Limit Calculation Using Poisson and Normal Approximation

We need to evaluate the limit: $L = \lim_{n \to \infty} e^{-n} \left( 1 + n + \frac{n^2}{2!} + \dots + \frac{n^n}{n!} \right)$

Understanding the Expression

The expression inside the parenthesis is a partial sum of the Taylor series expansion for $e^n$: $e^n = \sum_{k=0}^{\infty} \frac{n^k}{k!} = 1 + n + \frac{n^2}{2!} + \dots$

Let $S_n = \sum_{k=0}^{n} \frac{n^k}{k!}$. The limit can be rewritten as: $L = \lim_{n \to \infty} e^{-n} S_n = \lim_{n \to \infty} \frac{\sum_{k=0}^{n} \frac{n^k}{k!}}{e^n}$

Relating to Poisson Distribution

Recall the probability mass function (PMF) of a Poisson distribution with parameter $\lambda$: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ for $k=0, 1, 2, \dots$

If $X$ follows a Poisson distribution with parameter $n$ (i.e., $X \sim \text{Poisson}(n)$), then the probability of observing $k$ events is $P(X=k) = \frac{e^{-n} n^k}{k!}$.

The cumulative distribution function (CDF) is $P(X \le n) = \sum_{k=0}^{n} P(X=k) = \sum_{k=0}^{n} \frac{e^{-n} n^k}{k!} = e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$.

Therefore, the limit we want to evaluate is: $L = \lim_{n \to \infty} P(X \le n)$, where $X \sim \text{Poisson}(n)$.

Applying Normal Approximation

For large $n$, the Poisson distribution $X \sim \text{Poisson}(n)$ can be approximated by a normal distribution $Y \sim N(\mu, \sigma^2)$ with mean $\mu = n$ and variance $\sigma^2 = n$.

We approximate $P(X \le n)$ using the normal distribution: $P(X \le n) \approx P(Y \le n)$

To evaluate $P(Y \le n)$, we standardize the variable $Y$ by defining $Z = \frac{Y - \mu}{\sigma} = \frac{Y - n}{\sqrt{n}}$. $Z$ follows the standard normal distribution $N(0, 1)$.

The condition $Y \le n$ translates to: $\frac{Y - n}{\sqrt{n}} \le \frac{n - n}{\sqrt{n}}$ $Z \le 0$

So, $P(Y \le n) \approx P(Z \le 0)$.

Final Limit Calculation

The probability $P(Z \le 0)$ is the value of the standard normal CDF, $\Phi(z)$, evaluated at $z=0$. $L = \lim_{n \to \infty} P(Z \le 0) = \Phi(0)$

Since the standard normal distribution is symmetric about 0, the probability of being less than or equal to 0 is exactly 0.5. $\Phi(0) = 0.5 = \frac{1}{2}$

Thus, the value of the limit is $\frac{1}{2}$.