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Bama Gambler
08-15-2002, 10:36 AM
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)&gt;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
08-15-2002, 11:01 AM
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
08-15-2002, 11:04 AM
Thanks MathGuy. So LogNormal must have a long right-hand tail.

MathGuy
08-15-2002, 11:14 AM
Here's a random lognormal image I pulled off google:

It does appear to have a long right tail.

Howard Mahler
08-15-2002, 10:52 PM
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
08-16-2002, 10:16 AM
I just read section 2 of "Guide to Loss Models". It clearly states for a distribution with skewness &gt; 0 =&gt; mean &lt; median &lt; mode. Thanks again!!