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I am starting studying statistics and I have a task to find unestimated bias for Poisson distribution.
The task sounds like this: "Find unbiased estimator $\lambda_{s} = \lambda^{3}$ using sample $X_{1}...X_{n}$ from $Pois(\lambda)$ distribution".

I know that $\lambda$ is variance in Poisson distribution and that variance unbiased estimator is $$\bar{x}=\frac{1}{N}\sum_{i=1}^{N}X_{i}$$ But I see I need an answer correlated with $\lambda_{s}$, and I don't know how to use Poisson distribution formula to get my $X_{1}...X_{n}$ and then unbiased estimator.

Do you have any ideas?

P.S. Sorry, right problem statement is with UNbiased estimator...

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  • $\begingroup$ I'm not sure I'm following the question. If you have a sample from some distribution then you can estimate the variance by considering $N^{-1} \sum_i (X_i - \overline{X})^2$ where $\overline{X}$ is the average of your observations. This is typically biased in the sense that the expected value of this is not quite the variance of the original distribution. (To see this in action, consider the degenerate case $N = 1$.) Is that what you mean? $\endgroup$ Commented Jan 26, 2021 at 13:39
  • $\begingroup$ @JeroenvanderMeer, basically I also don't understand the question, that's all that I have in original task. Maybe I need to notice, that $\lambda_{s}$ could be $\lambda_{s} = \lambda^{2}$, for example. The sample is $X_{1}...X_{n}$ and I only know that these observations are from Poisson distribution, so I also don't understand how to use these conditions. The main question is, that role does specific $\lambda$ play. $\endgroup$
    – Kekov Kek
    Commented Jan 26, 2021 at 14:03
  • $\begingroup$ I'm sorry but I'm still not quite following. Could you quote the original problem verbatim perhaps? $\endgroup$ Commented Jan 26, 2021 at 16:00
  • $\begingroup$ @JeroenvanderMeer, I did it! Unfortunately that's all the information I have for this task :( $\endgroup$
    – Kekov Kek
    Commented Jan 26, 2021 at 18:21
  • $\begingroup$ bias(parameter, estimator)=var(estimator)^2 + (E(estimator)-parameter)^2 is the bias-variance decomposition. maybe this has something to do with it $\endgroup$
    – Vons
    Commented Jan 26, 2021 at 18:25

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First of all, notice that if $X\sim \text{Poisson}(\lambda)$ then $$E(X)=\lambda$$ $$E(X^2)=\lambda^2 + \lambda$$ $$E(X^3)=\lambda^3 +3\lambda^2 +\lambda$$ This implies $$E(X^3-3X^2+2X)=\lambda^3$$ That being said, define $$Y_{N}=\frac{1}{N}\sum_{i=1}^{N}\Big[X_i^3-3X_i^2+2X_i\Big]$$ Using linearity of expectation we have $E(Y_{N})=\lambda^3$ so $Y_{N}$ is an unbiased estimator for $\lambda^3$.

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