The question is as follows:
The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter $λ$. However, such a random variable can only be observed if it is positive, since if it is $0$ then we cannot know that such an insect was on the leaf. If we let $X$ denote a Poisson distributed random variable with parameter $λ$ and $Y$ denote the observed number of eggs, then $P(Y = i) = P(X = i|X > 0)$ where $X$ is Poisson with parameter $λ$. Find $E[Y]$.
Now the answer is:
$\frac{λ}{1-e^{-λ}}$
My question is: why did we have to consider the random variable $X$ in the first place; because if we were to calculate the expected value of $Y$ normally, we will start the summation from $i=1$ to infinity, so we shouldn't worry about $Y$ taking a value of $0$, I know my thinking is flawed so please help me understand, thanks in advance!
EDIT
I think it might be because the probability distribution has changed, because even though $0$ can be ignored when calculating the expected value, it still has a probability which is equal to $e^{-λ}$, having this probability equal $0$ (since $0$ isn't in the domain of $Y$) will have an effect on all other probabilities for all values of $Y$. Is this right?
The model being applied in this question is called a "zero-truncated" Poisson distribution.
There are two random variables described: $$X \sim \operatorname{Poisson}(\lambda), \\ \Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x \in \{0, 1, 2, \ldots \}$$ models the random number of eggs laid on a leaf, given that a (gravid) insect has visited that leaf. Theoretically, if you could watch such an insect--perhaps in a controlled enclosed environment--you might be able to observe landings in which no eggs were laid.
However, if we are observing eggs on leaves in nature, and we cannot observe or track the movements of such insects directly, then we cannot simply count all leaves with zero eggs, because we don't know if an insect has visited that leaf. So the observational data that we collect only contains data on the number of leaves we counted that had at least one egg, and for each such leaf, we count the number of eggs on it.
The random variable $Y$, which models the observed data, is not Poisson. It is given by $$Y = \begin{cases} X, & X > 0 \\ \text{undefined}, & X = 0. \end{cases}$$ In particular, the support of $Y$ is on the strictly positive integers $Z^+ = \{1, 2, 3, \ldots \}$. Note that although the variable $X$ exists in principle, we are not able to always observe it. We can only observe $Y$. The distribution that $Y$ follows is called a zero-truncated Poisson.
We can calculate $$\begin{align} \Pr[Y = y] &= \Pr[X = y \mid X > 0] \\ &= \frac{\Pr[(X = y) \cap (X > 0)]}{\Pr[X > 0]} \\ &= \frac{\Pr[X = y] \mathbb 1 (y \in \mathbb Z^+)}{1 - \Pr[X = 0]} \\ &= \frac{e^{-\lambda}}{1 - e^{-\lambda}} \frac{\lambda^y}{y!} \mathbb 1 (y \in \mathbb Z^+). \end{align}$$ This gives the PDF of the zero-truncated distribution of $Y$. You can easily use it to calculate the expectation $\operatorname{E}[Y]$. Is it equal to the expected value of $X$? Why or why not?
Now suppose we have observed a sample from $Y$, namely $$\boldsymbol Y = (y_1, y_2, y_3, \ldots, y_n),$$ where each $y_i$ is some positive integer that records the number of eggs on the $i^{\rm th}$ leaf that we saw that contained at least one egg, and we observed $n$ such leaves in total.
Under the zero-truncated Poisson model, how would we go about using this information to estimate the probability that an insect lands on a leaf but does not lay any eggs?