The RV X is lognormal with parameters µ and ?^2.  Use the moment
generating function to find the mean and the variance of the lognormal
variable X.

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Request for Question Clarification by mathtalk-ga on 20 Nov 2003 18:38 PST

Hi, brianct:

The question can most easily be answered if "the moment generating
function" you speak of is understood to be the moment generating
function of the normal distribution.

The moment generating function of the lognormal distribution is
infinite at all positive arguments, despite the finiteness of moments
of all orders for the lognormal distribution.

However random variable X having lognormal distribution "with
parameters µ and ?^2" means that ln(X) is normally distributed with
mean µ and variance ?^2.

If you wish a derivation of the mean and variance of the corresponding
lognormal distribution using the moment generating function for the
normal distribution, I'm sure that I or another of the Researchers
will be able to assist you.

regards, mathtalk-ga

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Request for Question Clarification by mathtalk-ga on 26 Nov 2003 15:24 PST

Hi, brianct-ga:

Just checking to see if you are still interested in an Answer to this.
 E-mail notifications were apparently off-line some earlier in the
month, and I wanted to make sure you'd seen my previos RFC.

Thanks,
mathtalk-ga

Let Y be lognormal and X=ln Y. So X is normal(m, s^2). Let MX(t) be
the moment-generating function of X and MY(t) be the moment-generating
function of Y. Then MY(t)=E[exp(tY)]=E[exp(t*exp(X))]. Take the
derivative of this equation with respect to t to obtain:
MY'(t)=E[exp(t*exp(X))*exp(X)]. Take another derivative to obtain:
MY"(t)=E[exp(t*exp(X))*exp(2X)]. Substitute t=0 in these two equations
to obtain: E[Y]=MY'(0)=E[exp(X)], which happens to be MX(1), and
E[Y^2]=MY"(0)=E[exp(2X)], which happens to be MX(2). It is well-known
that MX(t)=exp(m*t+(s*t)^2/2). So E[Y]=exp(m+s^2/2) and
E[Y^2]=exp(2m+2s^2). So VAR[Y]=E[Y^2]-E[Y]^2=exp(2m+s^2)*(exp(s^2)-1).