ASM Example Problem 16.7

[fishbowl525]

The exchange rate of euros in dollars follows geometric Brownian motion of the form

X(t) = X(0)e^(.005t + .1Z(t))

The current (time 0) exchange rate is 0.9 euros/dollar. Calculate the probability that at time 5, the exchange rate is less than 1 euro/dollar.

The solution uses an alpha of -.005. Simple question: where did the negative come from?

I'm using the 9th edition. Thanks!

[fishbowl525]

If anyone needs me to type the solution, I can do that too.

[Sub_Zero]

It's the conversion between arethmetic and geometric bm.

[Actuarialsuck]

Originally Posted by fishbowl525

If anyone needs me to type the solution, I can do that too.

You need to calculate probability. You are given the normal table. Remember that ABM : Normlas as GBM : LogNormal. If you had a lognormal table to look up probability values from, you wouldn't need a conversion.

[fishbowl525]

Right but when you calculate the mean for the corresponding arithmetic Brownian motion (which is normal), the mean should be (alpha - .5*(sigma^2))*s

In the solution, alpha is negative. I'm asking why it is negative. Isn't alpha just the coefficient of the t in the exponent of the X(0)e^.005t........Hence alpha should be positive and not negative?

[Abraham Weishaus]

I agree, it's a mistake.

A corrected solution is posted in the errata. For this particular question, no adjustment is made to 0.005 when logging. The adjustment referred to by the previous posters is only made when "d"ing the log.

[AdmiralWen]

I have another question regarding this very problem. If you go through the original solution exactly as it is except changing alpha to be positive, you'd eventually get 0.681. On the updated errata, a slightly different approach is used, resulting in the answer 0.3596.

Why is that? Can we not use this alpha value if it leads to a mean of 0?

[Abraham Weishaus]

To repeat, no adjustment should be made to 0.005. The original solution erroneously makes an adjustment.

[seishunrapport]

If that's the case, isn't the solution for Quiz 16.3 incorrect too?

[Abraham Weishaus]

Yes.