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Let $X$ be a real random variable and $f$ is its density w.r.t. the Lebesgue measure.

As a background, I was asked to show the density of $Y:=aX+b$ exists and is $g(x):=\frac{1}{|a|}f(\frac{x-b}{a})$. In order to do that, I eventually need to show that $\frac{1}{|a|}f(\frac{x-b}{a})$ is measurable. This is where I struggle to find why my alternative would be wrong.

Correct Answer: 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?)

Proposed Alternative Solution: 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 a constant function is continuous and therefore Borel-measurable) as well as $f(\frac{x-b}{a})$, and therefore the product is Borel-measurable. I have a feeling that this may however not be so simple because $f(x)$ being measurable does not imply that $f(\frac{x-b}{a})$ is indeed Borel-measurable.

Any ideas, as to why the alternative does not hold?

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Your alternative does work. Let $h(x)=(x-b)/a$. Since $$f:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal B(\mathbb R))$$ is measurable, and $$h:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal L(\mathbb R))$$ is measurable (since it is Borel measurable, and inverse images of Borel null sets are null), it follows their composition
$$f\circ h:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal B(\mathbb R))$$ is measurable.

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  • $\begingroup$ Do we indeed know $f$ is Borel? Or is it just Lebesgue measurable? $\endgroup$ – Robert Israel Dec 12 '18 at 2:16
  • $\begingroup$ @RobertIsrael See edit. Whatever level of measurability you have for $f$, the same will be true of $g$. $\endgroup$ – Mike Earnest Dec 12 '18 at 4:37
  • $\begingroup$ @RobertIsrael Oh sorry, I mistook you for OP! But to your question, I am not sure and might have misspoke before. $\endgroup$ – Mike Earnest Dec 12 '18 at 4:38
  • $\begingroup$ Careful: the composition of two Lebesgue measurable functions is not necessarily Lebesgue measurable. See e.g. here $\endgroup$ – Robert Israel Dec 12 '18 at 12:42
  • $\begingroup$ @RobertIsrael You are right, it is now fixed. $\endgroup$ – Mike Earnest Dec 12 '18 at 14:43
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The Radon-Nikodym theorem tells you that if $X$ is a random variable whose distribution is absolutely continuous wrt Lebesgue measure (i.e. $\mathbb P(X \in A) = 0$ for all sets $A$ of Lebesgue measure $0$), then this distribution has a density $f(x)$ which is a Lebesgue measurable function, i.e. $\mathbb P(X \in E) = \int_E f(x)\; dx $ for all Lebesgue measurable sets $E$.

It is clear that if $a \ne 0$, the distribution of $Y = aX+b$ is also absolutely continuous and its density is what you expect it to be: the change-of-variables formula works for Lebesgue integration.

I don't know why you have to drag Borel into this.

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  • $\begingroup$ Thank your for the insight, but we "dragged" Borel into this in our lecture, as we have not used Radon-Nikodym as of yet. And I am specifically looking for an answer on whether my alternative solution stands. And if not, why not. $\endgroup$ – SABOY Dec 11 '18 at 22:41

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