Question

Let L = lim$_{n \to \infty \sum_{k=0}^{n} \frac{e^{-n} }n^k}{k!}$ value of L

• A. 1.0
• B. 0.5
• C. 0
• D. e$^{-1}$

Hint:

This question connects several important concepts: recognizing the PMF of a distribution within a sum, understanding that the sum represents a CDF, and applying the Central Limit Theorem (in this case, the Normal approximation to the Poisson distribution). For any distribution with a large parameter, its CDF evaluated at its mean will approach 0.5.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are asked to find the limit of a sum. We need to recognize the expression inside the limit.
Step 2: Key Formula or Approach:
The term $\frac{e^{-\lambda} \lambda^k}{k!}$ is the probability mass function (PMF) for a Poisson random variable with parameter $\lambda$.
The expression $\sum_{k=0}^{n} \frac{e^{-n} n^k}{k!}$ represents the probability $P(X_n \le n)$, where $X_n$ is a random variable following a Poisson distribution with parameter $\lambda = n$. For a large parameter $\lambda$, the Poisson distribution Poisson($\lambda$) can be approximated by a Normal distribution $N(\mu, \sigma^2)$ where the mean $\mu = \lambda$ and the variance $\sigma^2 = \lambda$.
Step 3: Detailed Explanation:
We need to evaluate: \[ L = \lim_{n \to \infty} P(X_n \le n) \quad \text{where } X_n \sim \text{Poisson}(n) \] As $n \to \infty$, we can use the Normal approximation for the Poisson distribution.
Here, $\lambda = n$, so the approximating Normal distribution has:
- Mean: $\mu = n$
- Variance: $\sigma^2 = n$
- Standard Deviation: $\sigma = \sqrt{n}$
We want to find the probability $P(X_n \le n)$. To use the Normal approximation, we standardize the variable $X_n$ to a standard Normal variable $Z \sim N(0, 1)$ using the formula $Z = \frac{X - \mu}{\sigma}$.
\[ P(X_n \le n) \approx P\left(\frac{X_n - n}{\sqrt{n}} \le \frac{n - n}{\sqrt{n}}\right) \] \[ P(X_n \le n) \approx P(Z \le 0) \] For a standard Normal distribution, the mean is 0. The distribution is symmetric about its mean. Therefore, the probability of a value being less than or equal to the mean is exactly half of the total probability.
\[ P(Z \le 0) = 0.5 \] As $n \to \infty$, this approximation becomes exact. \[ L = \lim_{n \to \infty} P(X_n \le n) = P(Z \le 0) = 0.5 \] Step 4: Final Answer:
The value of the limit L is 0.5.