# How do you derive Mean and Variance of Log-Normal Distribution?

• Jun 24th 2008, 06:39 PM
892king
How do you derive Mean and Variance of Log-Normal Distribution?
Hey,

I found out the probability function

But I don't know how to derive E(X) and Var(X) without using integration.

Thank you, 892king
• Jun 24th 2008, 07:21 PM
tukeywilliams
Use the moment generating function of the normal distribution: $\displaystyle M_{X}(t) = e^{\mu t + \sigma^{2}t^{2}/2}$ to get the moments of the log normal. In particular, suppose that $\displaystyle X$ has a log normal distribution with parameters $\displaystyle \mu$ and $\displaystyle \sigma$. Show that $\displaystyle E(X^n) = e^{n \mu + \frac{1}{2}n^{2}\sigma^{2}}$. It then follows that $\displaystyle E(X) = e^{\mu + \frac{1}{2} \sigma^2}$ and $\displaystyle \text{Var}(X) = e^{2(\mu+\sigma^{2})}- e^{2 \mu + \sigma^{2}}$.

Also $\displaystyle X = e^{Y}$ where $\displaystyle Y$ is normally distributed with parameters $\displaystyle \mu$ and $\displaystyle \sigma$. You use the transformation theorem to get the pdf of the lognormal. Interestingly, the log normal distribution does not have a moment generating function.
• Jun 24th 2008, 08:05 PM
mr fantastic
Quote:

Originally Posted by 892king
Hey,

I found out the probability function

But I don't know how to derive E(X) and Var(X) without using integration.

Thank you, 892king

Of related interest: The Lognormal Distribution
• Jun 24th 2008, 11:16 PM
CaptainBlack
Quote:

Originally Posted by tukeywilliams
Use the moment generating function of the normal distribution: $\displaystyle M_{X}(t) = e^{\mu t + \sigma^{2}t^{2}/2}$ to get the moments of the log normal. In particular, suppose that $\displaystyle X$ has a log normal distribution with parameters $\displaystyle \mu$ and $\displaystyle \sigma$. Show that $\displaystyle E(X^n) = e^{n \mu + \frac{1}{2}n^{2}\sigma^{2}}$. It then follows that $\displaystyle E(X) = e^{\mu + \frac{1}{2} \sigma^2}$ and $\displaystyle \text{Var}(X) = e^{2(\mu+\sigma^{2})}- e^{2 \mu + \sigma^{2}}$.

Also $\displaystyle X = e^{Y}$ where $\displaystyle Y$ is normally distributed with parameters $\displaystyle \mu$ and $\displaystyle \sigma$. You use the transformation theorem to get the pdf of the lognormal. Interestingly, the log normal distribution does not have a moment generating function.

Can this be construed as not using integration?

RonL
• Nov 10th 2009, 04:39 PM
Diego
Use momentgenerating function
Actually using the moment generating fuction is quite easy to derive the moments of the lognormal:

E(exp(t*log(X)) = E(X^t)...(1)

also because log(X) is normal:

E(exp(t*log(X)) = exp(ut+st^2/2)...(2)

Then from (1) and (2) substituting with t = n:

E(X^n) = exp(un+sn^2/2)

From here you can easily obtain the mean and the variance.