Question

If \( A_{x:n} = \{k : |k - \tfrac{n}{2}| \le \tfrac{x\sqrt{n}}{2} \} \), then the value of

\( \lim_{n \to \infty} \sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n} \)

(where \( \Phi(\cdot) \) is the distribution function of the standard normal variate) is:

• A. $\Phi(-1)$
• B. $\Phi(1)$
• C. $1-2\Phi(-1)$
• D. $\frac{\Phi(1)}{2}$

Hint:

The De Moivre-Laplace theorem is a special case of CLT. For $p=0.5$, the interval $[\mu - \sigma, \mu + \sigma]$ always contains approximately $68\%$ of the distribution in the limit.

Answer 1

The Correct Option is C

This expression represents the probability of a Binomial random variable falling within a specific range, which converges to a Normal distribution.

Step 1: \color{redIdentify the underlying distribution
Let $X \sim Binomial(n, 1/2)$. Then $P(X=k) = \binom{n}{k} (1/2)^n$.
The sum $\sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n}$ is simply $P(X \in A_{1:n})$.

Step 2: \color{redAnalyze the set $A_{1:n}$
For $x=1$, $A_{1:n} = \{k : |k - \frac{n}{2}| \le \frac{\sqrt{n}}{2}\}$.
This is equivalent to $P(|X - \frac{n}{2}| \le \frac{\sqrt{n}}{2})$.

Step 3: \color{redNormal Approximation
For $X \sim Bin(n, 1/2)$, $E(X) = n/2$ and $Var(X) = n(1/2)(1/2) = n/4$.
Standard deviation $\sigma = \sqrt{n}/2$.
The probability is $P(|X - \mu| \le \sigma)$.

Step 4: \color{redCalculate the limit
$P(|X - \mu| \le \sigma) = P(-\sigma \le X - \mu \le \sigma) = P(-1 \le \frac{X-\mu}{\sigma} \le 1)$.
As $n \to \infty$, this becomes $P(-1 \le Z \le 1)$, where $Z \sim N(0,1)$.
$P(-1 \le Z \le 1) = \Phi(1) - \Phi(-1)$.
Using $\Phi(1) = 1 - \Phi(-1)$, we get $(1 - \Phi(-1)) - \Phi(-1) = 1 - 2\Phi(-1)$.
Thus, the limit is $1 - 2\Phi(-1)$.