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.
The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:
$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\lambda_{2}
\end{array}
$$
Now neither hypothesis says what the parameters are, so the actual values are nuisance parameters to be integrated out with respect to their prior probabilities.
$$P(H_{1}|D,I)=P(H_{1}|I)\frac{P(D|H_{1},I)}{P(D|I)}$$
The model likelihood is given by:
$$P(D|H_{1},I)=\int_{0}^{\infty} P(D,\lambda|H_{1},I)d\lambda=\int_{0}^{\infty} P(\lambda|H_{1},I)P(D|\lambda,H_{1},I)\,d\lambda$$
$$=\int_{0}^{\infty} P(\lambda|H_{1},I)\frac{\lambda^{x_1+x_2}\exp(-2\lambda)}{\Gamma(x_1+1)\Gamma(x_2+1)}\,d\lambda$$
where $P(\lambda|H_{1},I)$ is the prior for lambda. A convenient mathematical choice is the gamma prior, which gives:
$$P(D|H_{1},I)=\int_{0}^{\infty} \frac{\beta^{\alpha}}{\Gamma(\alpha)}\lambda^{\alpha-1}\exp(-\beta \lambda)\frac{\lambda^{x_1+x_2}exp(-2\lambda)}{\Gamma(x_1+1)\Gamma(x_2+1)}\,d\lambda$$
$$=\frac{\beta^{\alpha}\Gamma(x_1+x_2+\alpha)}{(2+\beta)^{x_1+x_2+\alpha}\Gamma(\alpha)\Gamma(x_1+1)\Gamma(x_2+1)}$$
And for the alternative hypothesis we have:
$$P(D|H_{2},I)=\frac{\beta_{1}^{\alpha_{1}}\beta_{2}^{\alpha_{2}}\Gamma(x_1+\alpha_{1})\Gamma(x_2+\alpha_{2})}{(1+\beta_{1})^{x_1+\alpha_{1}}(1+\beta_{2})^{x_2+\alpha_{2}}\Gamma(\alpha_{1})\Gamma(\alpha_{2})\Gamma(x_1+1)\Gamma(x_2+1)}$$
Now if we assume that all hyper-parameters are equal (not an unreasonable assumption, given that you are testing for equality), then we have an integrated likelihood ratio of:
$$\frac{P(D|H_{1},I)}{P(D|H_{2},I)}=
\frac{(1+\beta)^{x_1+x_2+2\alpha}\Gamma(x_1+x_2+\alpha)\Gamma(\alpha)}
{(2+\beta)^{x_1+x_2+\alpha}\beta^{\alpha}\Gamma(x_1+\alpha)\Gamma(x_2+\alpha)}
$$
Which you can see that the prior information is still very important. We can't set $\alpha$ or $\beta$ equal to zero (Jeffrey's prior), or else $H_{1}$ will always be favored, regardless of the data. One way to get values for them is to specify prior estimates for $E[\lambda]$ and $E[\log(\lambda)]$ and solve for the parameters - this cannot be based on $x_1$ or $x_2$ but can be based on any other relevant information. You can also put in a few different (reasonable) values for the parameters and see what difference it makes to the conclusion. The numerical value of this statistic tells you how much the data and your prior information about the rates in each hypothesis support the hypothesis of equal rates. This explains why the likelihood ratio test is not always reliable - because it essentially ignores prior information, which is usually equivalent to specifying Jeffrey's prior. Note that you could also specify upper and lower limits for the rate parameters (this is usually not too hard to do given some common sense thinking about the real world problem). Then you would use a prior of the form:
$$p(\lambda|I)=\frac{I(L<\lambda<U)}{\log\left(\frac{U}{L}\right)\lambda}$$
And you would be left with a similar equation to that above but in terms of incomplete, instead of complete gamma functions.
For the binomial case things are much simpler, because the non-informative prior (uniform) is proper. The procedure is similar to that above, and the integrated likelihood for $H_{1}:p_{1}=p_{2}$ is given by:
$$P(D|H_{1},I)={n_1 \choose x_1}{n_2 \choose x_2}\int_{0}^{1}p^{x_1+x_2}(1-p)^{n_1+n_2-x_1-x_2}\,dp$$
$$={n_1 \choose x_1}{n_2 \choose x_2}B(x_1+x_2+1,n_1+n_2-x_1-x_2+1)$$
And similarly for $H_{2}:p_{1}\neq p_{2}$
$$P(D|H_{2},I)={n_1 \choose x_1}{n_2 \choose x_2}\int_{0}^{1}p_{1}^{x_1}p_{2}^{x_2}(1-p_{1})^{n_1-x_1}(1-p_{2})^{n_{2}-x_{2}}\,dp_{1}\,dp_{2}$$
$$={n_1 \choose x_1}{n_2 \choose x_2}B(x_1+1,n_1-x_1+1)B(x_2+1,n_2-x_2+1)$$
And so taking ratios gives:
$$\frac{P(D|H_{1},I)}{P(D|H_{2},I)}=
\frac{B(x_1+x_2+1,n_1+n_2-x_1-x_2+1)}
{B(x_1+1,n_1-x_1+1)B(x_2+1,n_2-x_2+1)}
$$
$$=\frac{{x_1+x_2 \choose x_1}{n_1+n_2-x_1-x_2 \choose n_1-x_1}(n_1+1)(n_2+1)}{{n_1+n_2 \choose n_1}(n_1+n_2+1)}$$
And the choose functions can be calculated using the hypergeometric($r$,$n$,$R$,$N$) distribution where $N=n_1+n_2$, $R=x_1+x_2$, $n=n_1$, $r=x_1$
And this tells you how much the data support the hypothesis of equal probabilities, given that you don't have much information about which particular value this may be.
Best Answer
The power really is that low when you're dealing with such small numbers. If you multiplied your rates by ten, you'd have more observations and therefore more weight of evidence with which to reject the null hypothesis.
Examples using an exact Binomial test: (edit to note: my
rare yourlambda)You can reproduce this, but a tweak is needed to your code because it breaks for "large"
r1andr2owing to the way thatdis calculated. If you work with logs instead then you can reproduce these results. Specifically, change thedcalculation towhere
slog(x)=function(x)log(x+0.0001), then observe the (simulated) power with your original rates, e.g. ...and the (simulated) power for those
lambda*times ten: