# Scaling percentiles of log-normal distribution

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

$$exp(u+(s^2)/2)$$

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.

• In the lognormal family, $\exp(\mu)$ is the scale parameter and $\sigma$ is a shape parameter. Thus, you scale simply by changing $\mu.$
– whuber
Commented Dec 14, 2018 at 14:49
• @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. Commented Dec 14, 2018 at 14:55
• Can you clarify what exactly do you mean by "with mean A"? Is it $\mu$? Commented Dec 14, 2018 at 14:56
• @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 $\exp(\mu^\prime-\mu).$
– whuber
Commented Dec 14, 2018 at 15:06
• 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.
– whuber
Commented Dec 14, 2018 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 $$\exp\left(\mu_A + \frac{\sigma_A^2}{2}\right) = A$$.

• Almost right. However, you don't want to multiply $\mu_A$ by anything: you want to add a constant to it.
– whuber
Commented Dec 14, 2018 at 16:12
• @whuber But in practice is the same thing, $\mu_b = c*\mu_A$ or $\mu_B = c + \mu_A$, as long as you hit $B$. Commented Dec 14, 2018 at 17:27
• 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?
– whuber
Commented Dec 14, 2018 at 19:22
• @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. Commented Dec 14, 2018 at 19:37