Let $X_i \sim \text{IID Poisson}(\theta)$ for $i=1,2,...,n.$ The MLE of $\theta$ is written as


Suppose $\theta$ has a probability density function (pdf) $p(\theta) \propto \theta^{-1/2}$, which is Jeffrey's prior. The resulting posterior distribution is $\theta|\mathbf{x} \sim \text{Gamma}(\sum_{i=1}^{n} x_i +\frac{1}{2},\frac{1}{n})$. Under quadratic loss, it can be shown that the (minimum risk) Bayes estimator is

$$\hat{\theta}_\text{Bayes} = \bar{x}+\frac{1}{2n}.$$

Now, knowing the Bayes estimator in this scenario, if I were to compare the performances of the MLE and the Bayes estimator in terms of bias, can I simply simulate data from Poisson with parameter $\theta_0$ and estimate using $\hat{\theta}_\text{MLE}$ and $\hat{\theta}_\text{Bayes}$ directly? Then, repeat for say, $1000$ iterations and check the means of the estimated values of $\hat{\theta}$? Alternatively, should I use some sort of approximation techniques such as MCMC?

  • 2
    $\begingroup$ You know already that the sample mean is an unbiased estimator of the population mean, which in this case is $\theta$; therefore, the bias of the Bayes estimate is $1/2n$. No need to simulate or approximate! $\endgroup$
    – jbowman
    Commented Jul 31, 2022 at 19:29
  • $\begingroup$ I see. If the bias of MLE and Bayes estimator could not be proven theoretically, simulation is the way to go, right? $\endgroup$
    – RRMT
    Commented Aug 2, 2022 at 3:03

1 Answer 1


You can certainly use simulation to examine the performance of estimators, but the difficulty is that you would need to choose one or more values of $\theta_0$, and the performance of the estimators might (in principle) depend on this choice. In the present case you can get explicit distributional results for the estimators without simulation, so that is preferable. Specifically, you have the distributions:

$$\hat{\theta}_\text{MLE} \sim \frac{1}{n} \text{Pois}(n \theta) \quad \quad \quad \quad \quad \hat{\theta}_\text{Bayes} \sim \frac{1}{n} \bigg[ \text{Pois}(n \theta) + \frac{1}{2} \bigg].$$

Consequently, the biases of the two estimators are:

$$\text{Bias}(\hat{\theta}_\text{MLE}, \theta) = 0 \quad \quad \quad \quad \quad \text{Bias}(\hat{\theta}_\text{Bayes}, \theta) = \frac{1}{2n}.$$


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