Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X$ has a uniform $[-1,2]$ distribution. Find he density of $X^2$.... What I did was the change of variables, (and I know I'm supposed to do it because it's that section of the textbook) and I first found the range of $Y$ which is $[1,4]$ because were assuming $Y=X^2$ and $x$'s domain is from $-1$ to $4$ and the range would just be the end points. Next I found $f_Y(y)=\frac{f_X(x)}{|\frac{dy}{dx}|}$ which is pretty simple, it came out to be$$f_Y(y)=\frac{1}{6\sqrt y}\quad \forall \quad 1\le y \le 4$$ I know that this is correct because in the back of the book it lists this as the brief answer. However, there is also another part to it that says $$f_Y(y)=\frac{1}{3\sqrt y}\quad \forall \quad 0\lt y \lt 1$$ and I'm not sure where that second part came from. The interval [0,1] isn't even in the range of $Y$ so why would I be looking for it?

share|improve this question
1  
What is the value of $Y$ if $X = 0.1$? –  Dilip Sarwate Nov 5 '12 at 13:09
    
Sketch the plane with coordinate axes $x$ and $y$. Mark on it the points $(−1,0)$ and $(2,0)$, and then sketch the parabola$ y=x^2$ between these points. The point $(X,Y)$ always lies on this segment of the parabola. Then ask yourself: for fixed $y in [0,1]$, what values of the random variable $X$ result in the occurrence of the event $\{Y\leq y\}$? Can you find the probability $P\{Y\leq y\}$ which is $F_Y(y)$ for $y \in [0,1]$? Repeat for fixed $y \in (1,4]$ and note that the values of $X$ that result in $\{Y\leq y\}$ have a different form... Do not rely on magical formulas.... –  Dilip Sarwate Nov 5 '12 at 13:10
    
I see, thank you for that. I just remembered my professor mentioned this in lecture. –  TheHopefulActuary Nov 5 '12 at 13:32

1 Answer 1

The range of $Y$ is $[0,4]$, not $[1,4]$! As $x \mapsto x^2$ isn't monotonic on $[-1,2]$ you cannot just square the endpoints of the interval, you have to square all points in $[-1,2]$. So let's compute the distribution of $Y$ (to me, this seems often more direct to approach in contrast with the density), for $y \in [0,4]$ we have \begin{align*} P(Y \le y) &= P(X^2 \le y)\\ &= P(X \in [-\sqrt y, \sqrt y])\\ &= P(X \in [-\min\{\sqrt y, 1\}, \min\{\sqrt y, 1\}]) + P(X \in [1, \max\{\sqrt y, 1\}])\\ &= \frac 23 \cdot \min\{\sqrt y, 1\} + \frac 13\max\{\sqrt y - 1, 0\}\\ &= \begin{cases} \frac 23 \sqrt y, & y \le 1\\[3mm] \frac 13 + \frac 13\sqrt y & y \ge 1 \end{cases} \end{align*} Taking the derivative gives $Y$'s density \[ f_Y(y) = \begin{cases} \frac 1{3\sqrt y}, & y \in [0,1]\\[3mm] \frac 1{6\sqrt y}& y \in (1,4]. \end{cases} \]

share|improve this answer
    
could you explain the 3rd line of how you did that? I can kind of see what you did. Why did you use $[-min{\sqrt y,1}]$, etc.? –  TheHopefulActuary Nov 6 '12 at 16:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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