# Probability density of $Y = bX+c$ is $f_Y(x) = \frac{1}{b}f_X(\frac{x-c}{b})$ for $b>0$

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 $$b<0$$.

• 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 $t<x$. – Did Sep 3 '16 at 20:02
• You're right. I will fix it. – Michael Sep 3 '16 at 21:04

$$P(X\geq \frac{x-c}{b}) = \int_{\frac{x-c}{b}}^\infty f_X(\xi)d\xi = \text{(using \xi = (t-c)/b and reversing the limits)}=\int_{-\infty}^x -\frac{1}{b} f_X(\frac{t-c}{b})dt$$
So $$f_Y(x) = -\frac{1}{b} f_X(\frac{x-c}{b})$$.
Note: the expression is positive since $$b<0$$.