Exercise 1.36 (Probability density functions I)

Question

Let $f: {\bf R} \rightarrow [0,+\infty]$ be a measurable function with $\int_{\bf R} f(x)\ dx = 1$. If one defines $m_f(E)$ for any Borel subset $E$ of ${\bf R}$ by the formula

$$\displaystyle m_f(E) := \int_E f(x)\ dx,$$

show that $m_f$ is a probability measure on ${\bf R}$ with Stieltjes measure function $F(t) = \int_{-\infty}^t f(x)\ dx$. If $X$ is a real random variable with probability distribution $m_f$ (in which case we call $X$ a random variable with an absolutely continuous distribution, and $f$ the probability density function (PDF) of $X$), show that

$$\displaystyle \mathop{\bf E} G(X) = \int_{\bf R} G(x) f(x)\ dx$$

when either $G: {\bf R} \rightarrow [0,+\infty]$ is an unsigned measurable function, or $G: {\bf R} \rightarrow {\bf C}$ is measurable with $G(X)$ absolutely integrable (or equivalently, that $\int_{\bf R} |G(x)| f(x)\ dx < \infty$.

Elu Thingol · Accepted Answer

We shortly verify the probability axioms for $m_f$.
\begin{itemize}
\item We have $m_f(\emptyset) = \int_{\emptyset} f\ \mathrm{d}m = 0$ and $m_f(\mathbf{R}) = \int_{\mathbf{R}} f\ \mathrm{d}m = 1$ by definition.
\item For any disjoint collection $E_1,E_2,\ldots$ of measurable subsets of $\mathbf{R}$ we have
	$$m_f\left(\bigcup_{n=1}^{\infty}E_night) = \int_{\bigcup_{n=1}^{\infty}E_n} f\ \mathrm{d}m = \sum_{n=1}^{\infty} \int_{E_n} f\ \mathrm{d}m = \sum_{n=1}^{\infty} \mu_f(E_n).$$
\end{itemize}
The corresponding Stieltjes measure function is:
$$F(t) = m_f\left((-\infty,t]ight) = \int_{(-\infty,t]} f\ \mathrm{d}m.$$\par

Now let $X$ be a real-valued random variable characterised by the probability $m_f$. Applying Theorem 1.33 (Change of variables formula) it suffices to demonstrate that
$$\int_{\mathbf{R}}G\ \mathrm{d}m_f = \int_{\mathbf{R}} Gf\ \mathrm{d}m.$$
As usual for proves regarding integrals, we provide a nested argument. In other words, consider the following cases.
\begin{enumerate}
\item $G = 1_{E}$ is an indicator function. We then have
	$$\int_{\mathbf{R}}G\ \mathrm{d}m_f = \int_{\mathbf{R}}1_E\ \mathrm{d}m_f = m_f(E) = \int_{E} f\ \mathrm{d}m = \int_{\mathbf{R}}1_E f\ \mathrm{d}m = \int_{\mathbf{R}} Gf\ \mathrm{d}m.$$
\item $G = \sum a_i 1_{E_i}$ is a simple function.
ewline
	Then the theorem assertion holds as it is a linear combination of indicator functions.
\item $G$ is a non-negative function.
ewline
	Then by axiom of choice, there exists a sequence of non-negative increasing simple functions $(g_n)_{n\in\mathbb{N}}$ which converge to $G$. Thus, by applying the monotone convergence theorem (Theorem 1.18) twice we obtain:
	$$\int_{\mathbf{R}}G\ \mathrm{d}m_f
	= \int_{\mathbf{R}} \lim_{n	o\infty} g_n\ \mathrm{d}m_f
	\stackrel{MCT}{=} \lim_{n	o\infty}\int_{\mathbf{R}} g_n\ \mathrm{d}m_f
	= \lim_{n	o\infty}\int_{\mathbf{R}} g_n f\ \mathrm{d}m
	\stackrel{MCT}{=} \int_{\mathbf{R}} \lim_{n	o\infty} g_n f\ \mathrm{d}m
	= \int_{\mathbf{R}} G f\ \mathrm{d}m.
	$$
\item If, finally, $G$ is a real-valued or complex-valued function, then the assertion follows by splitting into positive and negative / real and imaginary parts.
\end{enumerate}

Exercise 1.36 (Probability density functions I)

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Exercise 1.36 (Probability density functions I)

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