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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 $X \sim \text{Uniform }(-1,1)$.

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    $\begingroup$ You're on the right track with trying to compute the CDF for Y. You know that the CDF for Y is given by $P(Y \leq y)$, and you know that $Y = 3-2X$, so you can substitute that into $P(Y \leq y)$ and rearrange the inequality to one involving $X$. Then you can relate this back to the CDF for $X$. To find the PDF for $Y$, remember the relationship between PDFs and CDFs. Hopefully that is enough of a push in the right direction. $\endgroup$ – TM Gallagher Mar 3 at 14:50
  • $\begingroup$ Thank you so much! When I solve for X, I get (Y - 3)/(-2), do I substitute that for Y or y? I'm not sure exactly what relating back to the CDF for X means, would you mind helping me out with that? @TMGallagher $\endgroup$ – Antonia C Mar 3 at 15:21
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    $\begingroup$ So you’re beginning with $$F_Y(y)=P(Y\leq y) = P(3-2X\leq y),$$ and solving for $X$ as you did leads to a formula for $F_Y$ in terms of a probability involving $X$ being greater than or equal to something involving $y$. This can be written in terms of the CDF for $X$, $F_X$. $\endgroup$ – TM Gallagher Mar 3 at 15:37
  • $\begingroup$ Regarding the relationship between PDFs and CDFs, think about how you would compute the CDF using the PDF. Then use calculus to think about how you could “undo” that computation—that is, how to begin with a CDF and use it to compute a PDF. $\endgroup$ – TM Gallagher Mar 3 at 15:45
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$$F_Y(y)=P(3-2X\leq y)=P(X\geq1.5-0.5y)=1-P(X\leq1.5-0.5y)=1-F_X(1.5-0.5y)$$ 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.

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