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I am trying to understand a very basic problem involving pdfs and cdfs for functions of random variables.

For example, my problem statement is as follows:

Find $F_{Y}\left ( y \right )$ and $f_{Y}\left ( y \right )$ if $Y=-4X+3$ and $f_{X}\left ( x \right ) = 2e^{-2x}U(x)$.

I know that I can start this problem by replacing the random variable $Y$ in terms of the RV $X$.

$F_{Y}\left ( y \right ) = P\left (\left \{ Y \leq y \right \} \right ) = P\left (\left \{ -4X+3 \leq y \right \} \right )$

and then I can write: $P\left (\left \{ -4X+3 \leq y \right \} \right ) = P\left (\left \{ X \geq \frac{3-y}{4} \right \} \right ) = 1-F_{X}\left ( \frac{3-y}{4} \right )$

My solution that I am learning from does the following. I am not sure how $f_{Y}\left ( y \right )$ is found in terms of $f_{X}$. And then I am not sure how the final answer is found. Are they plugging in the values from the cdf of $X$?

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    I suspect that $U$ denotes the indicator function of $(0,\infty)$. – 2017-02-26
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    Yes, that is correct. – 2017-02-26
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    $f_Y$ can be found by differentiating $F_Y(y)=1-F_X\left(\frac{3-y}4\right)$ on both sides. There are mistakes in the last two expressions. For $y\to\infty$ we get $F_Y(y)\to\infty$. The argument of $U$ must be $\frac{3-y}{4}$ (or simply $3-y$) so that we gain the indicator function of $(-\infty,3)$ – 2017-02-26

1 Answers 1

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In your solution the pdf of $Y$, $f_Y(y)$, is found by derivation of its CDF, $F_Y(y)$, which is obtained by substitution $x\gets \frac{3-y}{4}$ as indicated, and the CDF of $X$ is found by the integration of its pdf.

$\begin{align} f_X(x) & = 2e^{-2x}\mathsf U(x) \\[2ex] F_X(x) & = \mathsf U(x)\int_0^x 2e^{-2s}\mathrm d s \\[1ex] &= (1-e^{-2x})\mathsf U(x) \\[2ex]F_Y(y) & = 1-F_X(\dfrac{3-y}{4}) \\[1ex] &= 1-(1-e^{(3-y)/2})\mathsf U({3-y}{}) \\[1ex] & = e^{(y-3)/2}\mathsf U(3-y)+\mathsf H(y-3) \\[2ex]f_Y(y) &= \dfrac{\mathrm d ~}{\mathrm d y}F_Y(y) \\[1ex] &= \dfrac{\mathrm d ~}{\mathrm d y}\Big(e^{(y-3)/2}\mathsf U(3-y)\Big) \\[1ex] &= \tfrac 12 e^{(y-3)/2}\mathsf U(3-y)\end{align}$