Chapter 7: Problem 9
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0\) (a) Statistician \(A\) observes the sample to be the values \(x_{1}, x_{2}, \ldots, x_{n}\) with sum \(y=\sum x_{i} .\) Find the mle of \(\theta\). (b) Statistician \(B\) loses the sample values \(x_{1}, x_{2}, \ldots, x_{n}\) but remembers the sum \(y_{1}\) and the fact that the sample arose from a Poisson distribution. Thus \(B\) decides to create some fake observations which he calls \(z_{1}, z_{2}, \ldots, z_{n}\) (as he knows they will probably not equal the original \(x\) -values) as follows. He notes that the conditional probability of independent Poisson random variables \(Z_{1}, Z_{2}, \ldots, Z_{n}\) being equal to \(z_{1}, z_{2}, \ldots, z_{n}\), given \(\sum z_{i}=y_{1}\) is $$\frac{\frac{\theta^{x_{1}} e^{-\theta}}{z_{1} !} \frac{\theta^{x_{2}} e^{-\theta}}{z_{2} !} \cdots \frac{\theta^{x_{n}} e^{-\theta}}{x_{n} !}}{\frac{(n \theta)^{y_{1}} e^{n \theta}}{y_{1} !}}=\frac{y_{1} !}{z_{1} ! z_{2} ! \cdots z_{n} !}\left(\frac{1}{n}\right)^{z_{1}}\left(\frac{1}{n}\right)^{z_{2}} \cdots\left(\frac{1}{n}\right)^{z_{m}}$$ since \(Y_{1}=\sum Z_{i}\) has a Poisson distribution with mean \(n \theta .\) The latter distribution is multinomial with \(y_{1}\) independent trials, each terminating in one of \(n\) mutually exclusive and exhaustive ways, each of which has the same probability \(1 / n .\) Accordingly, \(B\) runs such a multinomial experiment \(y_{1}\) independent trials and obtains \(z_{1}, z_{2}, \ldots, z_{n} .\) Find the likelihood function using these \(z\) values. Is it proportional to that of statistician \(A ?\) Hint: Here the likelihood function is the product of this conditional pdf and the pdf of \(Y_{1}=\sum Z_{i}\)
Short Answer
Step by step solution
Key Concepts
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