Let . Using a probabilistic method, find: .

Probability Theory Solution – Problem 50: find:

Question

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

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 very 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 $q < 0.5$ .

$n - (2n+1)p = n - (2n+1)p = n - 2np - p = - (2p-1)n - p.$ Since $2p-1 > 0$ , for large $n$ this value 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 $\frac{n - \mu}{\sigma} \approx -\frac{\delta n}{\sqrt{2n pq}} = -\frac{\delta}{\sqrt{2pq}} \cdot \sqrt{n}.$ So it tends to $-\infty$ like $-c\sqrt{n}$ , 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 grading rubric for this probability limit problem (maximum 7 points).

1. Key Checkpoints (max 7 pts total)

Group 1: Probabilistic Model Conversion [additive]

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

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

*For different solution paths, select one of the following for grading; if a mixed approach is used, take the highest-scoring path.*

Path A: Normal Approximation and Asymptotic Analysis (Official Solution)
Standardization parameters: Writing or using the correct mean $\mu=(2n+1)p$ and variance $\sigma^2=(2n+1)p(1-p)$ , and attempting to standardize $Z = \frac{S-\mu}{\sigma}$ . (1 pt)
Boundary asymptotic behavior: Analyzing the standardized upper bound $x_n = \frac{n-\mu}{\sigma}$ and explicitly stating that its order of magnitude as $n\to\infty$ is $-c\sqrt{n}$ (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 item.*
Tail probability estimate: Citing the Mills Ratio or the normal distribution tail asymptotic formula ( $\Phi(x) \sim \frac{1}{|x|\sqrt{2\pi}}e^{-x^2/2}$ ), and explicitly obtaining that the probability term exhibits exponential decay (e.g., $e^{-cn}$ or $e^{-c^2 n/2}$ ). (2 pts)
Path B: Large Deviations / Concentration Inequalities (Hoeffding/Chernoff Bound)
Inequality setup: Correctly setting the inequality parameters, identifying the deviation quantity $\epsilon = p - 0.5 > 0$ (or noting that the distance between the expected mean and $n$ is linear $O(n)$ ). (2 pts)
Exponential upper bound establishment: Applying 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: Combining the linear growth of the prefactor $(2n+1)$ with the exponential decay of the probability term to conclude that the limit is 0. (1 pt)
*Requirement: Must demonstrate the logic that "exponential decay dominates polynomial/linear growth." Concluding merely from the probability tending to 0 without comparing rates earns no credit for this item.*

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

2. Zero-credit items

Merely copying the problem or listing the binomial expansion formula without any specific computation.
Only giving the answer "0" without any process.
Incorrectly using Chebyshev's inequality to prove the limit is 0: Chebyshev's inequality can only give an $O(1/n)$ upper bound, which when multiplied by $(2n+1)$ yields a constant limit, and cannot prove the limit is 0. If this path claims to have proved the result, it is treated as a logical error and only the model conversion point (if any) is awarded.
Assuming $p=0.5$ or other specific numerical values for the computation.

3. Deductions

Logical gap cap (Cap at 2/7):
If the student uses the CLT only to obtain $P(S \le n) \to \Phi(-\infty) = 0$ without analyzing the rate of convergence (i.e., without explaining why the $0 \cdot \infty$ indeterminate form yields 0 rather than something else), the total score cannot exceed 2 points (only the model and parameter points).
Computation/notation error (-1):
Coefficient errors in the mean or variance computation (e.g., omitting $2n+1$ ), provided they 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 $P \ge \dots$ ), but the subsequent logic assumes the correct direction.
Maximum deduction principle: Errors within the same logical chain are not penalized repeatedly; the total score after deductions cannot be less than 0.

Probability Theory – Problem 50: find: