$$p(x) = \frac{1 }{\theta} . \frac{(ln\theta)^x}{x!} $$
I have no idea what type of distribution is this? And what will be its first and second moment?
$$p(x) = \frac{1 }{\theta} . \frac{(ln\theta)^x}{x!} $$
I have no idea what type of distribution is this? And what will be its first and second moment?
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Assuming that $x$ is restricted to the non-negative integers and $\theta>1$ (so that $\ln(\theta)>0$), this is just a Poisson distribution with parameter $\ln(\theta)$. Note that $e^{-\ln(\theta)}=\frac{1}{\theta}$.