Consider random variable $Y$ with a Poisson distribution:

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

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:

$P(\theta |y)=likelihood\ast prior$

$P(\theta |y)\propto ({\theta}^{\sum _{i=1}^{n}{y}_{i}}{e}^{-n\theta})({\theta}^{-1})$

$P(\theta |y)\propto {\theta}^{(\sum _{i=1}^{n}{y}_{i})-1}{e}^{-n\theta}$