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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

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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
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.

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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

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