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I have a normal distribution of a variable y~N(f(x),bc2x), where x is the IDV and b and c are some estimated parameters. So, the SD of y is given by bcx.

Now I have a transformation of y:

z=g(y,n)=ey(n+1)/(1+ey)/n, where n is another IDV.

In this case, what is the best way to estimate the SD of z? Or, what is the "?" in z~(ef(x)(n+1)/(1+ef(x))/n,?) ?

I think it would be the fastest if I can have a closed form equation for the SD of z but I don't know how to do the transformation.

Meanwhile, I can think of the following alternatives: 1. Monte Carlo Simulations, but it is time-consuming, 2. I am wondering if 0.5[g(f(x)+bcx,n)-g(f(x)-bcx,n)] can be regarded as a reasonable estimator for SD of z

Any help is appreciated.

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  • $\begingroup$ It looks like $f$, $b$, $c$ and $n$ are irrelevant and potentially distracting. Ignoring the constant coefficients in $Z$, which affect the variance in simple obvious ways, aren't you just asking what the variance is of the logistic transformation $e^Y/(1+e^Y)=1-1/(1+e^Y)$ of a normal random variable $Y$? BTW, what is an "IDV" and what assumptions are you making about it? Is it a number, a random variable, something else? $\endgroup$
    – whuber ♦
    Commented Jun 1, 2017 at 14:46
  • $\begingroup$ I think so... Let's say I already have obtained a set of values for (b,c,x, n and f(x), then how do I transform the SD for y into the SD for z, noting that z = 1-1/(1+e^Y) ? $\endgroup$
    – Matthew Hui
    Commented Jun 1, 2017 at 17:33
  • $\begingroup$ It does't transform, because it depends on both the mean and variance of $Y$, but you can compute it numerically to high precision and, for certain prescribed regimes of the mean and variance of $Y$, you can develop approximations in various ways. $\endgroup$
    – whuber ♦
    Commented Jun 1, 2017 at 17:52

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