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We know that if $Y = bX+c$ where $b>0$, its density function with respect to $f_X$ (density of $X$) is: $f_Y(x) = \frac{1}{b}f_X(\frac{x-c}{b})$. Now how can I get the density of $Y$ when $b<0$?

Following the same approach as for case $b>0$, I got ($F_Y$ is the cumulative distribution function of $Y$):

$$F_Y(x) = P(Y\leq x) = P(bX+c\leq x) = (\text{since $b<0$}) =\\ P(X \geq \frac{x-c}{b}) = 1 - P(X< \frac{x-c}{b})=1 - P(X\leq \frac{x-c}{b})=\\ 1 - \int_{-\infty}^{\frac{x-c}{b}}f_X(\xi)d\xi=1 - \int_{-\infty}^{x}\frac{1}{b}f_X(\frac{t-c}{b})dt$$

If only I could write the last expression as the integral of something, then I could get $f_Y$.

Update: As pointed out in the comments, the last part should be $1-\int_{\infty}^x \frac{1}{b}f_X(\frac{t-c}{b})dt$.

Solution: I think I got it. We have $$P(X\geq \frac{x-c}{b}) = \int_{\frac{x-c}{b}}^\infty f_X(\xi)d\xi = \text{(using $\xi = (t-c)/b$ and reversing the limits)}=\int_{-\infty}^x -\frac{1}{b} f_X(\frac{t-c}{b})dt$$

So $f_Y(x) = -\frac{1}{b} f_X(\frac{x-c}{b})$. Note: the expression is positive since $b<0$.

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    $\begingroup$ The last = sign is wrong since the change of variable $\xi=(t-c)/b$ transforms the integral on $\xi<(x-c)/b$ into an integral on $t>x$, not $t<x$. $\endgroup$ – Did Sep 3 '16 at 20:02
  • $\begingroup$ You're right. I will fix it. $\endgroup$ – Michael Sep 3 '16 at 21:04
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$$P(X\geq \frac{x-c}{b}) = \int_{\frac{x-c}{b}}^\infty f_X(\xi)d\xi = \text{(using $\xi = (t-c)/b$ and reversing the limits)}=\int_{-\infty}^x -\frac{1}{b} f_X(\frac{t-c}{b})dt$$

So $f_Y(x) = -\frac{1}{b} f_X(\frac{x-c}{b})$.

Note: the expression is positive since $b<0$.

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