Question :

let $Z_n, n\geq 1$, be a sequence of random variables and $c$ a constant such that, for each $\epsilon \gt 0, P{|Z_n-c|\gt\epsilon}\rightarrow\ 0 \,as\, n \, \rightarrow \infty $. Show that, for any bounded continuous function g, $$E[g(Z_n)]\rightarrow g(c)\,as\, n\rightarrow\infty$$

Answer attempt:

I set $Z^n=Z_1+Z_2+...+Z_n$ to be my sequence of random variables.

Chebyshev's inequality states, $P{|X-\mu|\geq k}\leq \frac{\sigma^2}{k^2}$, so based on this inequality I set $c=n\mu$, where $\mu$ is the sample mean, and if $\epsilon = n\epsilon_0$, where $\epsilon_0$ is some arbitrary constant, then if I plug this into Chebyshev's inequality I get : $$p(|Z^n-\mu n|\geq\epsilon_0 n)\leq \frac{\sigma^2}{n^2 \epsilon^2_0}=p(|\frac{Z^n}{n}-\mu|\geq\epsilon_0)\leq \frac{\sigma^2}{n^3 \epsilon_0}$$ So this goes to $0$ as $n\rightarrow \infty$, so it makes sense that $c=n\mu$.

So then, to prove $ E(g(Z^n)) \rightarrow g(c) \, as \, n\rightarrow\infty$, I am not sure what to do. I think, intuitively, $E(g(z_i)) \,as \,n\rightarrow\infty = g(\mu)$, which would prove the inequality.

But, how do you prove $\sum_0^\infty g(z_i)p(z)=g(\mu)$? Anyway yeah, I am lost.

  • 3
    $\begingroup$ I suppose you mean $P(|Z_n-c|>\epsilon) \to \color{red}{0}$...? As it is written now it doesn't make any sense. $\endgroup$ – saz May 16 '18 at 19:25
  • $\begingroup$ It seems to me that this question is equivalent to the question: "Convergence in probability implies convergence in distribution". This is standard. See eg math.stackexchange.com/questions/236955/… $\endgroup$ – user52227 May 18 '18 at 12:53

I do not get your attempt sorry. Possible way:

What you are given says that for every $\epsilon >0$, $P(Z_n\notin B_\epsilon(c))\to 0$ as $n\to\infty$. For $f$ bounded and continuous and an arbitrary $\delta>0$, you can use the inequality \begin{align} |E[f(Z_n)]-f(c)|\le&\, E[|(f(Z_n)-f(c))|\mathbf 1_{B_\epsilon(c)}(Z_n)]+|E[(f(Z_n)-f(c))\mathbf 1_{B_\epsilon(c)^c}(Z_n)]|\\ \le &\, E[|(f(Z_n)-f(c))|\mathbf 1_{B_\epsilon(c)}(Z_n)]+ 2\sup_x|f(x)|P[Z_n\notin B_\epsilon(c)], \end{align} and choose $\epsilon $ small so that $E[|(f(Z_n)-f(c))|\mathbf 1_{B_\epsilon(c)}(Z_n)]\le \delta$ (independently of $n$), and then choose $n$ big to make the second term smaller than $\delta$.

(Generally, convergence in probability implies convergence in distribution which is equivalent to convergence testing against bounded continuous functions. For convergence to constants you can improve this result).


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.