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.

  • $\begingroup$ Are you asking for a name for it, or a functional form for the density? If the latter, is this homework or something? $\endgroup$ – Glen_b Apr 29 '14 at 3:28
  • 3
    $\begingroup$ 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 $\sigma$. $\endgroup$ – wolfies Apr 29 '14 at 6:57
  • $\begingroup$ I see your point, but sigma and mu are just symbols though. Anyway, I changed it to the canonical meaning. $\endgroup$ – Asker4567 Apr 29 '14 at 15:36
  • 2
    $\begingroup$ 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"? $\endgroup$ – whuber Apr 29 '14 at 16:32
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $-\mu$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.