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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 1.2012-12-04
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    @echoone And why not?2012-12-04
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    @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.2012-12-04
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    Oops. Yeah, you guys are right. What I am wondering now is how you are dealing with the $\sigma < 0$ case.2012-12-04
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    The formula is on wikipedia here http://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables. You just have to replace $g$ by the special case of a linear function!2012-12-04
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    Another version of the formula is at http://en.wikipedia.org/wiki/Location-scale_family#Converting_a_single_distribution_to_a_location-scale_family2012-12-04

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