I'm trying to figure out how to calculate the variance of some data I have. The basic throughput of the analysis is as follows:
1) Get data in triplicates for each group plus a control group & calculate the mean and variance
2) Subtract the experimental mean from the control mean. Call this value $X$
3) Calculate $2^{-X}$
Now this is easy enough (except for the data collection part ;) ). The more confusing point is calculating the variance for the new random variable $Y = 2^{-X}$. I tried looking around for some good references, but no luck. This is what I have so far:
First I want to figure out what the probability function is for Y:
$P(Y<t) = P(2^{-X}<t) = P(-Xln(2)<ln(t)) = P\left(X > -\frac{ln(t)}{ln(2)}\right) = 1 - P\left(X < -\frac{ln(t)}{ln(2)}\right)$
therefore:
$F_Y(t) = 1-F_X\left(-\frac{ln(t)}{ln(2)}\right)$
now I'm going to assume that $X$ is normally distributed around the mean and variance I calculated from my experimental triplicates. Therefore, the CDF for $Y$ would be:
$F_Y(t) = 1 - {\displaystyle \int_{-\infty}^{-\frac{ln(t)}{ln(2)}}{\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(t-\mu)^2}{2\sigma^2}}} }$
where $\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(t-\mu)^2}{2\sigma^2}} $ is the density function for a normal distribution centered around $\mu$ with variance $\sigma^2$
so, (I believe) the density function of random variable $Y$ is:
$f_Y(t) = \frac{d}{dt}F_Y(t) = {\displaystyle \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{\left(-\frac{ln(t)}{ln(2)}-\mu\right)^2}{2\sigma^2}}}$
However, I'm a bit confused on where to go from here (or if I'm even on the right track). Any help would be great!