How to show that simple random sample sensitivity is unbiased for population sensitivity In diagnostic testing, sensitivity $S$ is the probability that the test gives a positive result given that you have the condition being tested. From a simple random sample of people who take the test, an estimate for sensitivity is $s=n_{++}/n_{+}$, where $n_{++}$ is the number of people in the sample who test positive and have the condition and $n_{+}$ are the number of people who have the condition.
I want to get $E(s)$ and show that this is unbiased for sensitivity. That is,
$E(s)=E(n_{++}/n_{+})=S$
This follows easily if I can do $E(n_{++}/n_{+})=E(n_{++})/E(n_{+})$ but I know that in general, $E(X/Y)\neq E(X)/E(Y)$. Is there a reason why I might be able to do this for this setting or is there a different approach?
 A: Following the arguments of Duris, F., Gazdarica, J., Gazdaricova, I. et al. Mean and variance of ratios of proportions from categories of a multinomial distribution. J Stat Distrib App 5, 2 (2018). https://doi.org/10.1186/s40488-018-0083-x
Consider a sample of $n$ observations from a multinomial distribution with $k=4$, representing the number of True Positives, False Negatives, False Positives, and True Negatives in the sample. Let $X_1$ be the number of True Positives and $X_2$ be the number of False Negatives. Then the sample sensitivity is $X_1/(X_1 + X_2)$.
Define a function $f(X_1,X_2) = X_1/(X_1 + X_2)$ and consider a Taylor expansion of degree 2 around the point $u = (\mu_{X_1}, \mu_{X_2})$.
$$
f(X_1,X_2) \approx f(u) + (X_1 - \mu_{X_1})\frac{\partial f(u)}{\partial X_1} + (X_2 - \mu_{X_2})\frac{\partial f(u)}{\partial X_2} +\\
\frac{1}{2}(X_1 - \mu_{X_1})^2\frac{\partial^2 f(u)}{\partial X_1^2} + 
\frac{1}{2}(X_2 - \mu_{X_2})^2\frac{\partial^2 f(u)}{\partial X_2^2} + \\
(X_1-\mu_{X_1})(X_2 - \mu_{X_2})\frac{\partial^2 f(u)}{\partial X_1 X_2}
$$
Taking expected values leads to
$$
E[f(X_1,X_2)] \approx f(u) + \frac{1}{2}\frac{\partial^2 f(u)}{\partial X_1^2}\sigma_{X_1}^2 + \frac{1}{2}\frac{\partial^2 f(u)}{\partial X_2^2}\sigma_{X_2}^2 + \frac{\partial^2 f(u)}{\partial X_1 X_2}\sigma_{X_1X_2}
$$
Calculating the derivatives and substituting the values $\mu_{X_i} = np_i$, $\sigma_{X_i}^2 = np_i(1-p_i)$, $\sigma_{X_1X_2} = -np_1p_2$ leads to
$$
E[f(X_1,X_2)] \approx \frac{np_1}{np_1+np_2} - \frac{n^2p_1(1-p_1)p_2}{(np_1 + np_2)^3} + \frac{n^2p_1p_2(1-p_2)}{(np_1 + np_2)^3} + \frac{(np_1 - np_2)(-np_1p_2)}{(np_1 + np_2)^3}\\
= \frac{p_1}{p_1+p_2}
$$
So the sample sensitivity is (up to a 2nd order approximation) unbiased for the true sensitivity.
