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#1
05-10-2010, 09:47 AM
 DOPT Machine Member Join Date: Apr 2005 Location: Ellicott City, MD Studying for MFE College: Towson State University 1984 Posts: 73
SOA #27

This problem asks you to find the value of a put minus a call. I don't want to get involved with the matrices in the SOA's solution. (Markov Chains gave me enough fits last year). So I tried to work the problem the same way Abe did for a call option in problem 21 in PE #8. Where am I going wrong here?

Thanks.
Attached Files
 SOA #27.doc (32.0 KB, 85 views)

Last edited by DOPT Machine; 05-10-2010 at 10:00 AM.. Reason: typos
#2
05-10-2010, 10:05 AM
 JoeSpo Member Join Date: Sep 2009 Studying for life Favorite beer: Hoegaarden Posts: 237

I have difficulty doing it that way. If you like probabilities, like me, just try it this way:

For X (and Y), the risk-neutral value at t=1 has to match the expected value of the stock so:

$

X: 100e^{0.1}=200p+50q+0(1-p-q)
Y: 100e^{0.1}=0p+0q+300(1-p-q)
$

$
X: 100e^{0.1}=200p+50q
Y: 100e^{0.1}=300-300p-300q
$

$
X: 600e^{0.1}=1200p+300q
Y: 100e^{0.1}-300=-300p-300q
$

$
600e^{0.1}+100e^{0.1}-300=900p
$

$
p=\frac{600e^{0.1}+100e^{0.1}-300}{900}
$

$
q=\frac{100e^{0.1}-200(\frac{600e^{0.1}+100e^{0.1}-300}{900})}{50}
$

Then we can easily solve for the prices using the multinomial tree: If X goes to $200, payoff is 105. If Y goes to$0, payoff is $95. We buy one Py and Sell one Cx. Combined payoff is$-10 for probability p, and \$95 for probability q.

So

$P_y-C_x=(-10p+95q)e^{-0.1}$

You could also solve for the prices individually:

$C_x=e^{-0.1}(105p)$

$P_y=e^{-0.1}(95p+95q)$

$P_y-C_x=e^{-0.1}(95p+95q)-e^{-0.1}(105p)$

EDIT: I guess it's not liking "probabilities." It's liking statistics (expected values) over algebra (matrices/linear equations)!

And I didn't calculate the numbers out, so if there's an error, I apologize!

Last edited by JoeSpo; 05-10-2010 at 10:27 AM..
#3
05-10-2010, 11:47 AM
 DOPT Machine Member Join Date: Apr 2005 Location: Ellicott City, MD Studying for MFE College: Towson State University 1984 Posts: 73

That's great, thanks!

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