If a random variable is defined as $Y = 3 – 2X$, how do I find the CDF and PDF of $Y$ if $X$ follows a Uniform distribution of $X \sim (-1,1)$?
For CDF, am I trying to solve for $F(b) = F(-2)$ or $F(y)$? I tried setting $Y = 3 – 2X$ to y and solving for $X$ in the equality $\Pr(Y \le y) $ but if I don't have an actual value for $y$, I'm not sure where to go from here.
Sorry if this is messy, but ultimately, I'd like to find the CDF and PDF of $Y$ if $Y$ is a defined random variable $Y = 5 – 2X$ and $X \sim \text{Uniform }(-1,1)$.
Best Answer
$$F_Y(y)=P(3-2X\leq y)=P(X\geq1.5-0.5y)=1-P(X\leq1.5-0.5y)=1-F_X(1.5-0.5y)$$ Here the third equality rests on $P(X=1.5-0.5y)=0$ which is valid because $X$ has a continuous distribution.
The PDF $f_Y$ can be found by differentiating the CDF, leading to:$$f_Y(y)=0.5f_X(1.5-0.5y)$$
The CDF and PDF of $X$ are at your disposal.