Suppose that a random sample $X_1, ..., X_n$ is taken from a distribution where $X$ is a random variable that follows a Poisson distribution with mean $w$. Furthermore, $w$ is modeled with a random variable $W$ that follows a chi-square distribution with $r = 4$ degrees of freedom. (i) Determine the posterior distribution of $W$ given that $X_1 = x_1, X_2 = x_2, ..., X_n = x_n$ (ii) Find the Bayesian estimator for $W$ that minimizes the mean-square error (iii) Before the sample was taken, Jack tells you that $w$ lies in $(0, 1)$, however, no further information is provided about which values $w$ would most likely take in this interval. Do you pick a new prior pdf for $W$? Justify your answer, stating what new prior pdf you would take if so, or if not, why you would not pick a new prior pdf.
Hint: Integration is unnecessary. The posterior distribution should be familiar if you use proportionality! You should use this posterior distribution to find the mean-square error.
My attempt:
(i) From chi-square with $r = 4$, the prior pdf of $W$ is
$$f_W(w) = \frac{we^{-w/2}}{4}$$
The pdf of a Poisson with mean $w$ is:
$$f(x|w) = \frac{w^{x}e^{-w}}{x!}$$
The joint pdf of $X_1, X_2, ..., X_n$ is:
$$F(x_1, ..., x_n|w) = \frac{w^{\sum{x_i}}e^{-nw}}{\prod{x_i!}}$$
The posterior probability density function (pdf) of $W$ given $X_1 = x_1, X_2 = x_2, ..., X_n = x_n$ is:
$$f(w|x) = \frac{F(x_1, ..., x_n|w) \cdot f_W(w)}{f_X(x)} = \frac{F(x_1, ..., x_n|w) \cdot f_W(w)}{\int_{-\infty}^{\infty} F(x_1, ..., x_n|w) \cdot f_W(w) \,dw}$$
We apply proportionality:
$$\frac{F(x_1, ..., x_n|w) \cdot f_W(w)}{\int_{-\infty}^{\infty} F(x_1, ..., x_n|w) \cdot f_W(w) \,dw} \propto F(x_1, ..., x_n|w) \cdot f_W(w) = \frac{w^{1+\sum{x_i}}e^{-nw-\frac{w}{2}}}{4\prod{x_i!}}$$
From here, there are a few possible posteriors to choose from: Chi-square, Exponential or Gamma.
By mere inspection:
If Chi-square, we have parameter $r = 4 + 2\sum{x_i}$.
If Exponential, we have parameter $\theta = \frac{-2}{w(2n+1)}$ or $\theta = \frac{1}{w^{1+\sum{x_i}}}$.
If Gamma, we have parameters $\alpha = 2 + \sum{x_i}$ and $\theta = \frac{2}{2n+1}$.
I am unsure which one is the correct posterior distribution as they all seem likely.
(ii) Depends on (i) but am I supposed to use the fact that $E(W|X)$ is the least mean square here?
(iii) Not sure.
Any assistance especially with (i) and (iii) is much appreciated.
For (i), you must be careful of the assumptions made in the definitions of the density functions.
For the exponential distribution, we have the density function $$f_{X}(x) = \dfrac{1}{\theta} e^{- x/\theta}\text{, }\qquad x > 0$$ with $\theta > 0$ a constant - i.e., it cannot depend on $x$.
For the Gamma distribution, we have the density function $$f_{X}(x) = \dfrac{1}{\Gamma(\alpha)\theta^\alpha}x^{\alpha - 1}e^{-x/\theta}\text{, }\qquad x > 0$$ with $\alpha > 0$ and $\theta > 0$ constants - again, not depending on $x$.
The Chi-squared distribution with $\nu$ degrees of freedom is just the Gamma distribution with $\alpha = \nu/2$ and $\theta = 2$.
As far as I can tell, your computation of the posterior PDF is correct: we know that $$f(w \mid x) \propto \frac{w^{1+\sum{x_i}}e^{-nw-\frac{w}{2}}}{4\prod{x_i!}}$$ but we can take this one step further: note that $w$ is the variable of interest, so we can clean this up to end up with $$f(w \mid x) \propto \frac{w^{1+\sum{x_i}}e^{-nw-\frac{w}{2}}}{4\prod{x_i!}} \propto w^{1+\sum{x_i}}e^{-nw-\frac{w}{2}}\text{.}$$ Now, we observe that this is indeed similar to the exponential, Gamma, and Chi-squared distributions. However, for all three of these, we must separate the variable of interest in the exponent in the numerator - i.e., this suggests $$f(w \mid x) \propto w^{1+\sum{x_i}}e^{-w\left(n+\frac{1}{2}\right)}\text{.}$$ This has a $w$ term in it, which would not be the case for the exponential distribution (to match the notation of the exponential PDF above, replace $w$ with $x$). It is also not Chi-squared, since the factor of the exponent of $e$ should be $1/2$. So it's probably Gamma, and we can see this by writing $$-w\left(n + \dfrac{1}{2}\right) = -\dfrac{w}{\left(n + \dfrac{1}{2}\right)^{-1}}$$ which yields $\theta = \left(n + \dfrac{1}{2}\right)^{-1} > 0$ and $\alpha = 1 + \sum x_i + 1 = 2 + \sum x_i > 0$, thus we have a Gamma PDF.
Your thought for (ii) is correct; given (i), this should be easy to find since you know the posterior distribution.
I don't have enough experience with Bayesian model fitting to answer (iii).