LogNormal Question

Hi,

In the CT3 Q&A Bank-Part 1 questions, I am struggling to see the different approaches with the solution for 1.7(vi) and 1.11. 1.7(vi) has X~LogN(2,5), P(X<8)=>P(Z<(ln(8)-2)/sqrt(5)) - ie. it uses mean of 2 and sdev of sqrt(5). However in 1.11 we have X~N(10,4) and P(7.5<X<12.5) = P(X<12.5)-P(X<7.5) but instead of using mean of 10 and stdev of 2 for the Z transform, they use the mean and Var formulas for a LogNormal to convert them first such that mean=2.2830 and var=0.03922. Can someone please tell me the difference in approach - thanks.

Stewart.

 

The information given for the two lognormal variables are slightly different.

Q1.7: X ~ LN(2, 5) => mu = 2 and sigma^2 = 5
Here, the values of mu and sigma are given. These are the mean and variance of log(X) and not X.

Q1.11: X ~ LN with mean 10 and variance 4
Here, the values of mu and sigma are not given but those of the first two (central) moments of X are given.

 

This is a favourite trap of the examiners. Every time a lognormal distribution is mentioned, a little light should go off in your head that you need to check whether you are being given the parameters or the actual mean and variance.

 

Right - So that I can be clear, lets say we have a Random variable X with values {2,3,4,5,6} which is a skewed right distn (lets say), we then take the Ln of each value giving the Random variable Y with values {0.69,1.10,1.34,1.61,1.79) which normalises the distn (lets say). Would we then say that Y~LogN(1.31,0.15) with a mean of 4 and variance of 2?

Thanks.

 

Actually it does not work out like the way you have written ...

What you have done is simply equating sample means and variances.
Whereas the formula which links mu and sigma with mean and variance of a log-normal variable is a theoretical result holding true at the population level.

What you need to remember that if you see X ~ LN(a, b) then it means you are given the mu (= a) and sigma^2 ( = b) values. These incidentally are the population mean and variance of the corresponding Normal variate ie of Log(X). So, Log(X) ~ N(a, b)

If you see that you are given X ~ LN with mean 'm' and variance 'v', it means you need to derive mu and sigma by solving the following set of simultaneous equations (given in the ActEd notes):
m = exp(mu + 0.5 * sigma^2)
v = (exp(sigma^2) - 1) * exp(2*mu + sigma^2)

PS: LN(1.31,0.15) implies mu = 1.31 & sigma^2 = 0.15. This means mean = 3.99 and variance = 2.58 (using the above equations)

 

You're right - sorry for the confusion between population and sample. I guess what I am trying to ask is (and I think you have already answered it but I am keen to make sure I have completely got it) - if X is the skewed distribution which needs to be normalised which has a mean = 3.99 and variance = 2.58, then X~LogN(1.31,0.15). Does that sound right?

Thanks for your persistence - just getting my head back into this space again.

 

I do not think your statement is necessarily true for any skewed distribution.

In statistics when one say 'normalising a variable X' it just means you are transforming X --> (X - mean)/sd. Refer to this wiki page for further information.

The relation stated here only holds for Normal and LogNormal random variables:
X ~ N(mu, sigma^2) <=> Exp(X) ~ LN(mu, sigma^2)
OR
Y ~ LN(mu, sigma^2) <=> log(Y) ~ N(mu, sigma^2)

It just happens that when you apply log transformation to a symmetric Normal variate, the result is a known skewed distribution called Log-Normal.

PS: FYI, there is another pair of distributions that holds similar relationship. It is Gamma and LogGamma. The latter is used widely in risk theory

 

Thanks - I'm clear now.