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

1. ## 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

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

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

3. 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

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

Also $X = e^{Y}$ where $Y$ is normally distributed with parameters $\mu$ and $\sigma$. 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

5. ## 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.