Expected value of inverse? If I have a random variable $V$ that is normally distributed with some $\mu$ and $\sigma$, then what is the expected value of $1/V$? I tried doing by delta method, and I get expected value $1/\mu$, but I also read that this is not the correct result.
To clarify:
My random variable $V$ is a variable that can be measured. The data given back to us are mean and standard deviation of the measurements. (I understand that I should be using the t-distribution to best model it, as we only have an average of 4 reads, but I am trying to work it out for the Normal case first.)
The value $V$ is indicative of protein binding to ligand. I wish to understand fractional occupancy ($\alpha = \frac{current signal}{saturation signal}$), in which case I need to know the value of $V$ at saturation, which also can be estimated from the data (take the max value of all values present, this will be a consistent under-estimation). Therefore, if $V \sim N(\mu, \sigma^2)$, I expect $\alpha \sim N(\frac{\mu}{max}, \frac{\sigma^2}{max^2})$. I hope I did the math right.
From the Hill equation,
$$
\alpha = \frac{\mbox{ligand}^n}{\mbox{ligand}^n + K_d} ,
$$
therefore 
$$
K_d = \frac{\mbox{ligand}^n}{\alpha} - \mbox{ligand}^n.
$$
Therefore, if $V$ were a random variable, then alpha is a random variable, therefore Kd should be a random variable. I'm trying to find how to model $K_d$ here. My thought was to use the delta method to get an approximation, and that is how I led myself to this question I've posted here.
 A: Since your values are measurements, I assume the actual density close to 0 is not an issue and we can safely assume convergence for the Taylor expansion. 

(I understand that I should be using the t-distribution to best model it, as we only have an average of 4 reads, but I am trying to work it out for the Normal case first.)

This appears to be a common misconception. Having small samples doesn't cause approximately normal data (or its mean) to have a t-distribution; if your reasons for assuming approximate normality are good, small samples doesn't change that.
Your discussion doesn't make it clear how $V$ relates to $\alpha$. 
It's also not clear how finding the expectation of $1/V$ tells you about $K_d$. I feel like there are important details missing. 
In any case, I'll do the expansion:
\begin{align}
g(X) &= g(\mu_X + X - \mu_X)  \\
     &= g(\mu_X) + (X - \mu_X)\cdot g'(\mu_X) + \frac{(X - \mu_X)^2}{2!}\cdot g''(\mu_X) + \frac{(X - \mu_X)^3}{3!}\cdot g'''(\mu_X) + ...
\end{align}
Taking expectations term by term:
\begin{align}
E(g(X)) &= g(\mu_X) + g'(\mu_X)\cdot E(X - \mu_X) + \frac{g''(\mu_X)}{2}\cdot E[(X - \mu_X)^2] +  ...  \\
        &=g(\mu_X) + 0 + g''(\mu_X)/2\cdot \text{Var}(X) + ...   \\
        &\approx g(\mu_X) + g''(\mu_X)/2 \cdot\text{Var}(X) 
\end{align}
When $g(X) = 1/X$,
$\quad\quad\quad\quad\ \ \approx \frac{1}{\mu_X} - \frac{1}{2\mu_X^2} \sigma^2_X $
An approximation for the variance can also be obtained.
