Variance of MLE poisson distribution

Question

I am working on problems related to finding MLE from Mathematical Statistics with Applications, 7th Edition - Wackerly. Below is the exercise 9.80 that I'm a bit confused over. My concern is mostly regarding part B & C.

Suppose that $Y_1, Y_2,..., Y_n$ denote a random sample from the Poisson distribution with mean λ.

A) Find the MLE $\hatλ$ for $λ$.

B) Find the expected value and variance of $\hat λ$

C) Show that the estimator of part (a) is consistent for λ.

I have a table of discrete distributions that provides Probability function, mean and variance.

A) Given that we're working with a Poisson distribution, the estimator is the same as the sample mean. This yields $\hat λ = λ$.

B) $E(\hat λ) = λ$. So far so good. But for $Var(\hat λ)$; the variance for poisson distribution is $λ$. In the provided solution the answer to $Var(\hat λ) = λ/n$, why is this?

C) I don't know how to approach this.

Answer 1

From the beginning so it's easier to understand how everything falls together:

Given $Y \sim Poisson$; $p(y)= \frac{\lambda^y e^{-\lambda}}{y!}$

To find $E[Y]$ we need to take $\prod_{i = 1}^{n} p(y)$, which after some steps will lead us to our $\hat{\lambda}_{MLE}$.

Step 1 likelihood:

$p(y/\lambda) = \prod_{i = 1}^{n}\frac{\lambda^y e^{-\lambda}}{y!} = \frac{\lambda^{\prod_{i = 1}^{n} y_i} e^{-\lambda n}$}{\prod_{i = 1}^{n}y_i}$

Next we're taking logs, remember the following properties of logs:

• $log(a/b) = log(a)-log(b)$
• $log(ab) = log(a)+log(b)$
• $log(a^n) = nlog(a)$
• $log(e) = 1$

Step 2 logs:

$log(p(y/\lambda))=log(\lambda^{\sum_{i = 1}^{n}y_i})+log(e^{-\lambda n})-log(\prod_{i = 1}^{n}y_i) = \sum_{i = 1}^{n}y_i log(\lambda)-\lambda n$

Next we take the derivative and set it equal to zero to find the MLE. These properties of derivatives will often be handy in these problems:

• $\frac{d}{dx}log(x) = 1/x$
• $\frac{d}{dx} a log(x) = a/x$
• $\frac{d}{dx} a^n = n a^{n-1}$

Step 3 derivative (with respect to the parameter were interested in):

$\frac{d}{d\lambda}log(p(y/\lambda)) = \frac{\sum_{i = 1}^{n}y_i}\lambda -n = 0 => \frac{\sum_{i = 1}^{n}y_i}n = \lambda$

If we look at $\frac{\sum_{i = 1}^{n}y_i}n$ we can see that it is equal to ybar. This finally gives:

$\bar{Y}=\hat{\lambda}_{MLE}$

Great!

Now let's look at $E[\bar{Y}]$ and $V[\bar{Y}]$.

The following is important to note:

• $E[cY] = cE[Y]$
• $E[\sum_{i = 1}^{n}Y] = \sum_{i = 1}^{n}E[Y]$
• $V[cY] = c^2V[Y]$

First $E[\bar{Y}]$: $E[\bar{Y}] = E[\frac{\sum_{i = 1}^{n}y_i}n] = \frac{1}nE[\sum_{i = 1}^{n}y_i] = \frac{1}n n \lambda = \lambda$

And as $E[\hat{\lambda}] = \lambda$ we can conclude that it's unbiased.

Then $V[\bar{Y}]$: $V[\bar{Y}] = V[\frac{\sum_{i = 1}^{n} y_i}n] = \frac{1}{n^2} V[\sum_{i = 1}^{n} y_i] = \frac{1}{n^2} V[y_1 + y_2 + ... + y_n] = \frac{1}{n^2} \lambda n = \frac{\lambda}n$

To answer if it's consistent we can put it as following: $\lim_{x \to \infty}V[\hat{\lambda}_{MLE}]$, which given that we have $n$ in the denominator will make our expression $0$.

Answer 2

Hints:

• The random sampling assumption means that the r.v.s in your sample are i.i.d.
• Hence, they share the same distribution, hence each also have the same variance.
• By independence, the variance of a sum is just the sum of the variances. The estimator you derived is $\frac{1}{n}\sum_iY_i$. (Note that it is not correct to write $\hat\lambda=\lambda$ in A) - that says that the estimate is equal to the true unknown value. It would be great if we could claim that, but we generally can't!
• $Var(X/n)=1/n^2Var(X)$ for a constant $n$.
• As you note, the variance of a single Poisson random variable is $\lambda$.
• As to C, consider the law of large numbers.