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Let $X$ be a single observation from truncated Poisson distribution having probability mass function $P(X = x) =\frac{e^{-\theta} \theta^{x}}{x!(1-e^{-\theta})} ; x = 1,2,3,$ . The estimator $T = \left\{\begin{matrix} 2, x = 1,3,5...\\ 0, x = 2,4,6... \end{matrix}\right.$ $ \ \ \ \ \ \ $ is unbiased for :

So I a have been preparing for an examination and this question popped up. I do not even know where/how to begin.

EDIT: Taking @mlofton's advice I proceeded as follows:

$$E(T) = 2 * \left [\frac{e^{-\theta} \ \theta}{1!(1-e^{-\theta})} \ + \frac{e^{-\theta} \ \theta^{3}}{3!(1-e^{-\theta})} \ + \frac{e^{-\theta} \ \theta^{5}}{5!(1-e^{-\theta})} ..... \right ]$$

Using the Taylor series expansion of $e$ we can simplify the above as:

$$E(T) = \frac{ e^{-\theta} }{(1-e^{-\theta})} * \left [\frac{2\theta}{1!} \ + \frac{ 2\theta^{3}}{3!} \ + \frac{2\theta^{5}}{5!} ..... \right ] = \frac{ e^{-\theta}(e^{\theta} - e^{-\theta}) }{(1-e^{-\theta})} $$

$$= \frac{ (1 - e^{-2\theta}) }{(1-e^{-\theta})}$$

Is my answer correct?

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    $\begingroup$ This isn't a complete answer but an estimator is unbiased for its expectation so you need to calculate the expectation of your estimator $T$.. $\endgroup$
    – mlofton
    Commented Sep 10, 2021 at 12:47
  • $\begingroup$ @mlofton Okay thank you. I believe I got it now $\endgroup$
    – Kalvin
    Commented Sep 10, 2021 at 13:02
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    $\begingroup$ This seems correct to me, you can even simplify the last line into $1+e^{-\theta}$ $\endgroup$
    – periwinkle
    Commented Sep 10, 2021 at 13:49
  • $\begingroup$ @periwinkle Yes. Thank you $\endgroup$
    – Kalvin
    Commented Sep 10, 2021 at 13:52

1 Answer 1

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Your answer looks correct to me, but it can be further simplified. You should have:

$$\begin{align} \mathbb{E}(T) &= 2 \Bigg[ p_X(1) + p_X(3) + p_X(5) + \cdots \Bigg] \\[6pt] &= 2 \Bigg[ \frac{e^{-\theta} \theta}{1! (1-e^{-\theta})} + \frac{e^{-\theta} \theta^3}{3! (1-e^{-\theta})} + \frac{e^{-\theta} \theta^5}{5! (1-e^{-\theta})} + \cdots \Bigg] \\[6pt] &= \frac{e^{-\theta}}{1-e^{-\theta}} \Bigg[ \frac{2 \theta}{1!} + \frac{2 \theta^3}{3!} + \frac{2 \theta^5}{5!} + \cdots \Bigg] \\[8pt] &= \frac{e^{-\theta} (e^{\theta} - e^{-\theta})}{1-e^{-\theta}} \\[8pt] &= \frac{1 - e^{-2\theta}}{1-e^{-\theta}} \\[8pt] &= \frac{(1 + e^{-\theta})(1 - e^{-\theta})}{1-e^{-\theta}} \\[12pt] &= 1 + e^{-\theta}. \\[6pt] \end{align}$$

Your original question got cut off, so it's not clear what unbiasedness property the question is asserting. In any case, you can see that the statistic $T$ is an unbiased estimator for $1+e^{-\theta}$ and biased for other functions of $\theta$.

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