Krdytkyu

Let $X$ be a real random variable and $f$ the density belonging to $X$. Let $aneq0$, and $b in mathbb R$, while $Y:=aX+b$. Show:

i) The density of $Y$ wrt the lebesgue measure exists and is $g(x):=frac{1}{|a|}f(frac{x-b}{a}), xinmathbb R$

ii) Find the distribution of $Y$ if $X$ ~ $mathcal{N}(mu,sigma^{2})$

My steps:

i) I think I've got a good grasp of the it, but still got a few questions.
I will just do the case $a>0$ here: Let $c in mathbb R$

$P(Yleq c)=P(aX+bleq c)=P(Xleqfrac{c-b}{a})$ and we know the distribution of RV $X$, thus: $P(Xleqfrac{c-b}{a})=int_{-infty}^{frac{c-b}{a}}f(x)dlambda(x)=int_{-infty}^{frac{c-b}{a}}f(x)dx$

now I set $y=ax + b Rightarrow dy=adx$

Therefore $int_{-infty}^{frac{c-b}{a}}f(x)dx=int_{-infty}^{c}frac{1}{a}f(frac{x-b}{a})dx$

Using that case $a < 0$ too, we get $frac{1}{|a|}f(frac{x-b}{a}), xinmathbb R$

Now I have the proposed density function, I need to show that it is indeed a pdf. On measurability, it is clear since $int_{-infty}^{c}frac{1}{|a|}f(frac{x-b}{a})dx$ exists that $g(x):=frac{1}{|a|}f(frac{x-b}{a})$ is measurable on $(infty, c], forall c in mathbb R$. And since ${ (infty, c] |c in mathbb R}$ is a generator of the $mathcal{B}(mathbb R)$, therefore $g$ is borel measurable. (Is this correct reasoning?)

First Question: I would think that an alternative way of showing that $g$ is Borel-Measurable is simply stating that $g$ is the product of borel-measurable functions $frac{1}{|a|}$ as well as $f(frac{x-b}{a})$, and is therefore borel-measurable. I have a feeling that this my however not be so simple because $f(x)$ being measurable does not imply that $f(frac{x-b}{a})$ is indeed borel-measurable. Any notes, clarification on this would be of great help.

ii) I get $Y$ ~ $mathcal{N}(b+amu,(asigma)^2)$

In the language of measure theoretic probability $int, dx$ is same as $int , dlambda(x)$ where $lambda$ is Lebesgue measure. So there is no Riemann integral involved here.

$int f(frac {x-b} a), dx$ is not $1$. You have to make substitution $y= frac {x-b} a$ to evaluate this integral. If you do this you will get $int g(x), dx=1$.

The variance of $Y$ is $a^{2}sigma^{2}$ not $asigma^{2}$.

I am going to address just the measurability part: it is a fallacy to claim that since some integrals involving $g$ exist, then $g$ is measurable.
There would be no way to define the integral if $g$ was not measurable.

The sound reasoning, in this case, is to write before starting any kind of computation: let $g=frac{1}{|a|}f(frac{x-b}{a})$.
Then $g=h_1 circ f circ h_2$, where $h_1=|a|^{-1}Id$ and $f$ and $h_2= frac{Id-b}{a}$ are $(mathbb{R},mathcal{B}) rightarrow (mathbb{R},mathcal{B})$ measurable, therefore $g$ is measurable.

Then you can make the computations and change variables in your integrals and everything else.