I got a unbiased estimator but I don't know how to interpret it and adjust it to get estimate.
The original problem is to find out the unbiased estimator for $\lambda$ in Zero-truncated Poisson Distribution.
So, I am trying to find $T(X)$ such that $E(T(X)) = \lambda$.
Let, $X \sim ZTP(\lambda)$ then, the PMF of $X$ is $ \frac{\lambda^{k}}{k!(e^{\lambda}-1)}, \quad for \; k > 0$
\begin{align} E(T(X)) = \sum_{x=1}^{\infty}T(x)\cdot\frac{\lambda^{x}}{x!(e^{\lambda}-1)} \end{align} By expanding the above and Taylor Series then, \begin{align} T(1)\frac{\lambda^{1}}{1!} + T(2)\frac{\lambda^{2}}{2!} + T(3)\frac{\lambda^{3}}{3!} + \cdots = \lambda(e^{\lambda}-1) = \lambda(\lambda + \frac{\lambda^{2}}{2!} + \frac{\lambda^{3}}{3!} + \cdots) \end{align}
Two power series can be identical if their all coefficients match so,
\begin{align} T(1) = 0, \quad T(2) = 2, \quad T(3) = 3, \quad \cdots \end{align}
\begin{equation} T(X) = \begin{cases} 0 & X = 1 \\ X & X > 1 \end{cases} \end{equation}
Finally got unbiased estimator for $T(X)$ for $\lambda$.
And if I get some random samples, $X_{1}, X_{2}, X_{3}$ which are equivalent to $X_{1}=3, X_{2}=8, X_{3} = 1$ then how can I get the estimate from estimator $T(X)$?
Is it $3$? $8$? $1$? or maybe any value can be an estimator among those?
or maybe the mean of three values, which is equivalent to $4$ or maybe any other value?
To give some detail about this problem, actually this problem came from the online lecture I took.
And the lecturer used this example as a method to get an unbiased estimator by solving an equation.
So you don't need to consider whether the estimator is a proper value or not.
Btw, what makes me so confused is he does not make a clear notation between random variable such as $X$ and random sample $X_{i}$
so I got a trouble to distinguish them.
It seems like the lecturer used notation $T(X)$ instead of using $T(X_{i})$ but I am not sure.
Please help me!
Your method found an unbiased estimator $T$ for a single observation $X$. So if you take the sample mean of the estimates for $n$ iid observations $T(X_1), \ldots, T(X_n)$, then this will also be unbiased, since by linearity of expectation, $$\operatorname{E}\left[\frac{T(X_1) + \cdots T(X_n)}{n}\right] = \frac{1}{n} \sum_{i=1}^n \operatorname{E}[T(X_i)] = \frac{1}{n} (n \lambda) = \lambda$$ because each $\operatorname{E}[T(X_i)] = \lambda$ by your construction. So your unbiased estimator for a sample is $$T(X_1, \ldots, X_n) = \frac{1}{n}\left( \sum_{i=1}^n X_i - \mathbb 1(X_i = 1) \right),$$ in other words, take the sample total $\sum X_i$, subtract the number of observations in the sample that equal $1$, and then divide the result by $n$. For example, if you observe the sample $$(x_1, \ldots, x_n) = (3, 5, 2, 7, 1, 1, 3, 2, 4, 1, 10),$$ with $n = 11$, we would compute the total $39$, subtract $3$ because three of the observations are $1$, and divide by $11$ to obtain the estimate $T = \frac{36}{11}$.