Probability for a leading candidate to eventually win Two candidates contest a close election. Each of the $n$ voters votes independently with probability $\frac12$ each way. Fix $\alpha \in (0,1)$. Show that, for large $n$, the probability that the candidate leading after $\alpha n$ votes have been counted is the eventual winner is approximately
$$\frac{1}{2} + \frac{\sin^{-1}(\sqrt{\alpha})}{\pi}\;.$$
Hint: let Sm be the difference between the vote totals of the two candidates when m votes have been counted. What is the approximate distribution of Sαn (when appropriately rescaled)? What is the approximate distribution of $S_n - S_{\alpha n}$ (when appropriately rescaled)? What about their joint distribution? Finally, notice $\displaystyle\sin^{-1}(\sqrt{\alpha}) = \tan^{−1}\left( \frac{\alpha}{1 − \alpha}\right)$
 A: Define random variables $X_1, X_2,\ldots,X_n$ where $X_i$ equals $1$ if voter $i$ votes for Candidate A, and $-1$ otherwise; define $S_k=X_1 + \cdots + X_k$. Then $S_n>0$ means that Candidate A is the winner, and $S_{\alpha n}>0$ means Candidate A is leading after $\alpha n$ votes. Calculate for any $\alpha\in(0,1)$ that $S_{\alpha n}$ has mean $0$ and variance $\alpha n$. The covariance between $S_n$ and $S_{\alpha n}$ is $ \alpha n$:
$$
\operatorname{Cov}(S_n, S_{\alpha n}) = \operatorname{Cov} (S_{\alpha n} + (S_n - S_{\alpha n}), S_{\alpha n}) = \operatorname{Cov}(S_{\alpha n}, S_{\alpha n}) + 0
$$ by independence of $S_{\alpha n}$ and $S_n - S_{\alpha n}$. Hence the correlation between $S_n$ and $S_{\alpha n}$ is $\sqrt \alpha$. 
The aim is to calculate
$$P(S_n>0\mid S_{\alpha n}>0).$$
If $n$ is large, the joint distribution of $S_n$ and $S_{\alpha n}$ is approximately bivariate normal. (Reason: Apply the CLT to both $S_{\alpha n}$ and $S_n- S_{\alpha n}$ to deduce that each of them is approximately normally distributed. Since these two are independent, we find the joint distribution of $(S_n, S_{\alpha n})$ must be bivariate normal, using $S_n = S_{\alpha n} + (S_n-S_{\alpha n})$.)
The desired result follows from a fact(*) about bivariate normal variables:

If $(A,B)$ are bivariate normal with means $\mu_A$ and $\mu_B$ respectively, and correlation $\rho$, then
  $$P(B>\mu_B\mid A>\mu_A)=\frac12 + \frac{\arcsin\rho}\pi.$$

(*) The fact can be deduced from this result.
