Suppose \(X_{1}\) ...... \(X_{n}\) are independent observations from a Poisson distribution with probability mass function :
\[ \begin{aligned} f(x/\lambda) &= \frac{e^{-\lambda}\lambda^{x}}{x!}, \\ x &= 0, 1, 2, ......., \end{aligned} \]
\(\lambda \gt 0\) unkonwn.
Suppose the prior distribution of \(\lambda\) is desired as :
\[ g(\lambda) \propto \lambda,\ e^{-2\lambda},\ \lambda \gt 0. \]
Let \(n = 3\) and the data \((X_{1}, X_{2}, X_{3}) = (1, 1, 2)\). Then, suppose we estimate \(\lambda\) under squared error loss \(L(\lambda, a) = (\lambda - a)^{2}\). Then, the Bayes estimator based on the given data is :
MH SET - 2018, SET Exam - 2018,
Answer: D
The Bayes estimate is 1.2.
The prior, read as \(g(\lambda) \propto \lambda e^{-2\lambda}\) (the comma in the stem stands for multiplication), is a gamma density with shape 2 and rate 2. The Poisson likelihood of the data is proportional to \(e^{-3\lambda}\lambda^{1 + 1 + 2}\). Multiplying:
\[ \pi(\lambda|x) \propto \lambda^{5}e^{-5\lambda} \]
a gamma posterior with shape 6 and rate 5. Under squared error loss the Bayes estimator is the posterior mean, \(6/5 = 1.2\).
None of the other values is the posterior mean. Ignoring the prior would give the sample mean 4/3, and the posterior mode, \((6 - 1)/5 = 1\), is not offered either; the prior pulls the estimate below the sample mean because its own mean is 1.
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