{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4703-10-Notes-sensitivities

4703-10-Notes-sensitivities - Copyright c 2007 by Karl...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Copyright c 2007 by Karl Sigman 1 Estimating sensitivities When estimating the Greeks, such as the Δ, the general problem involves a random variable Y = Y ( α ) (such as a discounted payoff) that depends on a parameter α of interest (such as initial price S 0 , or volitility σ , etc.). In addition to estimating the expected value K ( α ) def = E ( Y ( α )) (this might be, for example, the price of an option), we wish to estimate the sensitivity of E ( Y ) with respect to α , that is, the derivative of E ( Y ) with respect to α , K 0 ( α ) = dE ( Y ) = lim h 0 K ( α + h ) - K ( α ) h . 1.1 Sample-path approach If the mapping α Y ( α ) is “nice” enough, we can interchange the order of taking expected value and derivative, dE ( Y ) = E h dY i . (1) Under this scenario, K 0 ( α ) itself is an expected value so we can estimate it by standard Monte Carlo: Simulate n iid copies of dY and take the empirical average. To dispense with the notion that such an interchange as in (1) is always possible (no, it is not!) one merely need consider the Δ of a digital option with payoff Y = Y ( S 0 ) = e - rT I { S ( T ) > K } , where S ( T ) = S 0 e X ( T ) = S 0 e σB ( T )+( r - σ 2 / 2) t . (The risk-neutral probability is being used for pricing purposes.) In this case, dY dS 0 = 0 since the indicator is a piecewise constant function of S 0 , thus E h dY dS 0 i = 0. But E ( Y ) = e - rT P ( S ( T ) > K ) is a nice smooth function of S 0 > 0, with a non-zero derivative. (Yes, the sample paths of Y are not differentiable (nor continuous even) at the value of S 0 for which S ( T ) = K , but P ( S ( T ) = K ) = 0, so this point can be ignored.) On the other hand, a European call option, with payoff Y = e - rT ( S ( T ) - K ) + satisfies dY dS 0 = e - rT e X ( T ) I { S ( T ) > K } which can be re-written as dY dS 0 = e - rT S ( T ) S 0 I { S ( T ) > K } , and indeed it can be proved that (1) holds. The basic condition needed to ensure that the interchange is legitimate is uniform integra- bility of the rvs { h - 1 ( Y ( α + h ) - Y ( α )) } as h 0. We present a sufficient condition for this next. Proposition 1.1 Suppose that Y ( α ) is 1. wp1, differentiable at the point α 0 , and satisfies 2. there exists an interval I = ( α 0 - , α 0 + ) , (some > 0 ), and a non-negative rv B with E ( B ) < such that wp1 | Y ( α 1 ) - Y ( α 2 ) | ≤ | α 1 - α 2 | B, α 1 , α 2 I. Then (1) holds. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Proof : For sufficiently small h , | Y ( α 0 + h ) - Y ( α 0 ) | ≤ hB , wp1, thus | Y ( α 0 + h ) - Y ( α 0 ) | h B, wp1 , and the result follows by the dominated convergence theorem (letting h 0). For the European call option, we have Y ( S 0 + h ) - Y ( S 0 ) e - rT he X ( T ) , so the above proposition applies with B = e - rT e X ( T ) . For the digital call option, Y ( S 0 ) is not differentiable at the point S 0 where S 0 e X ( T ) = K ; it is not even continuous there. A general rule is that if the mapping α Y ( α ) wp1 is continuous at all points, and differentiable except at most a finite number of points, then the interchange will be valid. Even if one can justify the interchange (1), it may not be possible to explicitly compute the derivative dY , thus rendering the sample-path approach impractical. So clearly other methods are needed for estimating K ( α ).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}