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.

Given $f_Y(y)=4y^3$, on $[0, 1]$, (where $Y$ is a continuous random variable), set $W=2Y$ and find $f_W(w)$:

I determined that $Y$ is the identity mapping on $[0, 1]$, thus $W:[0, 1]\rightarrow [0,2]$ is given by $2x$. I have no idea how the W affects the function. I guessed $f_W(w)=32w^3$ on $[0, 1/2]$ but I'm not exactly sure why. Help

share|improve this question
2  
People may be reluctant to answer when they see that the OP has an 0% accept rate on their questions. I noticed that on at least a few of your questions, there were some answers that you liked / found acceptable (at least, that's what it looked like from me stalking the comments!) Perhaps it might be a good idea to accept them? –  JakeR Mar 10 '12 at 22:39
1  
oops. Ok, I didn't know I had an "accept" option, let me look over some of my old questions. –  The Substitute Mar 10 '12 at 22:44
    
Quite all right. –  JakeR Mar 10 '12 at 22:50

1 Answer 1

up vote 1 down vote accepted

The probability distribution function of $Y$ is $P_Y(y) = y^4$, which you get by integrating $f_Y(y)$ from $0$ to $y$.

The probability distribution function of $W$ is $P_W(w) = P(W \leq w)$. Since $W = 2Y$, we have

\begin{equation} P_W(w) = P_Y(2Y \leq w) = P_Y(Y\leq w/2) = \frac{w^4}{16} ~~~~ \mbox{for } 0\leq w \leq 2 \end{equation}

The pdf of $W$ is obtained by differentiating $P_W(w)$ wrt $w$.

Therefore $f_W(w) = \frac{w^3}{4}$. As a check, when you integrate $p_W(w)$ from 0 to 2, you get 1.

share|improve this answer
    
Wow, I was really off track. Thanks for the help. I guess it will just take me some "getting used to" when it comes to integrating to get a probability function. –  The Substitute Mar 10 '12 at 22:59

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.