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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?

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  • $\begingroup$ 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
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What you say seems sensible.

Here is some R code to test the question empirically:

set.seed(1)
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.

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

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