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This is a question that a friend asked me (has the final answer too).

The pdf of a random variable $X$ is

$$ f(x) = 0.5,\quad -1 < x < 1 $$

The random variable Y is defined as

$$ Y = \begin{cases} -2X, & -1 < X < 0 \\ X+1, & 0 < X <1 \end{cases}$$

I tried using the inverse transform method but I'm unsure of how to go about this since $Y$ takes values in $[1, 2)$ in both intervals provided above. I get the fact that there should be some sort of an overlapping in this case but can someone provide me with a rigorous way to solve this problem.

The answer was given to be

$$ f(y) = \begin{cases} 0.25, & 0 < y < 1 \\ 0.75, & 1 < y <2 \end{cases}$$

Here is what I did:

$P(Y \leq y) = P(-2X \leq y) = \frac{1}{2} + \frac{y}{4}$.

Taking the derivative of the above CDF, I get $f_Y(y) = 0.25$ when $0 < y < 2$.

I carried out the same procedure for the other interval and obtained $f_Y(y) = 0.5$ when $1 < y < 2$.

Is it alright to conclude that the pdf is as provided in the solution because there is an overlap between the two intervals? Is there a more rigorous way of showing this?

Ad22
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2 Answers2

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Yes. It's called folding; when two disjoint intervals of the support of $X$ fold into the same interval in the support of $Y$, then the change of variables transformation folds the combined influence. $$\begin{align}f_Y(y)~=~& \Big\lvert\dfrac{\mathrm d x_1(y)}{\mathrm d y}\Big\rvert~f_X(x_1(y))+\Big\lvert\dfrac{\mathrm d x_2(y)}{\mathrm d y}\Big\rvert~f_X(x_2(y)) \\[1ex] ~=~& \tfrac 12 {f_X(-\tfrac 12y)}~\mathbf 1_{-1<-y/2<0} + {f_X(y+1)}~\mathbf 1_{0<y-1<1} \\[2ex] ~=~&\tfrac 1 4~\mathbf 1_{0<y<2}+\tfrac 12~\mathbf 1_{1<y<2} \\[2ex] ~=~&\tfrac 1 4~\mathbf 1_{0<y\leq 1}+\tfrac 34~\mathbf 1_{1< y<2}\end{align}$$

Where $x_1(Y), x_2(Y)$ are the two semi-inverses (the functions that map $Y$ back into the preimage).

Graham Kemp
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You are right intuitively, but I think it is not rigorous

how about this

for $[0,1)$, the inverse is $x=-\frac{y}{2}$ with absolute jacobian is $\frac{1}{2}$, hence

\begin{eqnarray*} f_Y(y)=f_X\left(-\frac{y}{2}\right)\frac{1}{2}=\frac{1}{4} \end{eqnarray*}

and for $[1,2)$, we get two inverses, that are $x=-\frac{y}{2}$ and $x=y-1$, with their respective absolute jacobians are $\frac{1}{2}$ and $1$, hence

\begin{eqnarray*} f_Y(y)=f_X\left(-\frac{y}{2}\right)\frac{1}{2}+f_X(y-1)1=\frac{3}{4} \end{eqnarray*}

why i used $+$? because when $y\in[1,2)$, then $y$ is $-2x$ OR $x+1$

Rizky Reza Fujisaki
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