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Let [itex]X_{1}, \ldots, X_{n} \; \mathtt{\sim} \; \textrm{Poisson} (\lambda)[/itex] and let [itex]\hat{\lambda} = n^{-1} \sum_{i = 1}^{n} X_{i}[/itex].
The bias of [itex]\hat{\lambda}[/itex] is [itex]\mathbb{E}_{\lambda} (\hat{\lambda}) - \lambda[/itex]. Since [itex]X_{i} \; \mathtt{\sim} \; \textrm{Poisson} (\lambda)[/itex], and all [itex]X_{i}[/itex] are IID, [itex]\sum_{i = 1}^{n} X_{i} \; \mathtt{\sim} \; \textrm{Poisson} (n \lambda)[/itex].
Thus, [itex]\mathbb{E} (\hat{\lambda}) = \sum_{nx = 1}^{\infty} x \exp{(-n \lambda)} \frac{(n \lambda)^{nx}}{(nx)!} = \lambda[/itex], and the indicator is unbiased (bias = 0).
However, I'm using [itex]\mathbb{E}_{\lambda}[/itex] as [itex]\mathbb{E}[/itex], and I don't know if I'm doing it right. I haven't seen any similar examples and this is the first time I'm calculating the bias, so I would really love some insight.
The bias of [itex]\hat{\lambda}[/itex] is [itex]\mathbb{E}_{\lambda} (\hat{\lambda}) - \lambda[/itex]. Since [itex]X_{i} \; \mathtt{\sim} \; \textrm{Poisson} (\lambda)[/itex], and all [itex]X_{i}[/itex] are IID, [itex]\sum_{i = 1}^{n} X_{i} \; \mathtt{\sim} \; \textrm{Poisson} (n \lambda)[/itex].
Thus, [itex]\mathbb{E} (\hat{\lambda}) = \sum_{nx = 1}^{\infty} x \exp{(-n \lambda)} \frac{(n \lambda)^{nx}}{(nx)!} = \lambda[/itex], and the indicator is unbiased (bias = 0).
However, I'm using [itex]\mathbb{E}_{\lambda}[/itex] as [itex]\mathbb{E}[/itex], and I don't know if I'm doing it right. I haven't seen any similar examples and this is the first time I'm calculating the bias, so I would really love some insight.