probability - Let $X$ be a continuous random variable with density function $f_X$. What is $Y=aX+b$?

Let $X$ be a continuous random variable with density function $f_X$ and let $a,b>0$.

What is $Y=aX+b$?

I need some help with this one. And I am quite sure it is not $af_X+b$.

Answer

We have that
$$

\Pr\{aX+b\le y\}=\Pr\{X\le(y-b)/a\}=\int_{-\infty}^{(y-b)/a}f_X(x)\mathrm dx.
$$
Using the substitution $x=(t-b)/a$, we obtain that
$$
\Pr\{Y\le y\}=\int_{-\infty}^{(y-b)/a}f_X(x)\mathrm dx=\int_{-\infty}^y\frac1af_X((t-b)/a)\mathrm dt
$$
and the density function $f_Y(y)=a^{-1}f_X((y-a)/b)$.

In general, if $Y=g(X)$ with a monotone function $g$, we have that
$$

f_Y(y) = \left| \frac{\mathrm d}{\mathrm dy} (g^{-1}(y)) \right| \cdot f_X(g^{-1}(y)),
$$
where $g^{-1}$ denotes the inverse function (see here for more details). In this particular case $g(x)=ax+b$ for $x\in\mathbb R$.