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.

If $X$ is uniform on the interval $[0, 2]$ find the PDF of $Y = X^2-2X$.

I solved for $X$ and got $x=1 + \sqrt{1+y}$ or $x=1 - \sqrt{1+y}$. Not sure what to do from here.

share|improve this question
    
This is likely a homework, please tag it appropriately if so. You can get an answer using Mathematica as well, try PDF[TransformedDistribution[x^2 - 2 x, Distributed[x, UniformDistribution[{0, 2}]]], x] –  Sasha Jan 17 '13 at 16:11
    
Thanks! Yes it is, and I do not have mathematica. –  Olivia Irving Jan 17 '13 at 16:23

1 Answer 1

Note that $Y=(X-1)^2-1$. Now let $W=X-1$. It is nearly obvious that $W$ is uniform on $[-1,1]$.

We find the distribution of $W^2$. Since $Y=W^2-1$, it is then easy to find the distribution of $Y$.

Let $Z=W^2$. Then (for $0\le z\le 1$), $$\Pr(Z\le z)=\Pr(W^2\le z)=\Pr(|W|\le \sqrt{z})=\frac{2\sqrt{z}}{2}=\sqrt{z}.$$

Remark: The problem can also be solved using a variant of your calculation. We have $Y\le y$ if and only if $X^2-2X\le y$. This inequality holds if and only if $$1-\sqrt{1+y}\le X\le 1+\sqrt{1+y}.$$ If $y\ge 0$, the above probability is $1$. So the cdf of $Y$ is $1$ for $y \ge 0$.

If $y\lt -1$, the above condition is an impossible one, so the probability is $0$. Thus the cdf of $Y$ is $0$ when $y\lt -1$. So the only interesting $y$ are between $-1$ and $0$.

For $-1\le y\le 0$, the length of the interval $[1-\sqrt{1+y},1+\sqrt{1+y}]$ is $2\sqrt{1+y}$, and the interval lies within $[0,2]$. So the probability we land in that interval is $\frac{2\sqrt{1+y}}{2}$, that is, $\sqrt{1+y}$.

Now that we have the cdf of $Y$, we can differentiate to get the density.

share|improve this answer
    
Where did the Y=W^2-1 go? I understand Z=W^2 but I am not sure where the -1 went. –  Olivia Irving Jan 17 '13 at 16:28
    
I showed how to find the cumulative distribution function for $W^2$. It was left to you to use this to find the cdf of $W^2-1$. –  André Nicolas Jan 17 '13 at 16:32

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.