# Probability density function of $\sigma X + \mu$.

I need a head check on this one. Suppose $\sigma,\mu \in \mathbb{R}$ and $\sigma \neq 0$. Let $X$ be a random variable with density $f_X(x)$. I think that the random variable $Z:= \sigma X + \mu$ has density $$f_Z(x) = \frac{1}{|\sigma|}f_X\left(\frac{x-\mu}{\sigma} \right).$$

Splitting the cases $\sigma > 0$ and $\sigma<0$ my proof comes down to a simple change of variables. However, I can't find any mention of a formula like this on wikipedia or google. Is it so simple that no one thought to mention it, or am I misunderstanding?

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For one thing, you expect that $\int_{\mathbb R} f_Z(x) \, dx = 1$, but if you use the formula for $f_Z(x)$ that you have, you don't get 1. – echoone Dec 4 '12 at 2:57
@echoone And why not? – Sasha Dec 4 '12 at 3:08
@echoone $\int_{\mathbb{R}} f_Z(x)dx = \int_{\mathbb{R}} f_X(x)dx = 1$. Don't forget the Jacobian when you change variables. – nullUser Dec 4 '12 at 3:09
Oops. Yeah, you guys are right. What I am wondering now is how you are dealing with the $\sigma < 0$ case. – echoone Dec 4 '12 at 3:37
The formula is on wikipedia here en.wikipedia.org/wiki/…. You just have to replace $g$ by the special case of a linear function! – Learner Dec 4 '12 at 3:44