Question

Using the central limit theorem for suitable Poisson random variables, prove that $$ \lim _{n \rightarrow \infty} e^{-n} \sum_{k=0}^{n} \frac{n^{k}}{k !}=\frac{1}{2} $$

Short answer

In summary, given a sequence of i.i.d Poisson random variables \(X_k\) with parameter \(\lambda = 1\), we defined \(Y_n = \sum_{k=1}^{n} X_k\). Using the Central Limit Theorem and standard normal distribution, we derived the probability \(P(Y_n \leq n)\) and took the limit as \(n \rightarrow \infty\), yielding \(\frac{1}{2}\). This result proves the given expression \(\lim _{n \rightarrow \infty} e^{-n} \sum_{k=0}^{n} \frac{n^{k}}{k!}=\frac{1}{2}\).

Step-by-step solution

Define the Poisson Random Variables

Let \(X_k\) be a sequence of i.i.d Poisson random variables, each with parameter \(\lambda = 1\). The probability mass function for each \(X_k\) can be defined as: $$ P(X_k = n) = \frac{e^{-1}(1)^n}{n!} $$ Now let's define another random variable \(Y_n\) which represents the sum of these \(n\) random variables. $$ Y_n = \sum_{k=1}^{n} X_k $$ The expected value of \(Y_n\) is \(E[Y_n] = n\lambda = n\).

Apply the Central Limit Theorem

By the Central Limit Theorem, as \(n \rightarrow \infty\), the standardized sum \((Y_n - E[Y_n])/ \sqrt{n}\) will converge in distribution to a standard normal distribution, \(Z \sim N(0, 1)\). So, we can write: $$ \lim_{n \rightarrow \infty} P\left(\frac{Y_n - E[Y_n]}{\sqrt{n}} \leq z \right) = \Phi(z) $$ where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.

Calculate the Probability

Now, let's calculate the probability of interest, \(P(Y_n \leq n)\): $$ P(Y_n \leq n) = P\left( Y_n - E[Y_n] \leq 0 \right) $$ Now, we standardize the sum to apply the CLT: $$ P\left( -\frac{- Y_n + E[Y_n]}{\sqrt{n}} \leq \frac{ E[Y_n] - Y_n }{\sqrt{n}} \right) = \Phi\left( - \frac{- Y_n + E[Y_n]}{\sqrt{n}} \right) $$ Take the limit as \(n \rightarrow \infty\): $$ \lim_{n \rightarrow \infty} P(Y_n \leq n) = \lim_{n \rightarrow \infty} \Phi\left( \frac{E[Y_n]-Y_n}{\sqrt{n}} \right) $$ By the symmetry of the standard normal distribution, we have: $$ \lim_{n \rightarrow \infty} P(Y_n \leq n) = \lim_{n \rightarrow \infty} \Phi\left( 0 \right) = \frac{1}{2} $$

Establish the Given Expression

Finally, we need to prove that the given expression is equal to the result we just derived: $$ \lim _{n \rightarrow \infty} e^{-n} \sum_{k=0}^{n} \frac{n^{k}}{k!}=\frac{1}{2} $$ Recall that \(Y_n\) is the sum of \(n\) independent Poisson random variables, each with parameter \(\lambda = 1\). Therefore, \(Y_n\) follows a Poisson distribution with parameter \(n\), and its probability mass function is: $$ P(Y_n=k) = \frac{e^{-n}n^k}{k!} $$ So, we can rewrite the given expression as: $$ \lim _{n \rightarrow \infty} P(Y_n \leq n)=\frac{1}{2} $$ This is the same expression we arrived at in step 3, proving the given expression.