By Kerry Back
"Deals with pricing and hedging monetary derivatives.… Computational tools are brought and the textual content comprises the Excel VBA workouts equivalent to the formulation and strategies defined within the publication. this can be beneficial due to the fact that machine simulation will help readers comprehend the theory….The book…succeeds in providing intuitively complicated spinoff modelling… it presents an invaluable bridge among introductory books and the extra complicated literature." --MATHEMATICAL REVIEWS
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Extra resources for A course in derivative securities : introduction to theory and computation
Another Risky Asset as the Numeraire When Y is the numeraire, Z(t) deﬁned as Z(t) = V (t) Y (t) must be a martingale. Using again the rule for ratios, we have dV dY dZ dV dY dY = − − + Z V Y V Y Y dY dV − − ρσs σy dt + σy2 dt = V Y dY dS − + (q − ρσs σy dt + σy2 ) dt . = S Y 2 We can apply our previous example to compute the dynamics of Y when Y is the numeraire. This shows that the drift of dY /Y is (r + σy2 ) dt. Because the drift of dZ/Z must be zero, it follows that the drift of dS/S is (r−q+ρσs σy ) dt.
17), using V as the numeraire, we have Y (0) = S(0)E V Y (T ) eqT S(T ) = e−qT S(0)E V [x] . As in the previous section, E V [x] = probV (x = 1), so we need to compute this probability of the option ﬁnishing in the money. We follow the same steps as in the previous section. 28) we have dS = (r − q + σ 2 ) dt + σ dB ∗ , S where now B ∗ denotes a Brownian motion when V is the numeraire. This is equivalent to 1 d log S = r − q + σ 2 dt + σ dB ∗ . 35), with α = r − q + σ 2 /2, we have probV S(T ) > K = N(d1 ) , where 52 3 Black-Scholes log d1 = S(0) K + r − q + 12 σ 2 T √ .
Consider the security that pays this number of dollars at time T . Because we obtained it with a trading strategy that required no investment at time t, its price at time 0 must be 0. We already observed that we can represent the price in terms of state prices, so we conclude that E φ(T )1A Y (T ) − Y (t) S(T ) S(t) =0 . When we divide by S(0), this will still equal zero. Factoring S(T ) outside the parentheses gives 11 The proof is due to Harrison and Kreps . See also Geman, El Karoui and Rochet .