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Notes On the Book:Paul Wilmott on Quantitative Financeby Paul WilmottJohn L. Weatherwax June 20, 2008IntroductionHere you’ll find some notes that I wrote up as I worked through this excellent book. I’veworked hard to make these notes as good as I can, but I have no illusions that they are perfect.If you feel that that there is a better way to accomplish or explain an exercise or derivationpresented in these notes; or that one or more of the explanations is unclear, incomplete,or misleading, please tell me. If you find an error of any kind – technical, grammatical,typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgmentsin later printings the name of the first person to bring each problem to my attention. wax@alum.mit.edu1
The Random Behavior of AssetsNotes on time-scaling returnsWhen we have prices sampled at fixed times, Si , the discrete model proposed for their returnsRi isSi 1 SiRi mean standard deviation φ ,(1)Siwhere φ is a random draw from a standard Gaussian distribution (mean zero and varianceone). We will call these mean and standard deviation estimates the measured estimates,since they explicitly depend on using the measured prices Si for their estimation. Theyalso correspond to a mean and standard deviation of the returns over the length of timerepresented by the amount of time between the prices Si 1 and Si or ti 1 ti . The questionwe then pose is: given the measured parameters mean and standard deviation how do wemodify these in the case we are interested in the mean return and the standard deviationor the return for timescales different than the measurement time scales. Let µ and σ be thenumerical values of these quantities for returns over the desired timestep length, which wewill take to be Ti 1 Ti . To express this difference in timescales the book defines δt whichis given byti 1 ti.(2)δt Ti 1 TiNote that the above fraction must be dimensionless, that is if the measurement timescalesis in days ti 1 ti 1 day and the desired timescale is in years Ti 1 Ti 1 year, then thevalue of δt should be1 day1 dayδt ,1 year252 daysince there are 252 trading days in one year. Thus if the desired timescale is over longera longer amount of time (where Ti 1 Ti ti 1 ti ) we expect δt 1 and if it is over ashorter amount of time we expect δt 1. Given this definition then we have that the meanreturn we want µ is given by Ti 1 Timeanµ mean .(3)ti 1 tiδtThis is the expression that shows how we scale the mean returns from one timescale toanother. The standard deviation is scaled in a similar manner. If σ is the standard deviationover the timescale of interest we havestandard deviation.(4)σ δt1/2This is the expression that shows how we scale the standard deviation of returns from onetimescale to another. Two simple examples will make this clear. Using the data from thebook we assume that we measure daily returns (using daily prices) and havemean 0.002916standard deviation 0.024521 .Then we want to compute2
The yearly mean and standard deviation of returns. In that case, as we talked aboutabove we have1 0.00396 .δt 252Thus using Equations 3 and 4 we getµ 0.7348 and σ 0.389242 . The hourly mean and standard deviation of returns. In that case we have6.5 hours1 day 6.5 .δt 1 hour1 hoursThen using Equations 3 and 4 we getµ 0.0004486 and σ 0.009617 .Typically we measure the returns over a daily timescale and then report yearly values for µand σ. In that case if we want statistics for returns over a shorter time period (less than ayear) then δt is what fraction of the longer time length the short time length is. For example,in going from the annual mean rate of return µyearly and standard deviation σyearly to adaily rate of return and uncertainty we scale by the appropriate fraction 1µyearlyµdaily 252 1/21σdaily σyearly .252Using these scalings, δt in the return model gives usSi 1 SiRi µδt σφδt1/2 ,Siwhere φ is a draw from a standard normal random variable. Solving for Si 1 we getSi 1 Si µSi δt σSi φδt1/2 .(5)(6)Notes on exponentially weighted volatility estimationFrom the definition of the exponentially weighted estimate of σi2 given by i 1 λ X i j 22λ Rj ,σi δtj (7)we can writeσi2# X "Xii 11 λ1 λλλi 1 j Rj2 λλi 1 j Rj2 λ 1 Ri2 δtδtj j 1 λδt2 λσi 1 λ 1 Ri2δt1 λ 1 λ2 λσi 1 Ri2 ,δt 3(8)
which is a recursive expression for exponentially weighted volatility estimation. If we com1. Ifpute Ri using daily prices and we want σ to be in units of yearly volatility then δt 252you want σ to be an estimate of daily volatility then δt 1.4
Elementary Stochastic CalculusNotes on the mean square limitTo evaluate the expectation after we expand the square ofneed to count how many terms we have in the double sumi 1n XXi 1 j 1Pnj 1 (X(tj ) X(tj 1 ))2 t, we(X(ti ) X(ti 1 ))2 (X(tj ) X(tj 1 ))2 .We can do this simply asi 1n XXi 1 i 11 nXi 1(i 1) nXi 1i nXi 1111 n(n 1) n n(n 1) .22Notes on functions of stochastic variables and Ito’s lemmaFrom the definition of a stochastic integral we have an expression likeW (t) Z0tf (τ )dX(τ ) limn nXj 1f (tj 1 )(X(tj ) X(tj 1)) ,(9)where tj j nt . When this is expressed as a differential relation we havedW f (t)dX .Thus we expect that sums of differences like X(tj ) X(tj 1) seen in Equation 9 play aprominent role in obtaining differential relationships. Given this observation and the factthat we want to evaluate the derivative of F (X) when X is stochastic variable motivates usto consider the following sum of differences (which we denote S) and where the time pointstj are spaced by h δxnS . [F (X(t h)) F (X(t))][F (X(t 2h)) F (X(t h))] [F (X(t 3h)) F (X(t 2h))] [F (X(t (n 1)h)) F (X(t (n 2)h))][F (X(t nh)) F (X(t (n 1)h))] .Note that one way to evaluate S is to note that since it is a telescoping sum that all of the“middle” terms cancel when summed and we are left withS F (X(t nh)) F (X(t)) F (X(t δt)) F (X(t)) .5
Another way to evaluate S is to use Taylor’s series to expand each difference in the functionF (·) in terms of a difference in terms of the stochastic variable X (shown here for the first)asd2 F (X(t))dF (X(t)) 1 (X(t h) X(t))2.F (X(t h)) F (X(t)) (X(t h) X(t))dX2dX 2Then each difference in S above turns into the sum of two terms and we getdF (X(t))dXd2 F (X(t))1(X(t h) X(t))2 2dX 2dF (X(t h)) (X(t 2h) X(t h))dX21dF(X(t h)) (X(t 2h) X(t h))22dX 2.dF (X(t (n 1)h)) (X(t nh) X(t (n 1)h))dX21dF(X(t (n 1)h)) (X(t nh) X(t (n 1)h))22dX 2nXdF (X(t (j 1)h)) (X(t jh) X(t (j 1)h))dXj 1S (X(t h) X(t))(10)nd2 F (X(t (j 1)h))1X(X(t jh) X(t (j 1)h))2. 2 j 1dX 2(11)We consider the two summation terms 10 and 11 above. The first sum above (Equation 10)is a discrete approximation toZ t δtdFdX .dXtFor each term in the second sum above (Equation 11), as argued in the text, we evaluate thesecond derivatives at the left-most end point X(t), so that it comes out of the summation.In addition, the quadratic sum that remains is a discrete approximation in the mean squaredsense ofZ t δt(dX)2 δt .tthus we getF (X(t δt)) F (X(t)) Ztt δt1dF(X(τ ))dX(τ ) dX2Ztt δtd2 F(X(t))dτ .dX 2Note that the argument of the first integral is evaluated at τ the variable of integration,while the argument of the second integral is evaluated at the left most end point t and is aconstant with respect to the variable of integration τ . If we desire to extend this expressiond2 Fto integration lengths t where we cannot just evaluate dX2 at the left-hand end point weneed to evaluate this expression at τ rather than t. This givesZZ tdF1 t d2 FF (X(t)) F (X(0)) (X(τ ))dX(τ ) (X(τ ))dτ .(12)2 0 dX 20 dX6
When we write this using the differential equation shorthand we getdF dF1 d2 FdX dt ,dX2 dX 2(13)for the stochastic differential equation satisfied by F (X).Notes on Ito from TaylorIf we have a variable, say S, that changes according to a stochastic dX and a continuousterm dt asdS a(S, t)dt b(S, t)dX .(14)If we have a function of S say V (S) then we can derive the expression for dV using Taylorseries and the heuristic dX 2 dt. Performing a two term Taylor expansion of V (S) we havedV 1 d2 V 2dVdS dS .dS2 dS 2Using the heuristics discussed in the book we finddS 2 (a(S, t)dt b(S, t)dX)2 a(S, t)2 dt2 2a(S, t)b(S, t)dtdX b(S, t)2 dX 2 b(S, t)2 dt .(15)So that with this dV becomes2dV12d VdV dS b(S, t)dt .dS2dS 2We could replace dS with adt bdXBrownian increment dX to get dVdV a(S, t) dS(16)in the above to get an expression in terms of the1d2 Vb(S, t)2 22dS dt b(S, t)dVdX .dS(17)As another slight generation if V V (S, t), so that V depends on the deterministic time tas well as the stochastic term S then by using Taylor’s series we getdV V1 2V 2 Vdt dS dS t S2 S 2as in Equation 15 we have dS 2 b2 dX 2 b2 dt so dV above becomesdV V V1 2Vdt dS b(S, t)2 2 . t S2 S7(18)
Notes on Ito in Higher DimensionsNow Taylor’s series for a function V V (S1 , S2 , t) of two variables S1 and S2 (and time t)would have first derivative terms for S1 , S2 and t, second derivatives terms for S1 and S2and a cross derivative term for S1 and S2 , given bydV Vdt t1 2V 2 VdSdS1 S12 S12 1 V1 2V 2 dSdS2 S22 S22 2 2VdS1 dS2 . S1 S2 Now as earlier we consider the heuristics dS1 b21 dt, dS22 b22 dt, for the squares of the randomterms anddS1 dS2 (a1 dt b1 dX)(a2 dt b2 dX2 ) b1 b2 dX1 dX2 b1 b2 ρdt ,for the cross product. In that case we thus get for dV the followingdV V V1 2V1 2 2V V 2VdS1 dS2 b21dt dt bdt bbρdt .12 t S1 S22 S1 2 S1 S22 2 S2 2(19)Pertinent Examples: The Lognormal Random WalkIn this case the differential equation for S satisfiesdS µSdt σSdX ,and we will use the heuristics thatdS 2 σ 2 S 2 dX 2 σ 2 S 2 dt .If we consider a function F defined as F (S) log(S) thendF1 dSSandd2 F1 2,2dSSso using the “Ito from Taylor” idea to compute the differential of this function givesdF1 d2 F 2dS dSdS2 dS 2 111 2 σ 2 S 2 dt (µSdt σSdX) S2S 1 µ σ 2 dt σdX .2d(log(S)) dF 8(20)
From the above expression for d log(S) we can integrate this to get 1 2log(S(t)) log(S(0)) µ σ t σ(X(t) X(0)) ,2or solving for S(t) we get1 2(21)S(t) S(0)e(µ 2 σ )t σ(X(t) X(0)) . Since X(t) is a Gaussian process we can write X(t) X(0) φ t where φ is a random drawfrom a N (0, 1) distribution.Pertinent Examples: A Mean Reverting Random WalkIn this case lets consider the stochastic differential equation for r given bydr (ν γr)dt σdX .If we let W r ξ where ξ is not yet determined. We then have thatdW dr (ν γr)dt σdX (ν γ(w ξ))dt σdX (ν γξ γW )dt σdX .Thus if we pick ξ such that ν γξ 0 or ξ νγthenW r ν,γanddW γW dt σdX .(22)Note the drift term in the above stochastic differential equation is γW dt and has a Was a factor while the stochastic term σdX does not have a W factor. This is like a partiallognormal random walk and is called a Ornstein-Uhlenbeck process. Let I eγt and considerd(IW ). We haved(IW ) IdW W dI eγt ( γW dt σdX) γW Idt σeγt dX .This we can integrate from 0 to t to getIW (t) IW (0) σMultiply by1IZteγs dX(s) .0 e γt on both sides to get γt γtZtW (0) σeeγs dX(s)Z t 0 e γt W (0) σeγ(s t) dX(s) .W (t) e09(23)
We can further simplify this by using integration by parts on the last term to getZ tZ ttγ(s t)γ(s t)edX(s) eX(s) s 0 γeγ(s t) X(s)ds00Z t γt X(t) e X(0) γeγ(s t) X(s)ds .0Since we assume that X(0) 0 the second term in the above vanishes. Thus we get forW (t) the following Z t γtγ(s t)W (t) e W (0) σ X(t) γeX(s)ds ,0and for r(t) we get Z tνννγ(s t) γt σ X(t) γr(0) eX(s)ds .r(t) W (t) eγγγ0Pertinent Examples: Another Mean Reverting Random WalkIn this case r is governed bydr (ν µr)dt σr 1/2 dX ,then the heuristics we use is dr 2 σ 2 rdX 2 σ 2 rdt. When we consider the function Fdefined as F r 1/2 we have that1dF r 1/2dr2andd2 F1 r 3/2 .2dr4Thus using Ito from Taylor to compute dF gives usdF1 d2 F 2dr dr2dr 2 dr 1 1/21 3/21 r(σ 2 rdt)rdr 224 14ν σ 2 1 µF dt σdX . 8F22d(r 1/2 ) dF Now the stochastic term is constant (a relatively simple expression) but the coefficient ofthe drift term is more complicated. This leads us to consider if we can find a function F (r)under the same stochastic differential equation for r such that with the help of Ito’s lemmahas a zero drift term for dF . This function F (r) would haved(r 1/2 ) dF dF1 d2 F 2dr drdr2 dr 21 d2 F 2dF((ν µr)dt σr 1/2 dX) σ rdt2 dr 2 dr dF1 2 d2 FdF (ν µr) σ r 2 dt σr 1/2dX .dr2drdr 10
Setting the coefficient of dt equal to zero gives(ν µr)dF1d2 F σ2r 2 0 .dr2dr(24)this equation becomesIf we introduce the function Y (r) defined by Y (r) dFdr dY2(ν µr)2ν2µ Y 2 2 dr .drσ2rσ r σIntegrating both sides giveslog(Y (r)) 2ν2µlog(r) r C,σ2σ2for a constant C. Thus solving for Y (r) we get2ν2µY (r) Ar σ2 e σ2 r11(25)
The Black-Scholes ModelNotes on solutions to the Black-Scholes equationIn this section we show that S and ert satisfy the Black-Scholes equation. The Black-Scholesequation is given by1 2V V V σ 2 S 2 2 rS rV 0 .(26) t2 S SFor V (S, t) AS, we can calculated each of the required derivative on the left hand side ofthis expression as follows V t V S 2V S 2 0 A 0.Thus substituting V AS into the left hand side of the Black-Scholes equation gives0 0 rSA rAs 0 ,showing that V AS is a solution. We note that this solution represents a pure investmentin the underlying. Note that also in this case V A. SIn the case when V Aert we again evaluate each derivative in tern and find that V t V S 2V S 2 rAert 0 0,so placing V Aert into the left hand side of the Black-Scholes equation we obtainrAert rAert 0 ,proving that V Aert is a solution. This solution represents an investment in a fixed interestrate account like a bank account. Note that when V Aert we have 0.12
Notes on options on futuresHere we will transform the Black-Scholes equation in the original variables (t, S) into thenew variables (v, F ) defined in terms of the original variables byv tF er(TF t) S .Note that the above transformation has an inverse given byt vS F e r(TF t) F e r(TF v) .Then the derivatives with respect to (t, S) transform as F v rer(TF t) S t t F t v F v F r F v F v er(TF t) S S F S v F er(TF v) F2 22r(TF v) e S 2 F 2Thus we put these two expressions into the Black-Scholes equation for V V (F, v) we get F r V V 1 2 2 2r(TF v) 2r(TF v) 2 V V (σ F e)e r(F e r(TF v) )er(TF v) rV 0 .2 F v2 F FWhen we cancel terms we get V 1 2 2 2 V σ F rV 0 , v2 F 2the pricing equation for an option on a future.13
Partial Differential EquationsTransformations to a constant coefficient diffusion equationIn this section of these notes we verify the transformations needed to change the BlackScholes equation into the diffusion equation. As suggested in the book lets define unitlessparameters x, τ , and U in terms of the given financial parameters T , t, S etc. asS ex x log(S)1τt T 1 2 τ σ 2 (T t)2σ2V (S, t) eαx βτ U(x, t) .Then with this transformation the derivatives needed in the Black-Scholes equation become τ x 1 σ2 t t τ t x2 τ τ x 1 e x S S τ S x S x x2 2 x 2x x 2x e e e e. S 2 x x x x2Using these the derivatives of V in the Black-Scholes Equation 26 becomes V U1 2 αx βτ1 2 αx βτβU σ(eU) σ e t2 τ2 τ U V e x (eαx βτ U) e(α 1)x βτ αU S x x 2 V U e(α 1)x βτ αU e x2 S x x U U 2U(α 1)x βτ x(α 1)x βταU (α 1)eα e e x x x2 2U U . e(α 2)x βτ α(α 1)U (2α 1) x x2When we put these expressions into the Black-Scholes Equation 26 we get 1 2 αx βτ1 2 αx βτ U U 2U σ e βU α(α 1)U (2α 1)σ e 2 τ2 x x2 U reαx βτ U 0 . reαx βτ αU xWhen we cancel the exponential factor, take the time derivative to one side of the equal sign,and group terms we get β 2 1 21 2 U1 2 2U1 2 Uσ σ σ α(α 1) rα r U σ (2α 1) r σ.2 τ222 x2 x214
If we take1α 2 2r 1σ2 ,(27)then 2α 1 σ2r2 and the coefficient of Uvanishes. With this value for α lets now look at xthe coefficient the U term. When we replace α with the above expression and simplify weget1β σ 2 2 (2r σ 2 )2 .28σTo make this vanish we must take β given by 211 2r2 2β 4 (2r σ ) 1 .(28)4σ4 σ2Using this value for β and canceling the common factor of 12 σ 2 we get the following purediffusion equation for U(x, τ ) 2U U , τ x2as claimed.Notes on Similarity ReductionsIf we consider the functionu(x, t) Zx/t1/21 2e 4 ξ dξ ,(29)0we can show that it satisfies ut uxx . To do this, we compute the needed derivatives. 1 x2x ut e 4 t 3/2 2t 1 x21ux e 4 tt1/2 1 x21x1 1 x2xuxx e 4 t 3/2 e 4 t ,2t2ttfrom which we see directly that ut uxx as claimed.Let us now look for solutions of a particular form motivated by the form of the abovexexpression for u(x, t). Consider u t 1/2 f (ξ) where ξ t1/2. Next we put t 1/2 f (ξ) intout uxx , to see what requirements this imposes on the function f (·). To do this we need tocompute ut and uxx . We find 11 x11 3/2 1/2 ′f (ξ) tf (ξ) 3/2 t 3/2 f (ξ) xt 2 f ′ (ξ)ut t22t22 1ux t 1/2 f ′ (ξ) 1/2 t 1 f ′ (ξ)t 1 1 ′′uxx t f (ξ) 1/2 t 3/2 f ′′ (ξ) .t15
We then put these expressions into the diffusion equation uxx ut we get11t 3/2 f ′′ (ξ) t 3/2 f (ξ) xt 2 f ′ (ξ) .22Multiply this equation by t3/2 and remember that ξ xt1/2to get11f ′′ (ξ) f (ξ) ξf ′(ξ) .22(30)d(ξf (ξ)), and so integrating both sidesNote that the right-hand-side of this equation is 12 dξgives1f ′ (ξ) ξf (ξ) C .2If we take C 0 (we just want to try and find any solution) we getf ′ (ξ)1 ξ,f (ξ)2or integrating both sides and solving for f (ξ) gives1 2f (ξ) De 4 ξ .If we take D 1 we have the function f (·) stated in the book. If we next replace ξ withwe see that1 x2u(x, t) t 1/2 e 4 t ,is a solution to ut uxx .16xt1/2
the Black-Scholes formula and the ’greeks’Derivations of the formula for Calls, Puts, and Simple DigitalsIn this section of these notes we perform and verify my understanding of the derivations suggested in simplifying the Black-Scholes equation. Defining V (S, t) as V (S, t) e r(T t) U(S, t)we have the time derivative of U(S, t) given by U U V re r(T t) U e r(T t) rV e r(T t). t t tWhen we put this into the Black-Scholes Equation 26 we get U1 2U U σ 2 S 2 2 rS 0. t2 S SIf we next let τ T t then the change in the time derivative from t to τ will introduce anegative sign and gives1 U 2U U σ 2 S 2 2 rS. τ2 S SWe next introduce the variable ξ log(S) so that S eξ and find that the S derivativestransform as ξ 1 e ξand S S ξS ξ ξ 2 2 ξ 2ξ ξ 2ξ e e e e. S 2 ξ ξ ξ 2 ξWith this transformation, the Black-Scholes equation becomes U1 2 2ξ 2ξ 2 U ξ Uξ 2ξ U re ee σ e e τ2 ξ 2 ξ ξ 2 U1 U1 σ2 2 r σ2.2 ξ2 ξWhich is a partial differential equation with constant coefficients. Next lets perform a changeof variables on this equation going from the variables (ξ, τ ) to new variables (x, τ ′ ) definedas 1 2x ξ r σ τ2′τ τ,so that the inverse of this transformation is given by 1 2 ′ξ x r σ τ2′τ τ .17
The derivatives in the old coordinates (ξ, τ ) transform to derivatives in the new coordinates(x, τ ′ ) using the chain rule as τ ′ x 1 2 ′ r σ τ τ τ ′ τ x τ2 x′ τ x ,′ ξ ξ τ ξ x xand our differential equation in the new variables (x, τ ′ ) is given by 1 2 U1 2 2U1 2 U U r σ σ r σ, τ ′2 x2 x22 xor dropping the prime on τ we get U1 2U σ2 2 . τ2 xTo solve this equation lets try a solution for U(x, τ ) of the form x x′αU(x, τ ) τ f.τβ(31)(32)Then to verify Equation 31 we need to evaluate τ and x derivatives of U. Defining η asη x x′,τβ(33)we find that the derivatives we need of U are βτ α η ′x x′α 1α 1α ′ ατf(η) f (η)Uτ ατf (η) τ f (η) βτ β 1τταUx β f ′ (η)ττ α ′′Uxx 2β f (η) .τThus Equation 31 becomes1 2 α 2β ′′σ τf (η) τ α 1 (αf (η) βηf ′(η)) .2(34)Equating the powers of η on both sides gives α 2β α 1 so that we getβ 1.2R When we require that U(x, τ ; x′ )dx to be independent of τ means that for the functionalform for U(x, τ ) we are considering and changing the x integration into one over η meansthat Z Z x x′αdx τ α β f (η) dη ,τ fβτ 18
must be independent of τ . This means that α β 0 or1α β .2Once we have α and β we can put these into Equation 34 to get an equation very similar toEquation 30 earlier. Following the same algebraic steps following Equation 30 and specifyingthe constant D so that the function f (·) over to integrates to one, we obtain thefunction form for f (η) given byη21f (η) e 2σ2 ,2πWhen we put η x x′τ 1/2into the expression for W (x, τ ) we finally end with (x x′ )21.exp W (x, τ ) 2σ 2 τ2πτ σSuperimposing fundamental solutions W (x, τ ) for various values of x′ weighted by the payofffunction Payoff(·), and then transforming back into the original S, t variables gives for thesolution V (S, t) of the Black-Scholes equationZ 1 2e r(T t)dS ′22′V (S, t) pPayoff(S ′ )e [log(S /S) (r 2 σ )(T t)] /2σ (T t) ′ .Sσ 2π(T t) 0When there is a dividend yield D the r in the expression above becomes r D orZ 1 2e r(T t)dS ′′22V (S, t) pPayoff(S ′ )e [log(S /S) (r D 2 σ )(T t)] /2σ (T t) ′ .Sσ 2π(T t) 0(35)From this point onward in these notes we will try to be consistent (in this chapter at least)in that we will always include a dividend yield term D in all of our expressions.Notes on the BS Formula for a European CallTo value a European Call recall that the payoff function in that case is given byPayoff(S) max(S E, 0) ,and so that when we put this expression into Equation 35, perform the required integrations,we getC(S, t) Se D(T t) N(d1 ) Ee r(T t) N(d2 )log(S/E) (r D 21 σ 2 )(T t) d1 σ T tlog(S/E) (r D 12 σ 2 )(T t) d2 σ T t d1 σ T tZ x1 21N(x) e 2 φ dφ .2π 19(36)(37)(38)(39)(40)
From the definition of N(x) we can see that1 21N ′ (x) e 2 x2πand N ′′ (x) xN ′ (x) .(41)With these results the expression for N ′ (d2 ) is given in terms of N ′ (d1 ) by 11 211 21 2112N ′ (d2 ) e 2 d2 e 2 (d1 σ T t) e 2 d1 ed1 σ T t e 2 σ (T t)2π2π2π 1 21 21 2 N ′ (d1 )ed1 σ T t e 2 σ (T t) N ′ (d1 )elog(S/E) (r D 2 σ )(T t) e 2 σ (T t)S ′ N (d1 )e(r D)(T t) .(42)ENotes on the BS Formula for a European PutSince we know the analytical expression for a European call we can use Put-Call parity toderive the analytic expression for a European put. From Put-Call parity we have thatC P Se D(T t) Ee r(T t) ,(43)which when we put in the known expression for C(S, t) and solve for P (S, t) we findP (S, t) Se D(T t) (N(d1 ) 1) Ee r(T t) (N(d2 ) 1) Se D(T t) N( d1 ) Ee r(T t) N( d2 ) ,(44)where we have used the fact thatN(d) N( d) 1 .(45)Notes on the BS Formula for a Binary Calls and PutsIf our payoff Payoff(S) is a step function at the strike E i.e. Payoff(S) H(S E), where His the Heaviside function, then from the general expression for the evaluation of the optionprice above in Equation 35 we see thatZ 1 2e r(T t)dS ′′22pH(S ′ E)e [log(S /S) (r D 2 σ )(T t)] /2σ (T t) ′V (S, t) Sσ 2π(T t) 0Z 1 2e r(T t)dS ′′22pe [log(S /S) (r D 2 σ )(T t)] /2σ (T t) ′ . Sσ 2π(T t) ETo evaluate this integral introduce an integration variable v (unrelated to the variable V foroption price) such that log(S ′ /S) (r D 12 σ 2 )(T t) v σ T t1dS ′dv′ dS .dv dS ′σ T t S′20and
′ σWith this our logarithmic differential becomes dST t dv and our integral above′Stransforms toZe r(T t) v2 /2V (S, t) edv e r(T t) N(d2 ) .(46)2πd2when we recall the definition of d2 . This is the formula for the value of a Binary calloption.If our payoff Payoff(S) is instead a step function that turns off at the strike E i.e. Payoff(S) H(E S), where H is the Heaviside function, then from the general expression for theevaluation of the option price above in Equation 35 we see thatZ 1 2dS ′e r(T t)22′pV (S, t) H(E S ′ )e [log(S /S) (r D 2 σ )(T t)] /2σ (T t) ′Sσ 2π(T t) 0ZE1 2dS ′e r(T t)22′pe [log(S /S) (r D 2 σ )(T t)] /2σ (T t) ′ . Sσ 2π(T t) 0To evaluate this integral we again introduce the integration variable v such that log(S ′ /S) (r D 12 σ 2 )(T t) andσ T tdvdS ′1′ dv dS .dS ′σ T t S′ ′T t dv and our integral aboveWith this our logarithmic differential becomes dS σ′Stransforms to ZZ e r(T t) d2 v2 /21 v2 /2 r(T t) V (S, t) edvedv e2π2π d2 Z d21 r(T t) v2 /21 eedv2π e r(T t) (1 N(d2 )) .(47)v again using the definition of d2 . This is the formula for the value of a Binary put option.Notes on the derivation of Delta for some common contractsThis section of the book introduces the notation of an options delta which is denoted as thesymbol . In this section of these notes we will derive all of the given delta expressionspresented in the book. To do this it will be helpful to have the S derivative of d1 and d2 .Using Equations 37 and 38 we see that these are equal and given by d21 d1 . S SσS T t21(48)
The expression for the delta of a European calls given by Equation 36 becomes d2 d1 Ee r(T t) N ′ (d2 ) S S D(T t) ′ r(T t) ′eN (d1 ) EeN (d2 ) e D(T t) N(d1 ) σ T tSσ T t E r(T t) 1 d221 D(T t) 12 d21 D(T t)e 2. eee eN(d1 ) pSσ 2π(T t) e D(T t) N(d1 ) Se D(T t) N ′ (d1 )when we use the expression for N ′ (·) given by Equation 41. Lets consider the twoterms in brackets above. From the definition of d1 and d2 we can replace d2 withd1 σ T t and then expand the square in the second exponent to get 1 2E r(T t) 1 ( 2σd1 T t σ2 (T t))e 2 d1 D(T t) D(T t)e e eN(d1 ) p.e 2Sσ 2π(T t) We now consider the exponent of the third term. Since the product d1 σ T t equals1log(S/E) (r σ 2 )(T t) ,2we get a third term with an exponent of 1 r(T t) d1 σ T t σ 2 (T t)211 r(T t) log(S/E) (r σ 2 )(T t) σ 2 (T t)22 log(S/E) .With this the expression the two terms in brackets becomee D(T t) E D(T t) log(S/E)ee 0.SThus we get for the delta of a European call e D(T t) N(d1 ) ,(49)the expression claimed in the book. The expression for the delta of a European put can be given by taking the S derivativeof Equation 44 or by taking the derivative of the put-call parity relationshipC P Se D(T t) Ee r(T t) ,and using the known delta for a European call. Taking the S derivative of this expression we see that C P e D(T t) , S Sor C P e D(T t) e D(T t) (N(d1 ) 1) ,(50) S Sthe expression claimed in the book.22
The delta for a binary call is given by taking the S derivative of Equation 46, wherewe find r(T t) d2 eN(d2 ) e r(T t) N ′ (d2 ) S S r(T t) ′eN (d2 ) .(51) σS T t The delta for a binary put is given by taking the S derivative of Equation 47, wherewe find r(T t)e r(T t) N ′ (d2 ) ,(52)(e(1 N(d2 ))) SσS T tthe negative of the binary call delta.Notes on the derivation of Gamma for some common contractsHere the book introduces the notation of an options gamma which is denoted as the symbolΓ. In this section of these notes we will derive the given gamma expressions presented in thebook. To derive Γ for a European call we take the S derivative of Equation 49. We find 2C D(T t) d1 eN(d1 ) e D(T t) N ′ (d1 )2 S S S D(T t) ′eN (d1 ) . σS T tΓ (53) To derive Γ for a European put we take the S derivative of Equation 50. We find D(T t) 2C 2P (e(N(d) 1)) 1 S 2 S S 2e D(T t) N ′ (d1 ) ,σS T tthe same as the Γ for a European call.Γ To derive Γ for a binary call using Equation 51 we have e r(T t) N ′′ (d2 )e r(T t) N ′ (d2 )1 e r(T t) N ′ (d2 ) Γ SSσ T tSσ T tσS T tS 2σ T t ′′N (d2 )e r(T t)′ N (d2 ) . σS 2 T t σ T t(54)But we can evaluate N ′′ (d2 ) in terms of N ′ (d2 ) using Equation 41 and we get i e r(T t)d2e r(T t) h′ Γ d2 σ T t N ′ (d2 ) 1 N (d2 ) 2 22σ S (T t)σS T tσ T t r(T t)′ed1 N (d2 ) .(55)22σ S (T t)23
To derive Γ for a binary put recall that since the delta’s for a binary put and a binarycall are the negatives of each other, the gamma for a binary put must be the negativeof the gamma of a binary call and we haveΓ e r(T t) d1 N ′ (d2 ).σ 2 S 2 (T t)(56)Notes on the derivation of Theta for some common contractsHere the book introduces the notation of an options theta which is denoted as the symbol θand defined as V. In this section of these notes we will derive the given theta expressions tpresented in the book. To derive these we will need the Black-Scholes equation with acontinuous dividend yield D which is given by1 V 2V V σ 2 S 2 2 (r D)S rV 0 . t2 S S(57)Thus an options theta can be obtained in terms of its value (V ), its delta ( V), and its S V 2Vgamma ( S 2 ) by solving for t or V1 2V V σ 2 S 2 2 (r D)S rV t2 S S1 σ 2 S 2 Γ (r D)S rV .2We now have everything we need to calculate θ for some options.θ (58) For a European call we haveV Se D(T t) N(d1 ) Ee r(T t) N(d2 ) e D(T t) N(d1 )Γ e D(T t) N ′ (d1 ) ,σS T tso Equation 58 above becomes 1 2 2 e D(T t) N ′ (d1 ) (r D)Se D(T t) N(d1 )θ σ S2σS T t r(Se D(T t) N(d1 ) Ee r(T t) N(d2 ))σSe D(T t) N ′ (d1 ) DSe D(T t) N(d1 ) rEe r(T t) N(d2 ) .2 T t For a European put we haveV Se D(T t) N( d1 ) Ee r(T t) N( d2 ) e D(T t) (N(d1 ) 1) e D(T t) N( d1 )Γ e D(T
which is a recursive expression for exponentially weighted volatility estimation. If we com-pute Ri using daily prices and we want σto be in units of yearly volatility then δt 1 252. If yo
