Question

Let X and Y be i.i.d. binomial random variables with parameters n and 0.5 and let Z be another binomial random variable with parameters 2n and 0.5. Then, \(P(X = Y)\) equals to:

• A. \(P(Z=0)\)
• B. \(P(Z=n)\)
• C. \(P(Z=2n-1)\)
• D. \(P(Z=n+1)\)

Hint:

Use Vandermonde's identity \(\sum_k \binom{n}{k}^2=\binom{2n}{n}\), or a coin-toss coupling argument with \(Y'=n-Y\).

Answer 1

The Correct Option is B

Step 1: \(X\) and \(Y\) are i.i.d. \(\text{Bin}(n, 0.5)\). We want \(P(X=Y)\):\[ P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k) = \sum_{k=0}^{n} \binom{n}{k}^2 (0.5)^{2n} \]
Step 2: Use Vandermonde's identity, \(\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}\):\[ P(X=Y) = \binom{2n}{n} (0.5)^{2n} \]
Step 3: Now consider \(Z \sim \text{Bin}(2n, 0.5)\). Its probability mass function at \(n\) is:\[ P(Z=n) = \binom{2n}{n} (0.5)^{n}(0.5)^{2n-n} = \binom{2n}{n} (0.5)^{2n} \]
Step 4: Comparing the two expressions, they are identical:\[ P(X=Y) = P(Z=n) \]So the answer is \(\boxed{P(Z=n)}\), option (B).