If $X$ is lognormally distributed, what is the distribution of $1 / (1 + X)$?

Let $X$ be lognormal with parameters $\mu$ and $\sigma$ (such that $\log(X)$ is Gaussian with mean $\mu$ and variance $\sigma^2$).

What is the distribution of $1 / (X + 1)$? I am wondering whether it is a "simple" parametric distribution. If the "+1" were replaced with zero, the problem becomes quite easy.

• Are you asking for a name for it, or a functional form for the density? If the latter, is this homework or something? – Glen_b Apr 29 '14 at 3:28
• If X is Lognormal with parameters $\mu$ and $\sigma$, then the mean is not $\mu$, and the standard deviation is not $\sigma$. Conversely, if the mean is $\mu$ and the standard deviation is $\sigma$, then the parameters are not $\mu$ and $\sigma$. – wolfies Apr 29 '14 at 6:57
• I see your point, but sigma and mu are just symbols though. Anyway, I changed it to the canonical meaning. – Asker4567 Apr 29 '14 at 15:36
• In one obvious sense it's a simple parametric distribution, because its PDF, which is $$\frac{1}{\sqrt{2\pi}\sigma(1-x)x}\exp\left({-\frac{1}{2\sigma^2}\left(\mu-\log\left(\frac{1-x}{x}\right)\right)^2}\right)$$ for $0\lt x\lt 1,$ is explicit and obviously parametric with two parameters $\mu$ and $\sigma \gt 0$. However, most of its basic properties, such as its moments, mode, characteristic function, and so on, cannot be written in simple closed forms. Moreover, its PDF can attain a rich and complex set of shapes. So in what sense do you mean "simple"? – whuber Apr 29 '14 at 16:32
• Since @whuber has already given the answer for the second option (which follows from the change of variable formula) I raised when asking about what you want, I'll give the answer for the first option - up to a change of sign of the argument, it's called the logit-normal. One or the other is probably an answer to the question, but it's hard to say which. – Glen_b Apr 30 '14 at 1:04

To close this one:

We want the pdf of the variable

$$Z = g(X) =\frac 1{X+1} \Rightarrow X = g^{-1}(Z)=\frac 1Z -1 \Rightarrow \frac {\partial g^{-1}(z)}{\partial z}=-\frac 1{z^2}$$

Note that by construction $0 \leq Z \leq 1$.

The density of the log-normal is known. Applying the change-of-variable formula we obtain

$$f_Z(z) = \left|\frac {\partial g^{-1}(z)}{\partial z}\right|\cdot f_X(g^{-1}(z))$$

$$=\frac 1{z^2}\cdot\frac{1}{[(1-z)/z]\sqrt{2\pi}\sigma} \exp{ \left\{-\frac{\left(\ln[(1-z)/z]-\mu\right)^2}{2\sigma^2}\right\}}$$

$$\Rightarrow f_Z(z) = \frac{1}{(1-z)z\sqrt{2\pi}\sigma} \exp{ \left\{-\frac{\Big(\ln[z/(1-z)]-(-\mu)\Big)^2}{2\sigma^2}\right\}}$$

which is the density of the "logit-normal" distribution with parameters $\sigma$ and $-\mu$.