 spockmonkey40

2022-07-07

Consider random variable $Y$ with a Poisson distribution:
$P\left(y|\theta \right)=\frac{{\theta }^{y}{e}^{-\theta }}{y!},y=0,1,2,\dots ,\theta >0$
$P\left(y|\theta \right)=\frac{{\theta }^{y}{e}^{-\theta }}{y!},y=0,1,2,\dots ,\theta >0$
Mean and variance of $Y$ given $\theta$ are both equal to $\theta$. Assume that $\sum _{i=1}^{n}{y}_{i}>1$.
If we impose the prior $p\propto \frac{1}{\theta }$, then what is the Bayesian posterior mode?
I was able to calculate the likelihood and the posterior, but I'm having trouble calculating the mode so I'm wondering if I got the right posterior:
$P\left(\theta |y\right)=likelihood\ast prior$
$P\left(\theta |y\right)\propto \left({\theta }^{\sum _{i=1}^{n}{y}_{i}}{e}^{-n\theta }\right)\left({\theta }^{-1}\right)$
$P\left(\theta |y\right)\propto {\theta }^{\left(\sum _{i=1}^{n}{y}_{i}\right)-1}{e}^{-n\theta }$ Marisol Morton

Expert

The posterior mode is just the maximizing value of the posterior, so this is essentially just a calculus problem. You have already correctly derived the posterior kernel:
$\pi \left(\theta |\mathbb{y}\right)\propto \mathrm{exp}\left(\left(n\overline{y}-1\right)\mathrm{ln}\theta -n\theta \right).$
So the log-posterior can be written as:
${F}_{\mathbb{y}}\left(\theta \right)\equiv \mathrm{ln}\pi \left(\theta |\mathbb{y}\right)=\left(n\overline{y}-1\right)\mathrm{ln}\theta -n\theta +\text{const}.$
We can maximise this via ordinary calculus techniques. Differentiating with respect to $\theta$ gives:
$\frac{d{F}_{\mathbb{y}}}{d\theta }\left(\theta \right)=\frac{n\overline{y}-1}{\theta }-n\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{{d}^{2}{F}_{\mathbb{y}}}{d{\theta }^{2}}\left(\theta \right)=-\frac{n\overline{y}-1}{{\theta }^{2}}.$
You are told that $n\overline{y}>1$ so the second derivative of the objective function is negative. This means that the objective function is strictly concave, so the maximizing value occurs at the unique critical point:
$0=\frac{d{F}_{\mathbb{y}}}{d\theta }\left(\stackrel{^}{\theta }\right)=\frac{n\overline{y}-1}{\stackrel{^}{\theta }}-n\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\stackrel{^}{\theta }=\overline{y}-\frac{1}{n}.$
So in this case we have mode (which is strictly positive since $\overline{y}>1/n$).

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