Mahler's Notes - Geometric Brownian Motion

[Bama Gambler]

I'll get straight to the point...

The price of a stock is 50 times a Geometric Brownian Motion with parameters 0 and 0.1.

What is the expected value of the stock price at time 9?
X(9)~LogNormal(ln50,0.1*sqrt(9))
E(X(9)) = 52.3 'using expected value of LogNormal formula

What is the prob. that the stock price will be higher than 50 at time 9?
Prob(X(9)>50) = 1 - F(50) = 1 - std. normal[(ln50 - ln50)/.3] = .5

Is this counter intuitive? We expect the stock price to be 52.3 at time 9, but the prob. of the stock price being higher than 50 at time 9 is only a coin toss (.5)? Please HELP!

Bama Gambler

[MathGuy]

I'll have to trust you that your math is correct.

The expected value at time 9, E[X] = 52.3 is the mean of the distrbution, while P(X >= 50) = .5 implies that 50 is the median. Depending upon the shape of the distribution, the mean and median can be in left-right order or right-left order (or they could be equal).

http://syllabus.syr.edu/STT/GRIFFITH/STT101/lec11-12.gif

[Bama Gambler]

Thanks MathGuy. So LogNormal must have a long right-hand tail.

[MathGuy]

Here's a random lognormal image I pulled off google:

http://www.aero.org/publications/crosslink/winter2001/images/02_05.gif

It does appear to have a long right tail.

[Howard Mahler]

Excellent graph Math Guy!

Yes the Lognormal is long-tailed.

See sections 14 and 21 of my Loss Dist. Study Guide. See Questions 14.27-14.29 Spring 2002 edition. (Q. 16.27 to 16.29 Fall 2002 Edition.)

Howard Mahler

[Bama Gambler]

I just read section 2 of "Guide to Loss Models". It clearly states for a distribution with skewness > 0 => mean < median < mode. Thanks again!!