This Solutions Manual includes answers to all of the end of chapter problems found in this book (except for a few coding problems where the numerical answer is provided in the text).
Answer: .
We have with . Since we have after substitution:
But and thus . Expanding the second squared bracket and cancelling terms we are left with as required.
In the dollar risk-neutral measure, 1/X is the price of the dollar-euro exchange rate quanto dollar. Following the notations of Section 1.2.4 and defining S = 1/X, we have ρ = −1 and η = σ; thus the drift of S quanto dollar under the dollar risk-neutral measure is r€ − r$ − ρση = r€ − r$ + σ2 as required.
We know the upper bound is:
and we know that so after rearranging terms we get:
We know the lower bound is:
and we know that so we get
From put-call parity: , and thus:
Putting them together we get:
Furthermore and
Thus:
as required.
Differentiating with respect to K:
as required.
Differentiating with respect to σ:
Using the chain rule:
But:
Substituting and :
Using the fact that , we get:
We have:
and:
Substituting back into :
Factoring by and recognizing d1 we obtain as required:
Substituting the identities from (a) we get the required result after further straightforward algebra.
Assume f(K1) = 0 for simplicity. From left to right: start with f′(K1) calls struck at K1 so as to be tangential to the payoff; add calls struck at K2 such that the portfolio matches the payoff f(K3) at K3; then add calls struck at K3 so as to be tangential to the payoff after K3; and so on (see Figure S.2).
From Section 3.2:
Splitting the integral at F and using terminal put-call parity we get:
But , and:
After substitutions and simplifications we get:
Thus:
because .
From we get and thus:
atm_vol = SVI(1); atm_vol1 = SVI(1/(1+epsilon)); price = 0; for i=1:n normal = randn; x = exp(atm_vol*normal*sqrt(T) - 0.5*atm_vol∧2*T); x1 = (1+epsilon)*exp(atm_vol1*normal*sqrt(T) − 0.5*atm_vol1∧2*T); price = (i-1)/i*price + Payoff(x)*ImpDist(x) … / lognpdf(x,-0.5*atm_vol∧2*T,atm_vol*sqrt(T)) / i; price1 = (i-1)/i*price1 + Payoff(x1)*ImpDist(x1) … / lognpdf(x1,-0.5*atm_vol1∧2*T,atm_vol1*sqrt(T)) / i; end delta = price − price1
Take and derive with respect to T:
We are given and so:
We then plug in and into Equation (4.1):
By dividing both numerator and denominator by , we obtain the desired Equation (4.2):
Now using the Hint:
Thus = and by integration we find that:
as required.
Near the money both denominators are close to 1; furthermore gives . After substitution and simplification:
whose leading term is .
But which yields the required result after substitution and simplifications.
Thus , that is, at time t the risk-neutral expected return on the delta-hedged portfolio is the risk-free interest rate plus a positive or negative risk premium, which is proportional to the risk of the portfolio, with proportionality coefficient λ. This result applies to any option and λ is independent from the particular option chosen, which is why it is called the market price of volatility risk.
Negative Path: See Figure S.5
(ii) We have . By iterated expectations:
As a sum of two independent normal distributions the aggregate distribution is normal with parameters:
This result sheds light on the diversification effect obtained by delta-hedging several options within an option book: the expected income is the sum of individual delta-hedging P&Ls, but it is less risky than delta-hedging a single position.
Thus generalized variance may be replicated by a combination of cash, two derivative contracts paying off g(ST) at maturity, and dynamically trading S to maintain a short position of 2g′(St) at all times. Taking expectations we find that the fair value is:
since (dWt) = 0.
(ii) Rewrite and make the approximation that for i = 1,…, n − 1:
which is justified by the fact that αn ≈ 1 since vn and e are assumed to form a tight angle. Proceed similarly for H.
After substitution, we have by Cauchy-Schwarz:
where .
At time 0 the portfolio value is , and the vega of each component is . Hence the portfolio vega is .
Implied correlation is constant iff , i.e., iff , whence the required result after simplifications.
When D = I and U = eeT Langnau's alpha is:
where . Define and divide both the numerator and denominator of the above expression by B to get:
It is easy to verify that .
Thus since diverges. A similar analysis shows that the bound 1 is also nonattracting.
Thus we must solve:
Taking ratios we get ; dividing both numerator and denominator by g2 on the left-hand side we obtain . Solving for p after substituting we find , that is, . Substituting in and simplifying we get , and thus .