To determine if $S = X_1 + X_2$ is a sufficient statistic for $\lambda$ when $X_1, X_2$ are independent random variables from a Poisson distribution $\text{Po}(\lambda)$, we can apply the definition.

Given the definition, a statistic $S$ is sufficient for $\lambda$ if the conditional distribution of any other statistic $T = T(\mathbf{X})$ given $S = s$ does not depend on $\lambda$.

Steps:

1. Joint Distribution: Since $X_1$ and $X_2$ are independent $\text{Po}(\lambda)$, their joint distribution is: $f(x_1, x_2; \lambda) = \frac{e^{-2\lambda} \lambda^{x_1 + x_2}}{x_1! \, x_2!}$

2. Distribution of $S$: $S = X_1 + X_2$ is also Poisson distributed with parameter $2\lambda$, i.e., $S \sim \text{Po}(2\lambda)$.

3. Factorization:

   • We can express the joint distribution in terms of $S$ and another statistic (e.g., $X_1$):
     • The distribution of $S$ is: $f_S(s; 2\lambda) = \frac{e^{-2\lambda} (2\lambda)^s}{s!}$
     • The conditional distribution of $X_1$, given $S = s$, is: $P(X_1 = x_1 \mid S = s) = \frac{\binom{s}{x_1}}{2^s}$
     • This distribution is a Binomial distribution with parameters $s$ and $0.5$, independent of $\lambda$.

4. Conclusion: Since the conditional distribution of $X_1$ given $S = s$ is independent of $\lambda$, $S = X_1 + X_2$ is a sufficient statistic for $\lambda$.

Thus, $S = X_1 + X_2$ is sufficient for $\lambda$.