Show that the density of Random Variable $Y=aX+b$ exists and if $X$ ~$\mathcal{N}(\mu,\sigma^{2})$ then how is $Y$ distributed

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

• Hint for ii): Apply i). – Did Dec 5 '18 at 22:46
• @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 $f(\frac{x−b}{a})$?? – SABOY Dec 6 '18 at 15:49

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 $$a\sigma^{2}$$.

• 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 $f(\frac{x-b}{a})$?? – SABOY Dec 6 '18 at 12:03

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.