LOTTERY TICKET PROBLEM
An urn contains N tickets numbered 1,2,…,N. A sample of size 2n+1 is drawn i) without replacement ii)with replacement and let M be the sample median. Find the pmf of M. Also find E(M) and V(M).
i) Without Replacement:
The pmf of M is given by:
p(m)=Pr[M=m] = \frac{\binom{m-1}{n} \binom{N-m}{n}}{\binom{N}{2n+1}} ; m=n+1,n+2,\dots , N-nIt is to be noted that,
\sum_{m=n+1}^{N-m} \binom{m-1}{n} \binom{N-m}{n} = \binom{N}{2n+1} Now,
\begin{align*}
E(M-n-1) &= \sum_{m=n+1}^{N-m} (m-n-1) p(m) \\
&= \sum_{m=n+2}^{N-m} (m-n-1) \frac{\binom{m-1}{n} \binom{N-m}{n}}{\binom{N}{2n+1}} \\
&= \sum_{m=n+2}^{N-m} \frac{(m-n-1)(m-1)!}{n!(m-n-1)(m-n-2)!} \frac{ \binom{N-m}{n}}{\binom{N}{2n+1}} \\
&= \sum_{m=n+2}^{N-m} (m-1) \frac{\binom{m-2}{n} \binom{N-m}{n}}{\binom{N}{2n+1}} \\
&= \sum_{m=n+1}^{N-m-1} m \frac{\binom{m-1}{n} \binom{N-m-1}{n}}{\binom{N}{2n+1}} \\
\end{align*}Also,
\begin{align*}
E(N-M-n) &= \sum_{m=n+1}^{N-m} (N-m-n) p(m) \\
&= \sum_{m=n+1}^{N-m-1} (N-m-n) \frac{\binom{m-1}{n} \binom{N-m}{n}}{\binom{N}{2n+1}} \\
&= \sum_{m=n+1}^{N-m-1} \frac{(N-m-n)(N-m)!}{n!(N-m-n)(N-m-n-1)!} \frac{\binom{m-1}{n} }{\binom{N}{2n+1}} \\
&= \sum_{m=n+1}^{N-m-1} (N-m) \frac{\binom{m-1}{n} \binom{N-m-1}{n}}{\binom{N}{2n+1}} \\
&= \sum_{m=n+1}^{N-m-1} m \frac{\binom{N-m-1}{n} \binom{m-1}{n}}{\binom{N}{2n+1}} = E(M-n-1)
\end{align*}Solving we get the Expectation of M as:
E(M) = \frac{N+1}{2}Again,
\begin{align*}
E\left[ (M-n-1) (N-n-M) \right] &= \sum_{m=n+1}^{N-m} (m-n-1)(N-m-n) p(m) \\
&= \sum_{m=n+2}^{N-m-1} (n+1)^2 \frac{\binom{m-2}{n+1} \binom{N-m-1}{n+1}}{\binom{N}{2n+1}} \\
&= (n+1)^2 \frac{\binom{N}{2n+3}}{\binom{N}{2n+1}} \\
&= (n+1)^2 \frac{N!}{(2n+3)!(N-2n-3)!} * \\
& \quad \frac{(2n+1)!(N-2n-1)!)}{N!} \\
&= (n+1)^2 \frac{(N-2n-1)(N-2n-2)}{(2n+3)(2n+2)} \\
&= (n+1) \frac{(N-2n-1)(N-2n-2)}{2(2n+3)} \\
\end{align*}Taking $z=M-E(M)$ we get,
\begin{align*}
E\left[ (M-n-1) (N-n-M) \right] &= E\left[ \left( \frac{N-2n-1}{2} +z \right) \left( \frac{N-2n-1}{2} - z \right) \right] \\
&= \left( \frac{N-2n-1}{2} \right)^2 - E(Z^2) \\
&= \left( \frac{N-2n-1}{2} \right)^2 - V(M)
\end{align*}Thus the variance of M is then given by:
\begin{align*}
V(M) &= \left( \frac{N-2n-1}{2} \right)^2 - (n+1) \frac{(N-2n-1)(N-2n-2)}{2(2n+3)} \\
&= \left( \frac{N-2n-1}{4} \right) \left( {N-2n-1} - \frac{2(n+1)(N-2n-2)}{(2n+3)} \right) \\
&= \left( \frac{(N-2n-1)(N+1)}{4(2n+3)} \right)
\end{align*}
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