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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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2 Answers 2

up vote 2 down vote accepted

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.

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Wonderful!! Thank you (both) for your time. –  lemon Apr 26 at 21:59

It seems that you would want to combine the standard errors of the two proportions, to obtain an estimate of the error of the combined proportion. See here for a good guide on how to obtain the combined standard error (note that the website provides instructions for the combined SE for the difference between two independent proportions, so it might not be applicable in your situation.)

Hope that helps.

edit: Forgot that you were looking for the 95% confidence interval. You'd get this by multiplying the SE with the appropriate critical value. For example t(df=10) = ~2.

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