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.

  • $\begingroup$ Please add the self-study tag and read its tag wiki,modifying your question as asked there (e.g. identify your specific problem). "I can't do it, show me how" is not adequate. $\endgroup$
    – Glen_b
    Apr 19, 2015 at 2:12
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Apr 19, 2015 at 2:50
  • 2
    $\begingroup$ I've updated my question again. $\endgroup$
    – Nicky_Ay
    Apr 24, 2015 at 4:06
  • $\begingroup$ $E(\bar X)\to\lambda$ and $V(\bar X)\to 0$ as $n\to\infty$ is a sufficient condition for $\bar X$ to be a consistent estimator of $\lambda$. $\endgroup$ Oct 29, 2018 at 7:09

1 Answer 1


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.

Start with definitions.

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.


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