Find probability density function of $Y=\sigma X+\mu$

Question

X has the pdf $f_x(x)$

Consider $Y=\sigma X+\mu$ where $\sigma$ and $\mu$ are constant. Find the PDF of $Y$?

I find that

$\mu_Y= \mu+\sigma \mu_x$

And $\sigma_Y^2=\sigma^2\sigma_x^2$

But I cannot further do this question.

Answer 1

Assuming that $\sigma >0$ you have that

$$ \begin{align*} f_Y(c)&=\frac{d}{d c}F_Y(c)\\&=\frac{d}{d c}\Pr [Y\leqslant c]\\&=\frac{d}{d c}\Pr [\sigma X+\mu\leqslant c]\\ &=\frac{d}{d c}\Pr [X\leqslant (c-\mu)/\sigma ]\\ &=\frac{d}{d c}F_X\left(\frac{c-\mu}{\sigma }\right)\\ &=\frac1{\sigma }f_X\left(\frac{c-\mu}{\sigma }\right) \end{align*} $$

where in the last step I used the chain rule.

Answer 2

If $\sigma>0$, $$P(a<Y\leq b)=P\left(\frac{a-\mu}{\sigma}<X\leq \frac{b-\mu}{\sigma}\right)=\int_{\frac{a-\mu}{\sigma}}^{\frac{b-\mu}{\sigma}}f_X(x)\mathrm dx\\ =\int_a^b\frac1\sigma f_X\left( \frac{x-\mu}{\sigma}\right)\mathrm dx$$ so that $f_Y(x)=\frac1\sigma f_X\left( \frac{x-\mu}{\sigma}\right)$.

By a similar computation, if $\sigma<0$, $$ P(a<Y\leq b) =-\int_a^b\frac1\sigma f_X\left( \frac{x-\mu}{\sigma}\right)\mathrm dx$$.

(note that if $\sigma=0$, there is no density; the random variable is discrete).

In summary, $f_Y(x)=\frac{1}{|\sigma|}f_X\left(\frac{x-\mu}{\sigma}\right)$.

Answer 3

A useful theorem that you can use to find the pdf of other transformations random variables easily is:

If $X$ is a continuous random variable and $g(X)$ is a differentiable function and for all $x$ in the range of $X$ you have either $g'(x)>0$ or $g'(x)<0$, the induced random variable $Y=g(X)$ has the pdf

$f_Y(y) = f_X(g^{-1}(y))|g^{-1}(y)|$.

In your question, $g(X)=Y=\sigma X + \mu$. Therefore, $X=g^{-1}(Y)=\frac{Y-\mu}{\sigma}$, so $f_Y(y)=f_X\bigg(\frac{y-\mu}{\sigma}\bigg)|\frac{1}{\sigma}|$