Linear Combination independent Random Variables Let $Y$ and $Z$ two independent real valued random variables with values in $\mathbb{R}$ and let $f_Y(x)$ and $f_Z(x)$ their distributions.
Let $a, b \in \mathbb{R}$. 
How can the distibution $f_{aY + bZ}(x)$ of the linear combination $aY + bZ$ be simplified? 
My ideas:
If $a, b = 1$ I know that we have the folding formula $f_{Y + Z}(x) = f_Y * f_Z(x)$.
Futhermore, what about $f_{aY}(x)$? 
Does $f_{aY}(x)= f_Y(x/a)$ or $f_{aY}(x)= f_Y(ax)$ hold?
Can I combine this results in some way? 
 A: *

*Compute the pdf of $aY$.  ($a\ne 0$)
CDF method:
$\begin{align} F_{aY}(x) &= P(aY\leq x) \\& = P(Y\leq \frac x a) \textbf 1_{a>0} + P(Y\geq \frac x a) \textbf 1_{a<0}  \\ &= F(\frac x a) \textbf 1_{a>0} + (1-F(\frac x a))  \textbf 1_{a<0}  \end{align}$
$f_{aY}(x) = \frac 1 a f(\frac x a) \textbf  1_{a>0} -\frac 1 a f(\frac x a)  \textbf 1_{a<0}   $
Finally,
$$f_{aY}(x) = \frac 1{|a|} f_Y(\frac x a) $$ 
Expectation method:
Let $U=aY$.
For all h bounded, 
 $$ \mathbb E(h(U))=\mathbb E(h(aY))=\int_{-\infty}^{\infty} h(aY)f_Y(y)dy $$
Now, perform the change of variable $x=ay$, $du=adx$. Note that the boundaries change when a is negative. 
$$ \mathbb E(h(U))=\int_{-\infty}^{+\infty} h(x) \frac 1 a f_Y(\frac x a) dx \textbf 1_{a>0}  + \int_{+\infty}^{-\infty} h(x) \frac 1 a f_Y(\frac x a) dx \textbf 1_{a<0} = \int_{-\infty}^{+\infty} h(u) \frac 1{|a|} f_Y(\frac u a)du  $$
Yet $\forall \ h$ bounded, $ \mathbb E(h(U))=\int_{-\infty}^{\infty} h(x)f_U(x)dx$, defines an unique $f_U$. Therefore $$f_U(x) = \frac 1{|a|} f_Y(\frac x a) $$


*Now let's compute the pdf of the sum of the independent random variable.
The pdf of $U=aY$ is $f_U(x)=\frac 1{|a|} f_Y(\frac x a)$. Likewise, $V=bZ$ pdf is $\frac 1{|b|} f_V(\frac x b)$. Then since $Y$ and $Z$ are independent, the pdf of the sum is the product of convolution:
$$f_{U+V}(x) = (f_U * f_V)(x)$$
where * means convolution. 
$$(f_U * f_V)(x) = \int_{-\infty}^{\infty}f_U(y)f_V(y-x)dy$$
Finally combining both results we get,
$$f_{aY+bZ}(x) = \int_{-\infty}^{\infty}\frac 1{|a|} f_Y(\frac y a) \frac 1{|b|} f_Z( \frac{y-x}{b})dy $$
A: if $X = aY, f_X(x) = f_Y(\frac {y}{a})$
if $X = Y + Z, f_X(x) = \int_{-\infty}^{\infty} f_Y(y)f_Z(x-y)\ dy$
if $X = aY+bZ, f_{X}(x) = \int_{-\infty}^{\infty} f_Y(\frac {y}{a}) f_Z(\frac {x}{b}-\frac {y}{a})\ dy$
