Extension of Poisson Parameter for Different Temporal Interval Suppose $Y$ follows a Poisson distribution with parameter $ \lambda $ that explains a temporal Poisson process over an interval of 30 seconds.
Now, it stands to reason that for an interval of 60 seconds, the rate parameter ($\lambda$) of the new distribution would be $2\lambda$.
Let’s generate a 95% confidence interval (Garwood) for both of these. When $ Y = 10 $ the CI is $ (4.80,18.40) $. When $ Y = 20 $ the CI is $ (12.22, 30.89) $.
Why is the second CI not double the first CI ($(9.60,37.20)$)?
 A: Let $Y_1$ be the number of events in the first 30 seconds and $Y_2$ be the number of events in the next 30 seconds.
The interval isn't twice as wide because the distribution of $Y_1+Y_2$ isn't twice as wide as that for $Y_1$. 
Note that waiting twice as long means you're observing two 30 second intervals and that (with a Poisson process) the events in those two disjoint intervals are independent.
Consider, for example, that $\text{Var}(Y_1+Y_2) = \text{Var}(Y_1) + \text{Var}(Y_2) = 2\lambda$, so $\text{sd}(Y_1+Y_2)=\sqrt{2}\,\text{sd}(Y_1)$.
[If you observed for 30 seconds and doubled the resulting count, that would have standard deviation twice that of the count from observing for 30 seconds. But two independent counts don't do that.]
Now even though the standard deviation increases by a multiple of $\sqrt 2$ (and in proportion to $\sqrt{t}$ more generally) the corresponding interval won't be quite $\sqrt{2}$ as wide either because the shape changes as well -- among other things the skewness is less pronounced.
