Convergence in Probability vs Central Limit Theorem

Let X1, X2, . . . be independent identically distributed random variables with P(X1 = 1) = P(X1 = −1) = 1/2. For any n ≥ 1, define

$S_n = \frac{X_1+...+X_2}{\sqrt{n}}$

The Central Limit Theorem says that Sn converges in distribution to a standard Gaussian random variable. We show that Sn does not converge in probability to any random variable. The intuition here is that if Sn did converge in probability to a random variable Z, thenwhen n is large, Sn is close to Z, $Y_n = \frac{\sqrt{2}S_{2n} - S_n}{\sqrt{2}-1}$ is close to Z, but Sn and Yn are independent. And this cannot happen. Proceed as follows. Assume that Sn converges in probability to Z.

• Let ε > 0. For n very large (depending on ε), we have P(|Sn − Z| > ε) < ε and P(|Yn − Z| > ε) < ε.

• Show that P(Sn > 0, Yn > 0) is around 1/4, using independence and the Central Limit Theorem.

• From the first item, show P(Sn > 0|Z > ε) > 1 − ε, P(Yn > 0|Z > ε) > 1 − ε, so P(Sn > 0, Yn > 0|Z > ε) > 1 − 2ε.

• Without loss of generality, for ε small, we have P(Z > ε) > 4/9.

• By conditioning on Z > ε, show that P(Sn > 0, Yn > 0) is at least 3/8, when n is large.

I understand in the first part that you multiple the 2 probabilities because they are independent. Don't know how to get the results I want though. I'm stuck with the rest too.

• It is not your fault: The instructions seem wrong. I think we agree on the first two bullets, but it is the third bullet that is strange: The true third instruction bullet should be to show $P[S_n<0, Z>\epsilon] < \epsilon$, $P[Y_n<0, Z>\epsilon] <\epsilon$, so $P[\{S_n<0, Z>\epsilon\} \cup \{Y_n<0, Z>\epsilon\}] < 2\epsilon$, so $P[\{S_n<0\} \cup \{Y_n<0\}|Z>\epsilon]< \frac{2\epsilon}{P[Z>\epsilon]}$ and $P[S_n\geq 0, Y_n\geq 0|Z>\epsilon] \geq 1 - \frac{2\epsilon}{P[Z>\epsilon]}$. Then you can use $P[Z>\epsilon] > 4/9$ to get a lower bound on $P[S_n>0, Y_n>0]$. – Michael Mar 2 '17 at 7:07
• Ok, I think I understood most of what you said, I'm now stuck at how to plug in the fact that P[Z>E] > 4/9 to the final equation we got from bullet point 3. – Lebron James Mar 2 '17 at 18:48
• My guess is that $P(S_n>0,Y_n>0|z>\epsilon) > 1- 2\epsilon = \frac{1}{4}$. Then solving the right side gives $\epsilon = \frac{3}{8}$? – Lebron James Mar 2 '17 at 19:39
• You can ignore the $3/8$ thing, that seems irrelevant (likely it was in your instructions because it was based on an incorrect step). Just multiply $P[S_n\geq 0, Y_n \geq 0 | Z>\epsilon] \geq 1 - \frac{2\epsilon}{P[Z>\epsilon]}$ on both sides by $P[Z>\epsilon]$ and use the fact that $\epsilon$ is small. – Michael Mar 4 '17 at 3:24