31 lines
1.6 KiB
Plaintext
31 lines
1.6 KiB
Plaintext
458 Part IV: Additional Considerations
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This approximation is quite accurate for option pricing purposes, since one is not
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really interested in thousandths of a point where option prices are concerned.
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Example: Suppose that XYZ is trading at 45 and we are interested in evaluating the
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July 50 call, which has 60 days remaining until expiration. Furthermore, assume that
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the volatility of XYZ is 30% and that the risk-free interest rate is currently 10%. The
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theoretical value calculation is shown in detail, in order that those readers who wish
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to program the model will have something to compare their calculations against.
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page:
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Initially, determine t, d1, and d2, by referring to the formulae on the previous
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t = 60/365 = .16438 years
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d _ In (45/50) + (.1 + .3 x .3/2) x .16438
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1-
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.3 X ✓.16438
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= -.10536 + (.145 X .16438) = __ 67025 .3 X .40544
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d2 = -.67025 - .3 ✓.16438 = -.67025 - (.3 x .40544) = -.79189
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Now calculate the cumulative normal distribution function for d1 and d2 by
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referring to the above formulae:
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dl = -.67025
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l 1
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y = l + (.2316419 I -.67025 I) = 1.15526 = ·86561
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z = .3989423e--(-.67025 X -.67025)/2
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= .3989423e-0·22462 = .31868
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There are too many calculations involved in the computation of the fifth-order
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polynomial to display them here. Only the result is given:
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X = .74865
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Since we are determining the cumulative normal distribution of a negative
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number, the distribution is determined by subtracting x from l.
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N(d1) = N(-.67O25) = l -x = l - .74865 = .25134
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In a similar manner, which requires computing new values for x, y, and z,
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N(d2) = N(-.79179) = 1- .78579 = .21421 |