{1}{2}$ for $0\le u\le 1$ and $=\frac{1}{4}$ for $1\lt u\le 3$. Thus the distribution function $F_U(u)=\frac{u}{2}$ for $0\le u\le 1$ and $=\frac{1}{2}+\frac{1}{4}(u-1)=\frac{1+u}{4}$ for $1\lt u\lt 3$. Since $U=\sqrt{Y}$ the final result is $F_Y(y)=\frac{\sqrt{y}}{2}$ for $0\le y\le 1$ and $=\frac{1+\sqrt{y}}{4}$. $\endgroup$ – herb steinberg Jul 6 '18 at 18:00
0
$\begingroup$

You need to consider it in 2 parts $ |X|\lt 1,\ X \ge 1$. For the first part alone $P( |X| \lt \sqrt{y})=\sqrt{y}$. For the second part alone $P(1\le X \le \sqrt{y})=\frac{1}{2}(\sqrt{y}-1)$. Since each part has probability $=\frac{1}{2}$, they combine to get $P(Y\le y)=\frac{1}{2}\sqrt{y}$ for $0\le y\le 1$ and $P(Y\le y)=\frac{1}{4}+\frac{1}{4}\sqrt{y}$ for $y\gt 1$.

$\endgroup$
  • $\begingroup$ @StubbornAtom My approach avoids taking unnecessary derivative for $\sqrt{y}$, by getting the density function first and putting $\sqrt{y}$ at the very end. $\endgroup$ – herb steinberg Jul 6 '18 at 18:06
  • $\begingroup$ What you call 'unnecessary' is baffling. You found the distribution function, I found the density directly. BTW, you cannot ping me by tagging me in a post I haven't commented on. $\endgroup$ – StubbornAtom Jul 6 '18 at 18:18
  • $\begingroup$ @StubbornAtom I was just looking over your post and you have expressions involving the derivative of $\sqrt{y}$. Also it was quite lengthy. I just feel my approach is more transparent. $\endgroup$ – herb steinberg Jul 6 '18 at 20:44
  • $\begingroup$ Well, the density has the derivative of $\sqrt y$. Regarding the length, I have nothing to say. You feel your approach is 'more transparent', that's absolutely finel! But please raise questions on others' posts on valid grounds. $\endgroup$ – StubbornAtom Jul 6 '18 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.