3
$\begingroup$

If $X$ is a real random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ then there exist several characterizations of the law of $X$ being $\mu$ :

$X \sim \mu$ if and only if for every $\{...\}$ function $f:\mathbb{R}$ $\to \mathbb{R}$ we have $\mathbf{E}[f(X)]=\int_{\mathbb{R}}f(x)d\mu$.

where $\{...\}$ may be measurable positive, measurable bounded, continuous bounded, continuous compactly supported.

Why we cannot take a test function $f$ that is only supposed to be continuous ? Is the integrability of $f(X)$ the only problem that arises ?

$\endgroup$
3
$\begingroup$

You could require $f$ to only be continuous, but the idea is to have as few test functions as possible to satisfy the definition. Another way to say this is, the above possible values for $\{...\}$ are equivalent to the requirement that the equality holds for continuous functions.

Here is a proof of that for the continuous bounded characterization. On one hand, if the equality holds for all continuous functions it holds for just bounded ones. In the other direction, suppose $f$ is continuous, we want to show: $$E[f(X)] = \int_{\mathbb{R}} f(x) d\mu$$ First let $f^+$ be the positive part of $f$: $f^+(x) = f(x)\chi_{\{f(x) \geq 0 \}}$. Define $f_n$ to be $f^+$ restricted to $[0, n]$ via: $$ f_n(x) = \min\{f(x), n\}$$ Then $f_n \to f$ and by the monotone convergence theorem: $$E[f^+(X)] = \int_{\mathbb{R}} f^+(x) d\mu$$ We can apply the same technique for the negative part of $f$.

$\endgroup$
  • $\begingroup$ What if both $\mathbf{E}[f^{+}(X)]$ and $\mathbf{E}[f^{-}(X)]$ are infinite ? You wouldn't be able to conclude by linearity since $\infty - \infty$ is not defined. $\endgroup$ – Nocturne Aug 30 '15 at 15:13
  • $\begingroup$ @Nocturne: For reference: en.wikipedia.org/wiki/Lebesgue_integration A general function is defined as being Lebesgue integrable only if both the positive and negative parts are integrable (as positive functions). (Or equivalently, $\int_\mathbb{R} |f| d\mu$ is integrable. Therefore, if either of them are infinite, $\int_\mathbb{R}$ is not well defined. But then, neither is $E[|f(X)|]$ by the above calculation. So the LHS is also in the same state. $\endgroup$ – muaddib Aug 30 '15 at 15:21
  • $\begingroup$ @Nocturne - Alternatively, you can just require that $$E[f(X)] = \int_\mathbb{R} f(x) d\mu$$ for any $f$ that is continuous and integrable. $\endgroup$ – muaddib Aug 30 '15 at 15:29
  • $\begingroup$ Thanks. That's why I asked if integrability of $f(X)$ was the only obstacle in considering test functions that are only supposed to be continuous. $\endgroup$ – Nocturne Aug 30 '15 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.