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Suppose I have that $Y_1, \ldots, Y_n$ are from a $Pois(\lambda)$ distribution, I am told that $\bar{Y} \pm 1.96\sqrt{\frac{\bar{Y}}{n}}$ a $95$% CI for $\lambda$.

However, I am not clear why. I know that for $n$ large, $\bar{Y} \sim N\left(\lambda, \frac{\lambda}{n}\right)$ approximately. However, how does this fact turn into the above? Can someone tell me where my derivations are incorrect below?:

$$ 0.95 = P\left(-1.96 \leq \frac{\bar{Y}-\lambda}{\sqrt{\frac{\lambda}{n}}} \leq 1.96\right) = P\left(\bar{Y}-1.96\sqrt{\frac{\lambda}{n}} \leq \lambda \leq \bar{Y}+1.96\sqrt{\frac{\lambda}{n}}\right) $$

However, at this point I CANNOT move the $\lambda$'s into the middle as it is not linear. Ideally, I would like $\lambda$ right in the middle, what am I doing wrong?

user321627
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    Many introductory textbooks approximate $\lambda$ in the denominator by the sample estimate $\bar Y$. Then you end up with the above interval. However, you _can_ solve the quadratic double inequality with respect to $\lambda$. This leads to an asymmetric interval with coverage closer to the nominal level. – Jarle Tufto Dec 03 '16 at 12:09
  • You should also look at this thread http://stats.stackexchange.com/questions/122836/poisson-confidence-interval-using-the-pivotal-method – Jarle Tufto Dec 03 '16 at 12:42

1 Answers1

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For the poisson distribution, the variance equals the expectation, so the variance can be estimated by $\bar{Y}$. Then the above interval follows by a normal approximation. But better intervals can be constructed, one mentioned by @Jarle Tufto in a comment. You can also see Poisson confidence interval using the pivotal method, Confidence interval for a small number of iid Poisson

kjetil b halvorsen
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