1.	What?s the value of a $100 call option.  (European Option)
2.	What?s the value of a $100 put option.  (European Option)

If:
-Option exercise price = $ 25
-Standard deviation of the continuously compounded return of the stock  = .8
-Time period = 2 years
-Current stock price = $ 20
-Interest rate = 5%

Please use Black-Scholes Model and show work.

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Request for Question Clarification by elmarto-ga on 29 Aug 2005 06:59 PDT

Hello,
Could you please clarify what do you mean by "a $100 call option"?
Does it mean that you bought it for $100 (not that this would have any
effect on its value)?

Best regards,
elmarto

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Clarification of Question by rr64519-ga on 29 Aug 2005 07:42 PDT

Typo.  It should be a $20 call and put option, which matches the stock price.

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Clarification of Question by rr64519-ga on 29 Aug 2005 07:46 PDT

I am incorrect.  Let me keep looking for clarification.

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Clarification of Question by rr64519-ga on 29 Aug 2005 14:26 PDT

Just ignore the $100 for both.  I should have never put them in there.

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Clarification of Question by rr64519-ga on 30 Aug 2005 16:43 PDT

Any update?

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Request for Question Clarification by elmarto-ga on 30 Aug 2005 18:07 PDT

Just one more clarification: the Black-Scholes model results in a
formula that shows the value of a put or call option given the
parameters you've supplied (exercise price, std. dev., etc). Do you
need me to explain how the Black-Scholes model is derived (which would
be a bit lengthy) or just show you the formula, plug your parameters,
and give you the value of the options?

Best regards,
elmarto

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Clarification of Question by rr64519-ga on 31 Aug 2005 07:12 PDT

No need for a derivation.  I know how ugly that is :).  Just plug and play is fine.

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Request for Question Clarification by elmarto-ga on 31 Aug 2005 07:31 PDT

Hello,
Sorry, I forgot to ask just one more thing in my previous request. Is
the standard deviation you provide annualized? Monthly? In any case it
seems quite high at 0.8 (80%). Could you confirm that this number is
right, and whether it is annual, semi-annual, etc?

Thank you very much,
elmarto

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Clarification of Question by rr64519-ga on 31 Aug 2005 09:14 PDT

80% is correct.  What would it normally be and how will the 80% effect
the numbers.  I will crunch the numbers once you are done to see as
well.

Hello!
At last, here's the answer :)

First of all, I'll assume that the standard deviation you provide is
for annual returns. That said, you can find the Black-Scholes formula
at the following site:

Wikipedia
http://en.wikipedia.org/wiki/Black-Scholes

Let's first compute "d1" and "d2" as described in that link. Using the
same notation, you have:

S = 20
K = 25
T = 2
r = 0.05
sigma = 0.8

Plugging these value into the formula gives:

d1 =  0.4568408747...
d2 = -0.6745299752...

Now, the formula for the options value makes use of the standard
normal cummulative distribution function (cdf). Since there is no
explicit formula for this function, I used the online calculator at
the following site:

Normal Calculator
http://cnx.rice.edu/content/m11328/latest/
 [make sure you select "Area left of" in this calculator]

Finally, plugging d1 and d2 in the calculator, and using these results
in the call value formula, we get that the theoretical value of the
call option , according to this model, is:

C = 7.86

Likewise, the theoretical value of the put option is:

P = 10.48

If you want to take a shortcut, you can just use the follwing
Java-based online European option value calculator

Option pricing
http://www.margrabe.com/OptionPricing.html

If you plug the parameters of your problem, you'll get the same
results as with the formula (probably a bit different because I did
some rounding).

Regarding the 80%, forget what I said in the clarification request
about it being quite high. There are lots of stocks with volatility
even higher than 80% annualy.


Google search terms
black scholes formula
://www.google.com.ar/search?hl=es&q=black+scholes+formula&meta=
option calculator black scholes
://www.google.com.ar/search?hl=es&q=option+calculator+black+scholes&meta=


I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
elmarto

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Request for Answer Clarification by rr64519-ga on 01 Sep 2005 13:48 PDT

Thx.  I am reviewing it now.

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Request for Answer Clarification by rr64519-ga on 01 Sep 2005 14:58 PDT

OK.  I am plugging the numbers into the formula for d1 and dont get
the same answer you do.  Can you tell me where I am messing up, see
below:


ln(20/25) = .2231

(.05 + 1/(2*.8*.8))2 =  1.66

sqrt(.8*.8*2) = 1.13

I come up with d1 = 1.88/1.13 = 1.66  ???

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Clarification of Answer by elmarto-ga on 01 Sep 2005 16:00 PDT

Hello!
There are some mistakes in your calculations:

ln(20/25) = -0.2231  (not 0.2231)

(r + (sigma^2)/2)*T = (0.05 + 0.64/2)*2 = 0.74 (not 1.66)

sigma*sqrt(T) = 0.8*sqrt(2) = 1.1313... (this one is correct)

Therefore, 

d1 = (-0.2231 + 0.74)/1.1313 = 0.45...


Please let me know if you have any other doubt regarding my answer.

Best wishes!
elmarto