# Can you show that $\bar{X}$ is a consistent estimator for $\lambda$ using Tchebysheff's inequality?

This question was taken from a practice exam in my statistics course.

Given a random sample $X_1, X_2, ... X_n$ from a Poisson distribution with mean $\lambda$, can you show that $\bar{X}$ is consistent for $\lambda$?

We are told to use Tchebysheff's inequality. Which is: $Pr(|X-\mu| \geq k\sigma) \le \frac{1}{k^2}$

Also, as far as I know, consistency of an estimator is the property that as we increase the sample size of $\bar{X}$, our estimator should return values closer and closer to the actual value we want to estimate.

So the first thing I did was find the variance for $\bar{X}$ as follows: $Var(\bar{X})=Var(\frac{\sum(X_i)}{n})=\frac{1}{n^2}Var(\sum(X_i))=\frac{\lambda}{n}$

I notice that as $n \rightarrow \infty$ the variance decreases to $0$, but how does this help me?

Now I guess we can use Tchebysheff's inequality where we need $Pr(|\bar{X}-\lambda| \geq \epsilon) = 0$ and that is where I get stuck...

Any help is appreciated.

• You haven't yet dealt with what consistency is. See steps 1 and 2 below - you haven't mentioned what it is you need to show to demonstrate consistency. (You also didn't write down the general form of Chebyshev - i.e. step 3 - before substituting details from this specific problem into it, so if you made a mistake there you would make it difficult for people to point out where you went wrong.) Please do steps 1, 2 and 3 properly, so that hints that relate parts of what you write can be offered. It appears such skipping over of essential steps is leading you into difficulty in the first place. Apr 19 '15 at 2:50
• I've updated my question again. Apr 24 '15 at 4:06
• $E(\bar X)\to\lambda$ and $V(\bar X)\to 0$ as $n\to\infty$ is a sufficient condition for $\bar X$ to be a consistent estimator of $\lambda$. Oct 29 '18 at 7:09

Edit: Since it seems the point didn't get across, I'm going to fill in a few more details; it's been a while, so maybe I can venture a little more.

step 1: give a definition of consistency

Like this one from wikipedia's Consistent estimator article:

Suppose $${p_θ: θ ∈ Θ}$$ is a family of distributions (the parametric model), and $$X^θ = {X_1, X_2, \ldots : X_i ~ p_θ}$$ is an infinite sample from the distribution $$p_θ$$. Let $${ T_n(X^θ) }$$ be a sequence of estimators for some parameter $$g(θ)$$. Usually $$T_n$$ will be based on the first $$n$$ observations of a sample. Then this sequence $${T_n}$$ is said to be (weakly) consistent if

$$\underset{n\to\infty}{\operatorname{plim}}\;T_n(X^{\theta}) = g(\theta),\ \ \text{for all}\ \theta\in\Theta$$

step 2: Note (hopefully!) that it relies on convergence in probability, so give a definition for that in turn (wikipedia article on Convergence of random variables).

A sequence $${X_n}$$ of random variables converges in probability towards the random variable $$X$$ if for all $$ε > 0$$

$$\lim_{n\to\infty}\Pr\big(|X_n-X| \geq \varepsilon\big) = 0.$$

step 3: Then write Chebyshev's inequality down:

Let $$X$$ (integrable) be a random variable with finite expected value μ and finite non-zero variance σ2. Then for any real number $$k > 0$$,

$$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.$$

(wikipedia article on Chebyshev's inequality)

step 4: now look at the rather strong similarity between two expressions in (2) and (3).
Does that not give you a huge clue about a way to approach this?

So let's start

From Chebyshev:

$$Pr(|\bar{X}-\lambda| \geq k\lambda/n) \le \frac{1}{k^2}$$ (you worked out all the parts of this but never actually wrote it down in your question. I can't fathom why you wouldn't)

Let $$\epsilon=k\lambda/n$$ (that's the obvious step from the "huge clue" you were supposed to see by comparing the two things I said to write down ... one had an $$\epsilon$$ where the other had a $$k\lambda/n$$ ... if you'd written them both down, as I'd been suggesting, I expect it would have been obvious as soon as you compared them).

So now, the only thing that's really left is to show that for any $$\epsilon$$, as $$n\to\infty$$, the RHS goes to 0.