GATE 2018 | MATHS | QI47

Question

Let \(\{X_i\}\) be a sequence of independent Poisson(\(\lambda\)) variables, and let \(W_n = \frac{1}{n} \sum_{i=1}^{n} X_i\). Then the limiting distribution of \(\sqrt{n}(W_n - \lambda)\) is the normal distribution with zero mean and variance given by

(A) \(1\)

(B) \(\sqrt{\lambda}\)

(C) \(\lambda\)

(D) \(\frac{\lambda}{2}\)

Answer 1

Central Limit Theorem:

The Central Limit Theorem (CLT) states that the sum or average of a large number of independent and identically distributed (i.i.d.) random variables, regardless of their underlying distribution, will tend to be normally distributed as the sample size increases.

Applying CLT to Poisson Distribution:

In this case, we have a sequence of i.i.d. Poisson(\(\lambda\)) variables.
The mean of a Poisson(\(\lambda\)) distribution is \(\lambda\), and the variance is also \(\lambda\).
We're considering the average of these variables, \(W_n = \frac{1}{n} \sum X_i\).
As \(n\) approaches infinity, the CLT tells us that the distribution of \(W_n\) will approach a normal distribution with mean \(\lambda\) and variance \(\frac{\lambda}{n}\).

Rescaling:

To get the limiting distribution of \(\sqrt{n}(W_n - \lambda)\), we rescale \(W_n\) by subtracting its mean \(\lambda\) and multiplying by \(\sqrt{n}\):
\[ \sqrt{n}(W_n - \lambda) = \sqrt{n} \left( \frac{1}{n} \sum X_i - \lambda \right) = \frac{1}{\sqrt{n}} \sum (X_i - \lambda) \]

This rescaling doesn't change the normality of the distribution, but it does affect the mean and variance:
\begin{align*}
\text{Mean:} & \quad \text{The mean of } \frac{1}{\sqrt{n}} \sum (X_i - \lambda) \text{ is } 0, \text{ as the mean of each } X_i - \lambda \text{ is } 0. \\
\text{Variance:} & \quad \text{The variance of } \frac{1}{\sqrt{n}} \sum (X_i - \lambda) \text{ becomes } \frac{\lambda}{n} \left( \frac{1}{n} \right) = \frac{\lambda}{n^2}.
\end{align*}

Applying \(\sqrt{n}\):

Finally, multiplying by \(\sqrt{n}\) changes the variance to \(\lambda/n^2 \times n = \lambda\):
\[ \text{Var}[\sqrt{n}(W_n - \lambda)] = \lambda \]