Consider truncated pdf $y=1,2,3...$ $$f(y,\theta)=\frac{\theta^y e^{-\theta}}{y! (1-e^{-\theta})}$$
I'm required to find expected information (likelihood), but in short I need to find $E(\bar y)=E(\sum y_i /n)$
According to my notes, the answer should be $$E(\bar y)=\frac {\theta}{(1-e^{-\theta})}$$
Generally, to find the mean I use $\int_{-\infty}^{\infty} x p(x) dx$ , but in this case I have 2 parameters.
Thanks in advance
On
Let $\boldsymbol y = (y_1, \ldots, y_n)$ be an IID sample drawn from a zero-truncated Poisson with unknown parameter $\theta > 0$. Then the likelihood given the sample is $$\mathcal L(\theta \mid \boldsymbol y) = \prod_{i=1}^n \frac{e^{-\theta}}{1-e^{-\theta}} \frac{\theta^{y_i}}{y_i!} \propto (e^{\theta} - 1)^{-n} \theta^{n \bar y},$$ where $\bar y$ is the sample mean, which we see is a sufficient statistic for $\theta$. Consequently, the log-likelihood is $$\ell(\theta \mid \boldsymbol y) = -n \log(e^{\theta} - 1) + n\bar y \log \theta,$$ and its derivative with respect to $\theta$ is $$\frac{\partial \ell}{\partial \theta} = \frac{e^\theta}{1-e^{\theta}} + \frac{\bar y}{\theta}.$$ Thus the MLE $\hat \theta$ satisfies $\partial \ell/\partial\theta = 0$, but this does not have a closed form expression: the best we can do is $$\hat \theta = \bar y + W(-\bar y e^{-\bar y}),$$ where $W(z)$ is a function that satisfies $z = W(z) e^{W(z)}$ and is denoted in Mathematica by ProductLog[]. A plot of $\hat \theta$ as a function of $\bar y$ is given below:
If $Y_1,\dots,Y_n$ are independent random variables having the given distribution (for some fixed $\theta>0$) and $\overline{Y}=\frac{1}{n}\sum_{k=1}^nY_k$, then $$ \mathbb{E}[\overline{Y}]=\frac{1}{n}\mathbb{E}[Y_1+\dots+Y_n]=\mathbb{E}[Y_1]=\frac{e^{-\theta}}{1-e^{-\theta}}\sum_{y=1}^{\infty}\frac{y\theta^y}{y!} $$ $$ =\frac{e^{-\theta}}{1-e^{-\theta}}\sum_{y=1}^{\infty}\frac{\theta^{y}}{(y-1)!}=\frac{\theta e^{-\theta}}{1-e^{-\theta}}\sum_{y=0}^{\infty}\frac{\theta^y}{y!}=\frac{\theta}{1-e^{-\theta}}$$