Exercise 0.30 (Properties of multivariate Stieltjes measure function)

Question

Let $\left(\mathbb{R}^n,\mathcal{B},Pight)$ be the Euclidean probability space, and let $F$ be the associated Stieltjes measure function. Establish the following properties of $F$:
\begin{enumerate}
	\item $F$ is non-decreasing: whenever $t_i < t_{i}'$ for all $i$, we have $F(t_1,\ldots,t_n) \leq F(t_1',\ldots,t_n')$.
	\item $\lim_{t	o-\infty}F(t) = 0$ and $\lim_{t	o+\infty}F(t) = 1$.
	\item $F$ is right continuous, i.e., for all $t\in\mathbb{R}^n$ we have $F(t_1,\ldots,t_n) = \lim_{(s_1,\ldots,s_n)	o (t_1,\ldots,t_n)^+} F(s)$.
	\item One has $$\sum_{(\omega_1,\ldots,\omega_n)\in\{0,1\}^n} (-1)^{\omega_1+\ldots+\omega_n+n} F(t_{1,\omega_1},\ldots,t_{n,\omega_n}) \geq 0$$ whenever $t_{i,0}\leq t_{i,1}$ are real numbers for $i=1,\ldots,n$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}
\item Let $\mathbf{t} \leq \mathbf{t}'$. Obviously, 
$$(-\infty,t_1] \times \dots \times (-\infty,t_n] \subseteq (-\infty,t_1'] \times \dots \times (-\infty,t_n'];$$
thus, the probability of the latter set is greater than or equal to the probability of the former set by \href{./exercise_23}{monotonicity of probability}.

\item By \href{./exercise_0-23}{continuity from above}, and \href{../analysis_I/exercise_9-3-1}{equivalently} looking at the convergence along a countable sequence $(t_k)_{k\in\mathbf{N}} \to - \infty$, we have 
\begin{align*}
\lim_{k\to\infty} F(t_k)
&= \lim_{k\to\infty} \mu\left( (-\infty,t_k^{(1)}] \times \dots \times (-\infty,t_k^{(n)}] \right)\
&= \mu\left( \bigcap_{k \in \mathbf{N}} (-\infty,t_k^{(1)}] \times \dots \times (-\infty,t_n] \right)\[5pt]
&= \mu\left( \emptyset \right) = 0.
\end{align*}
Similarly, by the \href{./exercise_0-23}{continuity from below}, and for $(t_k)_{k\in\mathbf{N}} \to + \infty$:
\begin{align*}
\lim_{k\to\infty}F(t_k)
&= \lim_{k\to\infty} \mu\left( (-\infty,t_k^{(1)}] \times \dots \times (-\infty,t_k^{(n)}] \right)\
&= \mu\left( \bigcup_{k \in \mathbf{N}} (-\infty,t_k^{(1)}] \times \dots \times (-\infty,t_n] \right)\[5pt]
&= \mu\left( \mathbf{R}^n \right) = 1.
\end{align*}

\item Let $(s_k)_{k\in\mathbf{N}}$ be an arbitrary sequence converging to $t$ from the right. We then have
\begin{align*}
\lim_{k\to\infty} F(s_k)
&= \lim_{k\to\infty} \mu\left( (-\infty,s_k^{(1)}] \times \dots \times (-\infty,s_k^{(n)}] \right)
\intertext{By additivity of the probability measure,}
&= \lim_{k\to\infty} \mu\left( (-\infty,t^{(1)}] \times \dots \times (-\infty,t^{(n)}]\right) + \mu\left( (t^{(1)},s_k^{(1)}] \times \dots \times (t^{(n)},s_k^{(n)}] \right)
\intertext{By \href{./exercise_0-23}{continuity from below},}
&=\mu\left( (-\infty,t^{(1)}] \times \dots \times (-\infty,t^{(n)}]\right) + \mu\left( \bigcap_{k \in \mathbf{N}} (t^{(1)},s_k^{(1)}] \times \dots \times (t^{(n)},s_k^{(n)}] \right)\
&=\mu\left( (-\infty,t^{(1)}] \times \dots \times (-\infty,t^{(n)}]\right)\
&= F(t).
\end{align*}

\item We induct on the dimension of the measure space $n$.
\begin{itemize}
\item \textit{(Induction base)}. In case of $n=1$ we have, by monotonicity of the Stieltjes measure function,
$$\sum_{\omega_1 \in \{0,1\}} (-1)^{\omega_1 + 1} F(t_{1\omega_1})
= (-1)^{1+1}F(t_{11}) + (-1)^{0+1}F(t_{10})
= F(t_{11}) - F(t_{10}) \geq 0$$
\item \textit{(Induction base)} Now suppose inductively that we have proven the theorem assertion for some $n\in\mathbf{N}$. Now we prove it for $n++$. Since every element of $\{0,1\}^{n}$ can be bijectively associated with an element of $\{0,1\}^{n+1}$ in either two ways, we have
\begin{align*}
&\sum_{(\omega_1,\dots,\omega_n,\omega_{n+1}) \in \{0,1\}^{n+1}} (-1)^{\omega_1+\dots+\omega_n+\omega_{n+1}+(n+1)} F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,\omega_{n+1}})\
&= 
\sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n + 0 + (n+1)} F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,0})\
&+
\sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n+ 1 + (n+1)} F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,1})\
&=
-\sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n + n} F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,0})\
&+
\sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n+n} F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,1})\
&=
\sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n+n} \left[ F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,1}) - F( t_{1,\omega_1}, \dots, t_{n,\omega_n}, t_{n+1,0})\right]
\intertext{Since Stieltjes measure function is multiplicative in each dimension, we can write this as:}
&=
\left[F(t_{n+1,1}) - F(t_{n+1,0})\right] \times \sum_{(\omega_1,\dots,\omega_n) \in \{0,1\}^{n}} (-1)^{\omega_1+\dots+\omega_n+n} F( t_{1,\omega_1}, \dots, t_{n,\omega_n})
\end{align*}
The former factor $\left[F(t_{n+1,1}) - F(t_{n+1,0})\right]$ is positive by the monotonicity of $F$, the latter - by induction hypothesis. This closes the induction.
\end{itemize}
\end{enumerate}

Exercise 0.30 (Properties of multivariate Stieltjes measure function)

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Exercise 0.30 (Properties of multivariate Stieltjes measure function)

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