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.

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    $\begingroup$ This question at math.SE may help, although it also only references Wikipedia and does not give a proof (if you have access to the book by Johnson, Kotz & Balakrishnan cited on the Wikipedia page, this may help): math.stackexchange.com/questions/646428/… $\endgroup$ – Stephan Kolassa Apr 4 '14 at 6:16
  • $\begingroup$ @Glen_b: Ah yes, I understand that the delta method always gives us an approximation. I think this is sufficient for us, since we are dealing with real-world data. We are modelling the binding coefficient of a ligand as a random variable, which is proportional to the inverse of the fluorescence signal produced. $\endgroup$ – ericmjl Apr 4 '14 at 20:00
  • $\begingroup$ Allow me to include additional information into the question. $\endgroup$ – ericmjl Apr 4 '14 at 20:06
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    $\begingroup$ ericmjl - I just noticed I typed the inequality the wrong way around in my previous comment. I decided to delete it so as not to mislead anyone with it. The correct inequality is here -- i.e. when $E(1/X)$ exists and variance is not zero $E(X)E(1/X)>1$. $\endgroup$ – Glen_b Apr 4 '14 at 23:48
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    $\begingroup$ Oh, I meant to say, since your values are measurements, you should be careful of specifying normality; normal distributions have density at 0, while measurements don't. $\endgroup$ – Glen_b Apr 4 '14 at 23:53

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.

  • $\begingroup$ alpha is proportional to V, but Kd is proportional to 1/alpha, that is how the variables are related to one another. V is the binding signal, alpha is the fractional binding, which is V/Vmax (where Vmax is also estimated from data), and Kd is proportional to 1/alpha by the Hill equation. In any case, your answer was really helpful. Thank you! $\endgroup$ – ericmjl Apr 6 '14 at 1:18
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    $\begingroup$ I don't see how those equations suggest $K_d$ is proportional to $1/\alpha$. Similarly if $\alpha=V/V_\text{max}$ and both are random variables, you can't really treat $V_\text{max}$ as a constant in computing expectations: for example $E(X/Y)\neq E(X)/E(Y)$ $\endgroup$ – Glen_b Apr 6 '14 at 3:48
  • $\begingroup$ Ah, I see. There was an error made when my question was edited to introduce LaTeX formatting. I have updated the equation properly. $\endgroup$ – ericmjl Apr 6 '14 at 4:07
  • $\begingroup$ If $\text{ligand}^n$ is a constant, say $c$, then $K_d=c(1/\alpha -1)$ which isn't $\propto 1/\alpha$ $\endgroup$ – Glen_b Apr 6 '14 at 4:09
  • $\begingroup$ I also see your point on Vmax... So how would you treat it in this case? $\endgroup$ – ericmjl Apr 6 '14 at 4:09

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