Discussion
Let \(X_{1}, X_{2}, \ldots X_{n}\) be a random sample from Poisson distribution with parameter \(\lambda\). Then covariance between \(\bar{X}\) and \((X_{1} - X_{2})\) is :
Answer: D
The covariance is 0.
Use bilinearity of covariance and independence of the sample values, each with variance \(\lambda\):
\[ \begin{aligned} \text{Cov}(\bar{X}, X_{1} - X_{2}) &= \tfrac{1}{n}[\text{Cov}(X_{1}, X_{1}) - \text{Cov}(X_{2}, X_{2})] \\ &= \tfrac{1}{n}(\lambda - \lambda) = 0 \end{aligned} \]
All the other terms \(\text{Cov}(X_{i}, X_{j})\) with \(i \ne j\) vanish by independence. The result holds for any iid sample, Poisson or not, since only the equal variances matter.
1, -1 and 1/2 come from keeping only one of the two variance terms, or forgetting the factor \(1/n\) from the mean.
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