Q. 8.4 - A First Course in Probability

Question

Let$Zn$, $n \ge 1$ be a sequence of random variables and$c$ a constant such that for each

$\epsilon >0,P|Zn-c|>\epsilon \to 0$as$n\to q$. Show that for any bounded continuous function$g$,

$E\left[g\right(Zn\left)\right]\to g\left(c\right)$ as$n\to q$.

Step-by-Step Solution

Answer

Split$E\left[g\left(Z_n\right)\right]$ into two parts: where$x$ is nearly close to$c$ and where $x$is far from$c$ and use assumptions to obtain the required.

Step 1 Given Information.

$Zn, n \ge 1,$ be a sequence of random variables and $c$ a constant such that for each

 $\epsilon >0,P|Zn-c|>\epsilon \to 0$as$n\to q$.

Step 2 Explanation.

We are given that$g$ is a continuous function, which means that for every $c\in \mathrm{ℝ}$ and$ϵ>0$ there exists $\delta >0$ such that

$|x-c|\le \delta \Rightarrow \left|g\right(x)-g(c\left)\right|\le ϵ$

Secondly, we are given that$g$ is a bounded function which means that there exists$B>0$ such that

$\left|g\right(x\left)\right|\le B$

for every$x\in \mathrm{ℝ}$. Using the theorem about the expected value of a function of a random variable, we have that

$E\left[g\left(Z_n\right)\right]={\int }_{\mathrm{ℝ}}g\left(x\right)d{P}_{Z_n}={\int }_{|x-c|\le \delta }g\left(x\right)d{P}_{Z_n}+{\int }_{|x-c|>\delta }g\left(x\right)d{P}_{Z_n}$

for some fixed$c\in \mathrm{ℝ}$. Now, for $x$such that $|x-c|\le \delta$we have that $g\left(x\right)\le g\left(c\right)+ϵ$and for$x$ such that$|x-c|>\delta$ we have that$g\left(x\right)\le B$. Therefore

$E\left[g\left(Z_n\right)\right]\le \left(g\right(c)+ϵ){\int }_{|x-c|\le \delta }d{P}_{Z_n}+B{\int }_{|x-c|>\delta }d{P}_{Z_n}$$=\left(g\right(c)+ϵ)P\left(\left|Z_n-c\right|\le \delta \right)+BP\left(\left|Z_n-c\right|>\delta \right)$

Now, apply lim sup$n$ to both sides and use the assumption that $P\left(\left|Z_n-c\right|>\delta \right)\to 0$as$n\to \infty$. It yields that $P\left(\left|Z_n-c\right|\le ϵ\right)\to 1$as$n\to \infty$. Therefore, we end up with 

 $\underset{n}{\mathrm{limsup}}E\left[g\left(Z_n\right)\right]\le g\left(c\right)+ϵ$

Step 3 Explanation.

On the other hand, we can make lower boundary on$E\left[g\left(Z_n\right)\right]$. Observe that for that same $ϵ$we have that for$|x-c|\le \delta$ holds $g\left(x\right)\ge g\left(c\right)-ϵ$ Also, for every $x$we have$-B\le$$g\left(x\right)$. Therefore, similarly to previous, we have that

$E\left[g\left(Z_n\right)\right]\ge \left(g\right(c)-ϵ){\int }_{|x-c|\le \delta }d{P}_{Z_n}-B{\int }_{|x-c|>\delta }d{P}_{Z_n}$

$=\left(g\right(c)-ϵ)P\left(\left|Z_n-c\right|\le \delta \right)-BP\left(\left|Z_n-c\right|>\delta \right)$

Applying limit inferior to both sides, we have that

$\underset{n}{\mathrm{liminf}}E\left[g\left(Z_n\right)\right]\ge g\left(c\right)-ϵ$

because of$P\left(\left|Z_n-c\right|>\delta \right)\to 0$ as$n\to \infty$. Finally, we have that

$g\left(c\right)-ϵ\le \underset{n}{\mathrm{liminf}}E\left[g\left(Z_n\right)\right]\le \underset{n}{\mathrm{limsup}}E\left[g\left(Z_n\right)\right]\le g\left(c\right)+ϵ$

and since$ϵ>0$ was arbitrary, we can get that $ϵ\to 0$to finally obtain that

$\underset{n}{\lim }E\left[g\left(Z_n\right)\right]=g\left(c\right)$

so we have proved the claim.