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(1) The statistic $T$ has Poisson($n\lambda$) distribution so the expectation of $\delta(T)$ must involve $n\lambda$: $$ E[\delta(T)] = \sum_{t=0}^\infty \delta(t)P(T=t)=\sum_{t=0}^\infty\delta(t)e^{-n\lambda}{(n\lambda)^t\over t!} $$ They then abbreviate $n\lambda$ as $\gamma$ for the sake of saving ink. (2) The equality comes from the line above it (which ...



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