Let $ (X_n)_{n \in \mathbb{N}}$ be i.d.d. random variables with $E{X_1}=0$, $Var(X_1)=1$ and $ S_n = X_1 + X_2 +...+ X_n $. Calculate $ \lim_{n \to +\infty}\Pr(S_n>\sqrt{n})$.

On the back page, it has as a solution that the limit equals to $\frac{1}{2}$ but I can't understand why.

What I did is to use the central limit theorem so i can show that $\frac{S_n -nE(X_1)}{\sqrt{nVar(x_1)}} = \frac{S_n}{\sqrt{n}}$ converges to $ Z \sim \mathcal{N}(0,1)$.

Then, $ \Pr(\frac{S_n}{\sqrt{n}}>1) $ converges to $\Pr(Z>1) = Φ(-1)$, where Φ the cumulative distribution function. Is there any fault on my solution that i cannot see?

  • $\begingroup$ Your answer is correct. Are you sure the question was not to compute $P(S_n > n^{-1/2})$ ? $\endgroup$ – Gautam Shenoy Aug 6 '14 at 18:44
  • $\begingroup$ Well it is written in Greek so i'm not sure whether my translation is accurate. In a raw translation it says to calculate that limit. $\endgroup$ – coldcoffee Aug 6 '14 at 18:48
  • $\begingroup$ Oh sorry i just understood your question, heh. Yes i just checked again, it asks to compute $\Pr(S_n>\sqrt{n})$ $\endgroup$ – coldcoffee Aug 6 '14 at 18:55

What you say seems sensible.

Here is some R code to test the question empirically:

n <- 1000
cases <- 10000
exampledata <- matrix(rnorm(n*cases, mean=0, sd=1), ncol=n)
Sn <- rowSums(exampledata)
mean(Sn > sqrt(n))

and this gives 0.1587 while $\Phi(-1) \approx 0.158655$, closer than might reasonably be expected.

  • $\begingroup$ Oh, thanks a lot. $\endgroup$ – coldcoffee Aug 6 '14 at 18:49

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.