Let . Using probabilistic methods, evaluate: .

Probability Theory Solution – Problem 70: Let

Question

Let $p \in (0.5,1)$ . Using probabilistic methods, evaluate: $\displaystyle \lim_{n \to \infty} (2n+1)\cdot \sum_{k = 0}^n \binom{2n+1}{k} p^k(1-p)^{2n+1-k}$ .

Step-by-step solution

Let $S_{2n+1} \sim \text{Binomial}(2n+1, p)$ . Then $\sum_{k=0}^n \binom{2n+1}{k} p^k (1-p)^{2n+1-k} = P(S_{2n+1} \le n).$ Since $p > 0.5$ , the expectation $E[S_{2n+1}] = (2n+1)p > n + 0.5$ (for large $n$ ), so $P(S_{2n+1} \le n)$ is small. Standardization: $\mu = E[S_{2n+1}] = (2n+1)p,$ $\sigma^2 = \mathrm{Var}(S_{2n+1}) = (2n+1)p(1-p).$ Let $Z_n = \frac{S_{2n+1} - \mu}{\sigma}.$ The event $\{S_{2n+1} \le n\}$ is equivalent to $Z_n \le \frac{n - (2n+1)p}{\sqrt{(2n+1)p(1-p)}}.$ Let $q = 1-p$ , so $p > 0.5$ implies $q < 0.5$ .

$n - (2n+1)p = n - 2np - p = - (2p-1)n - p.$ Since $2p-1 > 0$ , for large $n$ this quantity is negative with magnitude $\approx (2p-1)n$ .

More precisely: $\frac{n - (2n+1)p}{\sigma} = \frac{- (2p-1)n - p}{\sqrt{(2n+1)pq}}.$ Let $\delta = 2p-1 > 0$ . Then $\frac{n - \mu}{\sigma} \approx -\frac{\delta n}{\sqrt{2n pq}} = -\frac{\delta}{\sqrt{2pq}} \cdot \sqrt{n}.$ Thus it tends to $-\infty$ at rate $-c\sqrt{n}$ , where $c = \frac{\delta}{\sqrt{2pq}} > 0$ .

For the standard normal distribution, as $x \to -\infty$ , the Mills ratio inequality gives: $\frac{\phi(x)}{\Phi(x)} \sim -x, \quad x \to -\infty.$ More precisely, if $x \to -\infty$ , then $\Phi(x) \sim \frac{\phi(x)}{|x|}.$ Here $x_n \approx -c\sqrt{n}$ , so $\phi(x_n) = \frac{1}{\sqrt{2\pi}} e^{-x_n^2/2} \approx \frac{1}{\sqrt{2\pi}} e^{-c^2 n/2}.$ Thus $\Phi(x_n) \sim \frac{\phi(x_n)}{c\sqrt{n}}.$

Therefore $P(S_{2n+1} \le n) \sim \frac{1}{c\sqrt{n}} \cdot \frac{1}{\sqrt{2\pi}} e^{-c^2 n/2}.$

$(2n+1) P(S_{2n+1} \le n) \sim \frac{2n}{c\sqrt{n}} \cdot \frac{1}{\sqrt{2\pi}} e^{-c^2 n/2} = \frac{2\sqrt{n}}{c\sqrt{2\pi}} e^{-c^2 n/2}.$ As $n \to \infty$ , the exponential decay $e^{-c^2 n/2}$ dominates the $\sqrt{n}$ growth, so $(2n+1) P(S_{2n+1} \le n) \to 0.$ Therefore $\lim_{n \to \infty} (2n+1)\sum_{k=0}^{n}\binom{2n+1}{k}p^{k}(1-p)^{2n+1-k} =0.$

Final answer

Marking scheme

The following is the marking rubric for this probability limit problem (total: 7 points).

1. Checkpoints (max 7 pts total)

Group 1: Probabilistic Model Conversion [additive]

Recognize that the summation $\sum \binom{2n+1}{k} p^k(1-p)^{2n+1-k}$ equals the binomial cumulative probability $P(S_{2n+1} \le n)$ (or an equivalent random variable representation). (1 pt)
*Note: If no random variable is introduced but subsequent calculations correctly follow the normal approximation formulas, credit may be awarded retroactively.*

Group 2: Core Analysis and Decay Estimate (Score exactly one chain)

*For different solution paths, score whichever chain below yields the highest total; do not mix chains.*

Path A: Normal Approximation and Asymptotic Analysis (standard solution)
Standardization parameters: Write or use the correct mean $\mu=(2n+1)p$ and variance $\sigma^2=(2n+1)p(1-p)$ , and attempt standardization $Z = \frac{S-\mu}{\sigma}$ . (1 pt)
Boundary asymptotic behavior: Analyze the standardized upper bound $x_n = \frac{n-\mu}{\sigma}$ , explicitly identifying its order as $-c\sqrt{n}$ (i.e., tending to negative infinity proportionally to $\sqrt{n}$ ). (2 pts)
*If only the algebraic expression is written without simplification or without identifying the $\sqrt{n}$ relationship, no credit for this checkpoint.*
Tail probability estimate: Invoke the Mills ratio or the normal distribution tail asymptotic formula ( $\Phi(x) \sim \frac{1}{|x|\sqrt{2\pi}}e^{-x^2/2}$ ), explicitly obtaining exponential decay of the probability term (e.g., $e^{-cn}$ or $e^{-c^2 n/2}$ ). (2 pts)
Path B: Large Deviations / Concentration Inequalities (Hoeffding/Chernoff Bound)
Inequality setup: Correctly set up the inequality parameters, identifying the deviation $\epsilon = p - 0.5 > 0$ (or noting that the distance between the expected mean and $n$ is linear $O(n)$ ). (2 pts)
Exponential upper bound: Apply the inequality to obtain an exponential upper bound of the form $P(S_{2n+1} \le n) \le C e^{-kn}$ . (3 pts)

Group 3: Limit Conclusion [additive]

Resolving the indeterminate form: Combine the linear growth of $(2n+1)$ with the exponential decay of the probability term to conclude the limit is 0. (1 pt)
*Requirement: Must demonstrate that "exponential decay dominates polynomial/linear growth." If the conclusion is drawn solely from the probability tending to 0 without comparing rates, no credit for this checkpoint.*

Total (max 7) check: 1 + 5 + 1 = 7.

2. Zero-credit items

Merely copying the problem statement or listing the binomial expansion formula without any concrete computation.
Stating only the answer "0" without any supporting work.
Incorrectly using Chebyshev's Inequality to prove the limit is 0:
Chebyshev's inequality only yields an $O(1/n)$ upper bound; multiplying by $(2n+1)$ gives a constant limit, which is insufficient to prove the limit is 0. If this path claims to complete the proof, it constitutes a logical error; award only the model conversion point (if applicable).
Assuming $p=0.5$ or substituting other specific numerical values.

3. Deductions

Logic gap cap (Cap at 2/7):
If the student uses the Central Limit Theorem (CLT) only to conclude $P(S \le n) \to \Phi(-\infty) = 0$ without analyzing the rate of convergence (i.e., without explaining why the $0 \cdot \infty$ indeterminate form resolves to 0 rather than something else), the total score shall not exceed 2 points (model point and parameter point only).
Computational/symbolic error (-1):
Coefficient errors in the mean or variance computation (e.g., omitting $2n+1$ ) that do not affect the core qualitative conclusion of "exponential decay."
Inequality direction error (-1):
In Path B, writing the inequality in the wrong direction (e.g., writing $P \ge \dots$ ) while the subsequent logic assumes the correct direction.
Maximum deduction principle: Errors within the same logical chain are not penalized repeatedly; the total score shall not fall below 0.

Probability Theory – Problem 70: Let