I need help with this basic question. A study found that a variable is log-normal, with mean A and percentiles p1, p2 and p3 (could be 10%, 50% and 90%).

Another study for a different group found that the mean is B, but provides no data on percentiles. I want to assume the two distributions are the same, except for scaling the mean. So, if I want to know the percentiles p1, p2 and p3 of this new distribution, is it just the scaling of p1 p2 and p3 by B/A?

For the log-normal, the mean is


and the quartile (which gives the percentiles ...?) is

$exp(u + s*sqrt(2)*erfinv(2F-1)$

If I scale the mean by a given factor, it seems from the above that it is not enough to scale the quartile by the same factor. Then, how can I do it?

Thank you for the help.

  • $\begingroup$ In the lognormal family, $\exp(\mu)$ is the scale parameter and $\sigma$ is a shape parameter. Thus, you scale simply by changing $\mu.$ $\endgroup$ – whuber Dec 14 '18 at 14:49
  • $\begingroup$ @whuber Oh. I understood the mean to be the log-normal mean (as that's the equation shown in the question). Surely, if by mean the OP means the mean of the underlying normal (mu), things are different. $\endgroup$ – luchonacho Dec 14 '18 at 14:55
  • $\begingroup$ Can you clarify what exactly do you mean by "with mean A"? Is it $\mu$? $\endgroup$ – luchonacho Dec 14 '18 at 14:56
  • $\begingroup$ @Lucho I believe you might be misinterpreting some things. The expectation of the lognormal distribution is $\exp(\mu + \sigma^2/2).$ Nevertheless, $\exp(\mu)$ (which is the geometric mean) is still a scale parameter. Thus, one easily rescales the distribution simply by changing $\mu$ to another value $\mu^\prime.$ The scale factor, as you can verify, is $\exp(\mu^\prime-\mu).$ $\endgroup$ – whuber Dec 14 '18 at 15:06
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    $\begingroup$ Actually, you do scale all quantiles uniformly when you alter $\mu.$ If that's not perfectly clear, look at the last equation in this question and notice that the quantiles are directly proportional to $\exp(\mu),$ QED. $\endgroup$ – whuber Dec 14 '18 at 15:17

As you say, the mean of a log-normal is $ \exp\left(\mu + \frac{\sigma^2}{2}\right)$. If you want to be this to be equal to $B$, there are two parameters which you can change, and only one equation. The system is undetermined. In other words, there is a whole array of log-normals which fulfil your condition. Which one to choose?

Well, you can impose some properties of the original log-normal on the second one. Your suggested "solution" is to just scale all the quantiles. As Whuber suggested, this is possible by scaling $\mu_A$ by an additive (or multiplicative) factor $c$, such that

$$\exp\left(c + \mu_A + \frac{\sigma_A^2}{2}\right) = B$$

To do the above, you need to estimate $\mu_A$ and $\sigma_A$ (the parameters of the original log-normal). This is possible using the percentiles. For instance, see this question. In the first solution, you could also add to the optimisation the empirical constraint to the mean, i.e. that $ \exp\left(\mu_A + \frac{\sigma_A^2}{2}\right) = A$.

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  • $\begingroup$ Almost right. However, you don't want to multiply $\mu_A$ by anything: you want to add a constant to it. $\endgroup$ – whuber Dec 14 '18 at 16:12
  • $\begingroup$ @whuber But in practice is the same thing, $\mu_b = c*\mu_A$ or $\mu_B = c + \mu_A$, as long as you hit $B$. $\endgroup$ – luchonacho Dec 14 '18 at 17:27
  • $\begingroup$ Consider the case $\mu=0.$ Then think about the different interpretations of $c\mu_A$ and $c+\mu_A.$ The latter is easy to interpret: the scale factor is $\exp(c).$ It's not so easy to interpret the former, is it? $\endgroup$ – whuber Dec 14 '18 at 19:22
  • $\begingroup$ @whuber I find easier to interpret the multiplicative case (which of course, doesn't work in the unlikely case the OP gets $\hat\mu_A=0$). But it's true. Additive is a more general transformation. $\endgroup$ – luchonacho Dec 14 '18 at 19:37

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