This is a particularly ill-formed question.
If by "alpha" you mean Type I error, you need to go back to Square One and get definitions straight. Type I error is not something inherent in the data, or even in the hypothesis; it's a subjectively and externally applied measure of risk. And without the Type I error, you have no reference point from which to calculate Type II error, the complement of power.
Worse yet, it's not clear--after Adam's question--whether you have JUST TWO OBSERVATIONS (10 and 25), or two distributions with means of 10 and 25, and you're looking for a suitable sample size for a balanced test comparing the means. In the first case, all you can do is a likelihood ratio test that gives an approximate p-value; there's no more information to be had in two observations. In the second case, simulation can give some useful results, but you still need a value for the Type I error to get started.
You can do this using simulation.
Write a function that does your test and accepts the lambdas and sample size(s) as arguments (you have a good start above).
Now for a given set of lambdas and sample size(s) run the function a bunch of times (the replicate function in R is great for that). Then the power is just the proportion of times that you reject the null hypothesis, you can use the mean function to compute the proportion and prop.test to give a confidence interval on the power.
Here is some example code:
tmpfunc1 <- function(l1, l2=l1, n1=10, n2=n1) {
x1 <- rpois(n1, l1)
x2 <- rpois(n2, l2)
m1 <- mean(x1)
m2 <- mean(x2)
m <- mean( c(x1,x2) )
ll <- sum( dpois(x1, m1, log=TRUE) ) + sum( dpois(x2, m2, log=TRUE) ) -
sum( dpois(x1, m, log=TRUE) ) - sum( dpois(x2, m, log=TRUE) )
pchisq(2*ll, 1, lower=FALSE)
}
# verify under null n=10
out1 <- replicate(10000, tmpfunc1(3))
mean(out1 <= 0.05)
hist(out1)
prop.test( sum(out1<=0.05), 10000 )$conf.int
# power for l1=3, l2=3.5, n1=n2=10
out2 <- replicate(10000, tmpfunc1(3,3.5))
mean(out2 <= 0.05)
hist(out2)
# power for l1=3, l2=3.5, n1=n2=50
out3 <- replicate(10000, tmpfunc1(3,3.5,n1=50))
mean(out3 <= 0.05)
hist(out3)
My results (your will differ with a different seed, but should be similar) showed a type I error rate (alpha) of 0.0496 (95% CI 0.0455-0.0541) which is close to 0.05, more precision can be obtained by increasing the 10000 in the replicate command. The powers I computed were: 9.86% and 28.6%. The histograms are not strictly necessary, but I like seeing the patterns.
Best Answer
You find the MLE of $\lambda$ and then compute the log-likelihood at that value of $\lambda$ (and multiply by -2)
[However, some simplifications may be made.]