Zero-Truncated Poisson Distribution

The zero-truncated Poisson distribution arises when a Poisson process is observed conditional on at least one event occurring. Let $Y \sim \text{Poisson}(\lambda)$ with $\lambda > 0$, and define $X = (Y \mid Y \ge 1)$.

(a) Derive the PMF of $X$: show that $P(X = k) = \frac{\lambda^k}{k!(e^\lambda - 1)}$ for $k = 1, 2, 3, \ldots$, and verify it sums to 1.

(b) Compute $E[X]$ by relating it to $E[Y]$ via the truncation. Specifically, show that $$E[X] = \frac{\lambda}{1 - e^{-\lambda}}.$$

(c) Derive $\text{Var}(X)$ using the identity $\text{Var}(X) = E[X^2] - (E[X])^2$. Show that $$\text{Var}(X) = \frac{\lambda(1 + \lambda)}{1 - e^{-\lambda}} - \frac{\lambda^2}{(1 - e^{-\lambda})^2}.$$

(d) Evaluate $P(X = 1)$, $E[X]$, and $\text{Var}(X)$ numerically for $\lambda = 0.5$.