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$ X \sim N(\mu,\sigma^2) $

$ Y = \frac{\exp(X)}{1+\exp(X)} $

Y has a logit-normal distribution.

When I have a correct estimate of the mean, say $\bar{Y}$, of the logit-normal distribution, how can use it to get a correct estimate of $\mu$ ?

I don't have full sample from logit-normal distribution, but only the correct estimate of the mean. I know that $ \log(\dfrac{\bar{Y}}{1-\bar{Y}}) $ is not the right answer due to Jensen's inequality.

user67275
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  • You say: "I have a correct estimate of the mean of $Y$", /// Do you actually intend to say that you "have a correct estimate of the mean": i.e. $E[Y]$, OR that you have an unbiased estimator of the mean of $Y$? Not the same. – wolfies Apr 19 '14 at 19:28
  • Yes, I meant that I have an unbiased estimator of the mean of Y. – user67275 Apr 19 '14 at 19:40
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    You are basically asking to solve for $\mu$ given $\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty \frac{e^x}{1+e^x} e^{-(x-\mu)^2/(2\sigma^2)} dx = \bar{Y}$. According to mathematica, there is no closed form solution of the integral in terms of $\mu$ and $\sigma$ in general, so your best shot is probably to use some numerical solver, such as Newton's method. – John Jiang Apr 19 '14 at 20:58
  • A common approach to what I think you're asking about would be to approximately unbias the estimate via use of Taylor expansion. But on the other hand, I think too much is made of bias. If you have ML estimates on the transformed scale, they'll remain ML after transformation back. – Glen_b Apr 20 '14 at 01:42

1 Answers1

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The integral appearing here is the so-called logistic-normal integral. Denoting it as \begin{equation} \varphi(\mu,\sigma) = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty \frac{1}{1+e^{-x}} e^{-\frac{1}{2\sigma^2}(x-\mu)^2} \end{equation}

then one needs to solve the equation $\varphi(\mu,\sigma) = \bar Y$. One must assume that $\sigma$ is known in order to be able to solve this equation.

One can show that $\varphi(\mu,\sigma)$ can be computed exactly at $\mu = 0, \pm \sigma^2, \pm 2 \sigma^2, \cdots $. This was shown in Pirjol (2013).

(The integral is a limiting case of the Mordell integral from analytic number theory, and has many surprising symmetry properties.) Then it is just a matter of bracketing $\bar Y$ between two points on the grid of spacing $\sigma^2$, and solving for $\mu$.

References

Pirjol, D. (2013). "The logistic-normal integral and its generalizations." Journal of Computational and Applied Mathematics, 237(1), 460-469.

Silverfish
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Dan
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  • Welcome to our site! It's a good idea to put full citations/references to an article rather than just a bare link, as you would in a paper or similar. That gives a degree of protection if publishers ever move links around, and also makes it easier for readers to spot a paper or author they recognise than the URL does! – Silverfish Nov 26 '16 at 01:47