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Let $X$ be a real random variable and $f$ the density belonging to $X$. Let $a\neq0$, and $b \in \mathbb R$, while $Y:=aX+b$. Show:

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

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 $c \in \mathbb R$

$P(Y\leq c)=P(aX+b\leq c)=P(X\leq\frac{c-b}{a})$ and we know the distribution of RV $X$, thus: $P(X\leq\frac{c-b}{a})=\int_{-\infty}^{\frac{c-b}{a}}f(x)d\lambda(x)=\int_{-\infty}^{\frac{c-b}{a}}f(x)dx$

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 $\frac{1}{|a|}f(\frac{x-b}{a}), x\in\mathbb R$

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 $Y$ ~ $\mathcal{N}(b+a\mu,(a\sigma)^2)$

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    $\begingroup$ Hint for ii): Apply i). $\endgroup$ – Did Dec 5 '18 at 22:46
  • $\begingroup$ @Did Is my reasoning for as to why g(x) is measurable sound? Surely, as an alternative, why can I not say $\frac{1}{|a|}f(\frac{x−b}{a})$ is measurable simply as the product of two measurable functions $\frac{1}{|a|}$ and $f(\frac{x−b}{a})$?? $\endgroup$ – SABOY Dec 6 '18 at 15:49
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In the language of measure theoretic probability $\int\, dx$ is same as $\int \, d\lambda(x)$ 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 $a\sigma^{2}$.

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  • $\begingroup$ Is my reasoning for as to why $g(x)$ is measurable sound? Surely, as an alternative, why can I not say $\frac{1}{|a|}f(\frac{x-b}{a})$ is measurable simply as the product of two measurable functions $\frac{1}{|a|}$ and $f(\frac{x-b}{a})$?? $\endgroup$ – SABOY Dec 6 '18 at 12:03
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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.

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