A MOVER Confidence interval for the ratio of two Poisson rates If X and Y be two independent Poisson rates with $\hat{\lambda}_1$ and $\hat{\lambda}_2$ how do we obtain this confidence interval for the ratio of $\hat{\lambda}_2/\hat{\lambda}_1$?
Where $(l_i,u_i)$ are seperate confidence itervals for  $\hat{\lambda}_1$ and $\hat{\lambda}_2$
Then the lower and upper bound is:
$L=\hat{\lambda}_2 / \hat{\lambda}_1-\sqrt{(\hat{\lambda}_2 / \hat{\lambda}_1 )^2-l_1 (2\hat{\lambda}_2-l_1 )[u_2 (2\hat{\lambda}_1-u_2 )] )}/(u_2 (2\hat{\lambda}_1-u_2 ) )$
$U=\hat{\lambda}_2 / \hat{\lambda}_1+\sqrt{(\hat{\lambda}_2 /  \hat{\lambda}_1 )^2-u_1 (2\hat{\lambda}_2-u_1 )[l_2 (2\hat{\lambda}_1-l_2 )] )}/(l_2 (2\hat{\lambda}_1-l_2 )) $
We have obtained a confidence interval for the difference between two independent Poisson rates. Then by using MOVER method we obtain these bounds:
$L'=λ ̂_2-λ ̂_1-√((λ ̂_2-l_2 )^2+(u_1-λ ̂_1 )^2 )$
$U'=λ ̂_2-λ ̂_1+√((u_2-λ ̂_2 )^2+(λ ̂_1-l_1 )^2 )$
Now we want to obtain a confidence interval for the ratio of them. So we want to get $Pr(L≤λ_2/λ_1 ≤U)=1-α$ or equality $Pr(λ_2-Uλ_1≤0≤λ_2-Lλ_1 )=1-α$
Now we can apply $L'$ and $U'$ for  $λ_2-Lλ_1$ and $λ_2-Uλ_1$ and then obtain the confidence intervals which is above. But how?
 A: Your formula does not seem to match the actual Fieller's Theorem, which according to Wiki, has a correction due to the correlation of the two variables.
If your goal is to construct an approximate CI for a ratio of poisson RVs using the upper and lower bounds of the CIs for two Poisson means, this is impossible: you must assume those variables are independent or you must know their covariance.
If they are independent, there is no reason to appeal to Fiellers theorem, the CI is a direct result of the $\delta$-method. In either case, your only task is to identify the SE from the CI which I describe below.
The first task, for poisson RVs, is to identify the scale on which a normal approximation is made. On the linear scale, the SE of lambda is $\sqrt{\hat{\lambda}/n}$, on a log scale the SE of the log lambda is $\sqrt{2 \hat{\lambda}/n}$ which results in a better coverage 95% non-symmetric CI when re-exponentiated. 
In either case, it is only algebra to transform the upper and lower bounds of a CI into the SE for the estimate (either lambda or log-lambda). Then you can use the general form of Fieller's theorem to obtain the SE and CI for the ratio or set the covariance to 0 for the special case. Or you can derive the SE yourself with $\delta$ method.
A: In a Bayesian framework, you can obtain posterior samples for $\lambda_1$ and $\lambda_2$, and divide them to get posterior samples for their fraction. Then get a Bayesian credibility interval from those samples. 
You can use pymc3 in Python or stan in R to obtain these samples, and Poisson models are easy and straightforward to sample for. 
