1997 Exam Question 16

Lupica, Paul ( (no email) )
Wed, 31 Mar 1999 13:29:12 -0500

I sat for this exam and I have a copy of it, but I don't have the published
answers. I know I didn't work it out right when I took the exam because I
didn't use the fact that interest was compounded continuously. (Plus it was
marked wrong on my analysis.)

The way this problem is worked out in the CSM study manual is to just use
the shortcut formula for the probability of a rise p:

p = (annual interest rate - downward change) / (upward change - downward
change)

= (16 - (-22.1)) / (28.4 - (-22.1)) = .754

So their answer is 1 - .754 = .246

Not only does this solution not use any continuous compounding, it didn't
answer the question: "What is the probability of a decrease over the next 3
months?"

After doing some preliminary studying, here is how I would now work out the
problem:

First, calculate the standard deviation d of annual returns:

1 + upward change = e^(d * sqrt(t)) ===> 1.284 = e^(d*1) for annual
figures ===> d = .25

Now calculate the 3-month upward and downward changes:

1 + upward change = e^(.25 * sqrt(.25)) ===> upward change = .1331
1 + downward change = 1/1.1331 ===> downward change = -.1175

quarterly interest rate = .04

So the probability of a decrease in the next 3 months is 1 - p, where p =
(.04 - (-.1175)) / (.1331 - (-.1175)) = .628.

Ans: .372

Am I overlooking something?

Thanks,
Paul Lupica
paul.lupica@atlantacasualty.com