1
$\begingroup$

$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $\mu_n=E(X_n I(X_n \le n)),S_n=\sum_{i=1}^n X_i$.

Proof:$$\frac{S_n}{n}-\mu_n\overset{p}{\to }0$$

My answer is:

$$\frac{S_n}{n}-\mu_n=(\frac{S_n}{n}-E(X_i))+(E(X_i)-\mu_n)$$

According to the law of large numbers, $\frac{S_n}{n}-E(X_i)\overset{p}{\to }0$, and it is easy to proof that $E(X_i)-\mu_n \to 0$, so the proposition is proved.

Because its form is also very close to the strong law of large numbers, I wonder if $\frac{S_n}{n}-\mu_n\overset{a.s.}{\to }0$, too.

Assumptions: The following proposition has been proven that $$X_n\overset{p}{\to }X,Y_n\overset{p}{\to }Y \Rightarrow X_n+Y_n\overset{p}{\to }X+Y$$ It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.

$\endgroup$
  • $\begingroup$ Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem. $\endgroup$ – Song Dec 8 '18 at 9:59
  • $\begingroup$ If $X_n \to X$ almost surely and $Y_n \to Y$ almost surely then $X_n+Y_n \to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$. $\endgroup$ – Kavi Rama Murthy Dec 8 '18 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.