Prove that

Unbaised estimator of $\ e^{- 2 \lambda } $ is t(x) = $ \ (-1)^x $ ,

when x follows poisson distribution with parameter $\lambda $

Sorry to ask the direct question but I need a starting hint to solve the question , actually last time I studied statistical inference was 1 year back , I would be great if anyone could just give some starting steps .

  • 4
    $\begingroup$ This is likely to be a duplicate. The interesting point is not so much that this absurd estimator is unbiased (it gives the wrong sign for odd $x$ and is too high for even $x$) but that it is the only unbiased estimator. $\endgroup$
    – Henry
    Commented Jul 8, 2021 at 9:24

3 Answers 3


I'll provide a step by step answer. In case you want to try it yourself, please stop at each point and try from there.

  1. Define what you are looking for. What is an unbiased estimator?

An unbiased estimator, $t(x)$, of a quantity $\theta$ (which in this case is $\theta = e^{-2\lambda}$), is such that: $$ \mathbb{E}t(x) = \theta = e^{-2\lambda} $$

  1. Plug-in information available to start solving the problem

we know $t(x)=(-1)^{x}$ and $x\sim \text{Poisson}(\lambda)$. Then we want to compute the expected value of this quantity: $$ \mathbb{E}\left[(-1)^x\right]. $$ Recall that expected value (for a discrete random variable) is obtained by summing all possible values the random variable can take multiplied by their probability. Recall further that if $x\sim \text{Poisson}(\lambda)$ then $P(x=i)=\frac{\lambda^{i}e^{-\lambda}}{i!}$. Thus, just using definitions: $$ $$ $$ \mathbb{E}\left[(-1)^x\right] = \sum_{i=0}^{\infty} (-1)^{i}P(x=i) $$

  1. Solve the above.

Using the Poisson law, we have $$ \sum_{i=0}^{\infty} (-1)^{i}P(x=i) = \sum_{i=0}^{\infty} (-1)^{i}\frac{\lambda^{i}e^{-\lambda}}{i!}=e^{-\lambda}-\lambda e^{-\lambda}+\lambda^{2} \frac{e^{-\lambda}}{2!}-\lambda^{3} \frac{e^{-\lambda}}{3!}+\ldots$$

  1. What is the right-hand side term equal to?

Note that we can further simplify by taking common factor $e^{-\lambda}$ $$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right) $$ Recall Taylor series? You can use them to rewrite $e^{-\lambda}$ around $0$ (which is actually the definition of exponential): $$ e^{-\lambda}= e^{-0} +\frac{\partial e^{-\lambda}}{\partial\lambda}|_{\lambda=0}\frac{(\lambda-0)}{1!}+\frac{\partial^{2} e^{-\lambda}}{\partial\lambda^{2}}|_{\lambda=0}\frac{(\lambda-0)^{2}}{2!} \ldots$$ solving the derivatives, you see that the above is: $$ e^{-\lambda}= 1 -\lambda+\frac{\lambda^{2}}{2!}-\frac{\lambda^{3}}{3!} \ldots$$ So, we use this fact to rewrite: $$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda}. $$

Remember where we started from? $$ \mathbb{E}t(x) = e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda} $$ Done!


If you want to show it's unbiased then you need to show it has the correct expectation. Then your hint is that you're going to want to think about the probability generating function of a Poisson. Hopefully that's enough.

  • $\begingroup$ @dan_phillips i answered the question according to your hint , but I had one more doubt , the second part of the same question asks the suitability of the estimator , so shall I write the advantages of unbaised estimator and its limitations ?? , if you could tell $\endgroup$
    – simran
    Commented Jul 8, 2021 at 9:25
  • 1
    $\begingroup$ I think what they're getting at is basically Henry's comment. The estimator can only be 1 or -1. It has the wrong sign much of the time as $exp(-2\lambda)$ is positive. So it is (as Henry notes) a rather absurd estimator, and you'd be silly to use it. $\endgroup$ Commented Jul 8, 2021 at 15:31

E $\ (-1)^{x} $ = $ \frac {\sum_{x=0}^{ \infty} (-1)^{x} \ e^{- \lambda } \lambda^{x} }{\ x! } $

= $ e^{- \lambda } $ $ \frac {\sum_{x=0}^{ \infty } (- \lambda )^{x}}{ x !} $ { this is the property of exponential function }

= $ e^{- \lambda } $ $ e^{- \lambda } $

= $ e^{- 2\lambda } $


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