I had a (truncated at zero) Poisson distribution with pmf $f(x;\theta)=\frac{e^{-\theta}\theta^{x}}{x!(1-e^{-\theta})}$ and $x=1, 2, ...$ and $\theta>0$ where $\theta$ is the parameter and $X$ is the r.v.
My lecturer asked how we'd find the ML estimate of $f(k;\theta)$ with fixed integer $k>0$, and I'm really not sure how you'd do that.
The likelihood function generally involves a product of pmfs for each random variable in a sample, whereas this has a random sample of $1$. So does that mean the likelihood function would look like $L(\theta)=f(k;\theta)=\frac{e^{-\theta}\theta^{k}}{k!(1-e^{-\theta})}$? And then you could find the derivative as $\frac{\partial}{\partial \theta}L(\theta)=\frac{1}{k!}\frac{\partial}{\partial \theta}(\frac{\theta^k e^{-\theta}}{1-e^{-\theta}})$. But, I can see that (after using a calculator) $\frac{1}{k!}\frac{\partial}{\partial \theta}(\frac{\theta^k e^{-\theta}}{1-e^{-\theta}})=0$ has no solution for $\theta$, so I must have gone wrong somewhere.
Any tips would be appreciated.
Most of the times it is easier to work with the log-likelihood function, so in this case that would be $l(k, \theta) = - \theta + k \log \theta - \log k! - \log(1 - e^{-\theta}).$ We differentiate $l$ with respect to $\theta$ and equate it to $0$ to obtain $$\frac{\partial l}{\partial \theta} (k, \theta) = -1 + \frac{k}{\theta} - \frac{e^{-\theta}}{1 - e^{- \theta}} = 0,$$ so $\frac{k}{\theta} = \frac{1}{1 - e^{-\theta}}$ or $\frac{\theta}{1 - e^{- \theta}} = k.$ A little analysis helps us conclude that for all $k$ there is a unique solution to this equation, call it $\theta_k.$ Moreover, the graph of the function $g(\theta) = \frac{\theta}{1 - e^{-\theta}}$ has the asymptote $y = x$ at $\infty,$ so $\theta_k$ is approximately $k$ if $k$ is large enough. I hope this helps. :)
Edit: Note that $\theta = k - k e^{-\theta}$ means that $\theta e^\theta = k e^\theta - k,$ so $e^{\theta}(\theta - k) = -k,$ from which we get $(\theta - k) e^{\theta - k} = - k e^{-k}.$ Then $\theta - k = W(-k e^{-k}),$ so $\theta_k = k + W(-k e^{-k}),$ where $W$ denotes the Lambert W-function.