lognormal

**stat_RA** · 01-14-2011 03:09 PM

How do I find the p50 value in a lognormal distribution given p10 and p90 using excel/@RISK ??

**squareandrare** · 01-14-2011 07:20 PM

$e^{\frac{ln(p_{90})+ln(p_{10})}{2}}$

Basically, take the average the logarithms of the 10th and 90th percentiles, then take e to that power.

**Dason** · 01-14-2011 07:22 PM

Nice answer.

**BGM** · 01-14-2011 10:39 PM

Originally Posted by squareandrare

$e^{\frac{ln(p_{90})+ln(p_{10})}{2}}$

Basically, take the average the logarithms of the 10th and 90th percentiles, then take e to that power.

Which is also the geometric mean $\sqrt{p_{90}p_{10}}$

**Dason** · 01-14-2011 10:55 PM

Good point BGM. But I still like squareandrare's answer because it can shed some light on the process of how to figure it out yourself.

**squareandrare** · 01-15-2011 01:31 AM

I actually got a lot further in this problem than I needed to.

I took the cdf for the lognormal from wikipedia:

$F_X(x;\mu{},\sigma{}) = \Phi{}\left({}\frac{ln(x)-\mu{}}{\sigma{}}\right){}$

I set up this system of equations:

$\Phi{}\left({}\frac{ln(p_{90})-\mu{}}{\sigma{}}\right){}=.9$
$\Phi{}\left({}\frac{ln(p_{10})-\mu{}}{\sigma{}}\right){}=.1$

Therefore:

$\frac{ln(p_{90})-\mu{}}{\sigma{}}=\Phi{}^{-1}(.9)=-\Phi{}^{-1}(.1)\approx{}1.2816$
$\frac{ln(p_{10})-\mu{}}{\sigma{}}=\Phi{}^{-1}(.1)=-\Phi{}^{-1}(.9)\approx{}-1.2816$

I solved for $\mu{}$ and $\sigma{}$ :

$\mu{}=\frac{ln(p_{10})+ln(p_{90})}{2}$
$\sigma{}=\frac{ln(p_{90})-\frac{ln(p_{10})+ln(p_{90})}{2}}{\Phi{}^{-1}(.9)}=\frac{ln(p_{10})-\frac{ln(p_{10})+ln(p_{90})}{2}}{\Phi{}^{-1}(.1)}$

I started to plug $\mu{}$ and $\sigma{}$ into the cdf, setting it equal to .5. Then the answer suddenly hit me...

The logarithm function is monotonic. Therefore, the median of the normal distribution with parameters $\mu{}$ and $\sigma{}$ will be the logarithm of the median of the lognormal distribution with parameters $\mu{}$ and $\sigma{}$ (remember that normal distributions are symmetric, so the mean and median are both equal to $\mu{}$ ).

Thus, the median for the lognormal distribution is:

$e^{\mu{}}=e^{\frac{ln(p_{10})+ln(p_{90})}{2}}$

Then the logic of the formula made sense. The 10th and 90th percentiles are symmetric about the mean of any normal distribution, so the average of the logarithms of the 10th and 90th percentiles of the lognormal distribution with parameters $\mu{}$ and $\sigma{}$ must be equal to the mean (and median, by symmetry) of the normal distribution with parameters $\mu{}$ and $\sigma{}$ . To find the median of the lognormal distribution with parameters $\mu{}$ and $\sigma{}$ , you just need to take the antilog of $\mu{}$ .

It didn't occur to me that $e^{\frac{ln(p_{10})+ln(p_{90})}{2}}$ can be simplified to $\sqrt{p_{90}p_{10}}$ . Good catch, BGM.

**Dason** · 01-15-2011 01:52 AM

Originally Posted by squareandrare

The logarithm function is monotonic. Therefore, the median of the normal distribution with parameters $\mu{}$ and $\sigma{}$ will be the logarithm of the median of the lognormal distribution with the same parameters for $\mu{}$ and $\sigma{}$ (remember that the mean and median are equal in normal distributions).

Exactly. That's why I liked your answer. My initial thought was to do something similar to what you set up. Then when I saw your answer it made sense. Because of the monotonicity we can take logs of the percentiles to get back to a normal distribution. Since the normal is symmetric we can average the 10th and 90th percentile to get the 50th percentile and this will be the same as the mean. Once again by the monotonicity just backtransform. It's all there in your answer. BGM is right in that we could simplify it to the geometric mean but like I said I think the form of your answer sheds some light on the situation.

**BGM** · 01-15-2011 04:59 AM

Yes I totally agree that using the normal distribution is the most natural way
to tackle the problem, because we are much more familiar with the properties
of the normal distribution

The geometric mean is just a little point to add that there is some analogy
from the lognormal distribution.

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