# Combining confidence intervals for a particular equation

I perform a Bernoulli experiment to obtain a binomially distributed probability $p_1$ with a 95% confidence interval $\delta p_1$. I perform a second, independent Bernoulli experiment to obtain a new probability $p_2$ and a 95% confidence interval $\delta p_2$.

I now combine these two probabilities as follows $$p=\frac{p_1}{1-(1-p_1)p_2}$$ What will the 95% confidence interval be for $p$? Do the standard rules for combining errors apply here?

Thank you.

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I think you could use the delta-method to calculate the approximate standard error (and thus a confidence interval) of the combined probabilities. Let's call your combination $g(p_1, p_2)$. First, we need the derivatives of $g$ wrt $p_1$ and $p_2$. These are: \begin{align} \frac{\partial}{\partial p_1} g(p_1, p_2) &=\frac{1-p_2}{\left(1+\left(p_1 -1\right)p_2\right)^{2}}\\ \frac{\partial}{\partial p_2} g(p_1, p_2) &= -\frac{\left(p_1-1\right)p_1}{\left(1+\left(p_1-1\right)p_2\right)^{2}}\\ \end{align} For the approximate variance of your combination, we get $$\mathrm{Var}(g)\approx \frac{(1-p_2)^{2}}{\left(1+\left(p_1 -1\right)p_2\right)^{4}}\mathrm{Var}(p_1) + \frac{\left(p_1-1\right)^{2}p_1^{2}}{\left(1+\left(p_1-1\right)p_2\right)^{4}}\mathrm{Var}(p_2) + 2 \frac{(p_1-1)(p_2-1)p_1}{(1+(p_1-1)p_2)^{4}}\mathrm{Cov}(p_1,p_2)$$ Assuming that the experiments are independent means that the covariance is zero, so we can drop the last term: $$\mathrm{Var}(g)\approx \frac{(1-p_2)^{2}}{\left(1+\left(p_1 -1\right)p_2\right)^{4}}\mathrm{Var}(p_1) + \frac{\left(p_1-1\right)^{2}p_1^{2}}{\left(1+\left(p_1-1\right)p_2\right)^{4}}\mathrm{Var}(p_2)$$ Also, the variances of $p_1$ and $p_2$ would simply be the squared standard errors of them: $$\mathrm{Var}(p_1)= \frac{p_1(1-p_1)}{n_1}$$ To the the standard error of the combination, just take the square root of the above expression. With the standard error, the confidence interval can easily be calculated.