Exercise 1.5.19 (Convergence in distribution)

Question

Let $(X,\mathcal{F},P)$ be a probability space. Given any real-valued function $f:X	o\mathbb{R}$ we define the cumulative distribution function of $f$ to be the function
$$F: \mathbb{R}	o [0,1], \quad F(\lambda) := P\left(\left\{ x \in X: f(x) \leq \lambdaight\} ight).$$
Given a sequence $\left(fight)_{n\in\mathbb{N}}:X	o\mathbb{R}$ of real valued measurable functions, we can say that $\left(fight)_{n\in\mathbb{N}}$ converges in distribution to $f$ if the cumulative distribution functions $\seq{F}$ of $\left(fight)_{n\in\mathbb{N}}$ converges pointwise to the cumulative distribution $F$ of $f$ at all $\lambda \in \mathbb{R}$ for which $F(\lambda)$ is continuous.
\begin{enumerate}[label=(oman*)]
\item Show that if $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in any of the seven senses discussed in this chapter, then it converges in distrubition to $f$.
\item Give an example in which $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in distribution, but not in any of the above seven senses.
\item Give an example in which $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in distribution, and $\left(gight)_{n\in\mathbb{N}}$ converges to $g$ in distribution, but $(f_n+g_n)_{n\in\mathbb{N}}$ does not converge to $f+g$.
\item Give an example in which a sequence $\left(fight)_{n\in\mathbb{N}}$ can converge in distribution to two different limits $f,g$ which are not equal almost everywhere.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
\item We will demonstrate this for the weakest modes of convergence: a.e. pointwise convergence and convergence in measure. In case of the measure convergence we have for any $\lambda \in \mathbb{R}$:
\begin{align*}
\lim_{n	o\infty}|F_n(\lambda) - F(\lambda)| &= \lim_{n	o\infty} |P\left(\left\{f_n\leq \lambdaight\} ight) - P\left(\left\{f\leq \lambdaight\} ight) \
\intertext{Depending on which of the both probabilities is bigger, we can apply the rule $P(A/B) = P(A) - P(A\cap B) \geq P(A) - P(B)$, and look at the quantities $P(\{f_n \leq \lambda, f > \lambda\}) = P(\{f_n \leq \lambda, f\geq \lambda + c\})$, or $P(\{f \leq \lambda, f_n > \lambda\}) = P(\{f \leq \lambda, f_n \geq \lambda + c\})$ for some $c\in [0,+\infty)$. In both cases we have:}
&\leq \lim_{n	o\infty} P\left(\left\{|f-f_n|\geq cight\} ight) = 0
\end{align*}
Now consider the a.e. pointwise convergence. On a finite measure space, pointwise convergence is equivalent to almost uniform convergence by \href{./exercise_1-5-18}{Exercise 1.5.18}, and almost uniform convergence implies convergence in measure by \href{./exercise_1-5-2}{Exercise 1.5.2}, and we are done.
			
\item Consider the unit Lebesgue measure space $([0,1],\mathcal{B},m)$, and define
$$f_n:[0,1]	o[0,1], \quad f_n(x) = \left\{ \begin{array}{ll} x, &	ext{if } n|2\ 1-x, & 	ext{if } n
mid 2 \end{array} ight. $$
We then have
$$F_n(\lambda) = P\left(\left\{x \in [0,1]: f_n(x) \leq \lambdaight\} ight) = \left\{ \begin{array}{ll} P\left(\left\{x \in [0,1]: x \leq \lambdaight\} ight) = \lambda\ P\left(\left\{x \in [0,1]: 1-x \leq \lambdaight\} ight) = 1 - P\left(\left\{x \in [0,1]: x \geq 1-\lambdaight\} ight) =\lambda \end{array} ight.$$
Yet when it comes to the convergence in probability, then the two subsequences diverge from each other:
$$P\left(\left\{x \in [0,1]: |f_{2n}(x)-f_{2n+1}(x)| \geq \epsilon ight\} ight) = P\left(\left\{x \in [0,1]: |x-(1-x)| \geq \epsilon ight\} ight) = P\left(\left\{x \in [0,1]: 2x \geq \epsilon ight\} ight) = 1 - \frac{\epsilon}{2} \stackrel{\epsilon 	o 0}{
ot	o} 0.$$
Thus $\left(fight)_{n\in\mathbb{N}}$ does not converge to a single limit in probability.

\item Consider a measurable function $f(x) = x$ and $-f$ on $([-1,1], \mathcal{B}, m)$. It is easy to verify that both functions have the distribution function $F(\lambda)=P(\{f\leq \lambda\}) = (\lambda+1)/2$ symmetric around the 0. Their sum $-f+f = 0$, however, has the distribution $F_0(\lambda) = P(\{0\leq \lambda\}) = 0$ if $\lambda <0$ and $=1$, if $\lambda \geq 0$.

\item As in the example above, the constant sequence $\left(fight)_{n\in\mathbb{N}}=f$ can converge in distribution both to $f$ and $-f$, both not equal anywhere except at 0.
\end{enumerate}

Exercise 1.5.19 (Convergence in distribution)

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Exercise 1.5.19 (Convergence in distribution)

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