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# Greeks

#### ShaktiRathore

##### Well-Known Member
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Hello all,
I am starting a new thread for greeks. I shall discuss all the greeks in coming days. So lets start with the delta.
Delta is the rate of change of option price with respect to the underlying asset.
Delta hedging is maintaining delta neutral portfolio. delta=dC/dS so delta hedging assumes that option price change linearly w.r.t. the underlying without considering the convexity of the option price w.r.t. the underlying. As delta keeps on changing with the price of the underlying in order to maintain a delta neutral portfolio the hedge position must be re-balanced continuously.
If portfolio is currently x dollars of stocks and we wish to hedge it against negative movement of stock prices than we can take (1/delta) position in put options. If portfolio position changes negatively to x-k where k is the downward movement of portfolio than the corresponding payoff for put options position is -delta*change in portfolio position(delta=-dP/dS) or equal to delta*-(-k)=delta*k.So net payoff is total position in put options*delta*k=(1/delta)*delta*k=k. so overall change in position of combination of put options and portfolio is zero(k+(-k)=0).
The delta neutral hedging is ideal for very small changes in the value of the underlying but for good changes in underlying value the hedge position needs to be maintained continuously.
The delta of a European call with on a non dividend paying stock is N(d1) and delta of the European put on the stock is N(d1)-1.

thanks

#### RiskNoob

##### Active Member
Hi ShaktiRathore, David, BT folks,

I do have a question for deriviation of delta for futures.

Let’s start with a definition of delta (for option) delta=dC/dS = N(d1), where C, the (present) value of call option, was derived from the solution of BSM PDE.

So we can apply the same idea for deriving delta for forwards and futures.
The (present) value of forward of non-dividend paying stock S, risk-free rate r is
Value_forward f = S – Ke^(-rT)

Take the derivative with respect to S, df/dS = 1 – 0 = 1, so delta of 1 is obtained for forwards.

From my understanding, due to daily settlement, the above (present) valuation cannot be used for deriving delta for futures. Hull instead used the price of future (F = S*e^(rT)) to derive the delta (dF/dS = e^(rT)) for futures. I am not so convinced about the derivation above.Wonder if any of us could help me out?

RiskNoob

#### ShaktiRathore

##### Well-Known Member
Subscriber
As far I can explain i have done:
consider a future contract with time t with exercise price K,
value of contract for t=1, v1=S-K*e^-r
value of contract for t=2, v2=e^r(S-K*e^-r)
value of contract for t=3, v3=e^2r(S-K*e^-r)
...value of contract for t=T, vT=e^Tr(S-K*e^-r)
so for t=T maturity the value of contract is ,
V=e^Tr(S-K*e^-r)
dV/dS=delta=d/dS(e^Tr(S-K*e^-r))=e^Tr
which is the delta of future contract with maturity T.
You can imagine futures as rollover positions of a series of forward contacts with maturity 1 with value S-K*e^-r. So that the forward contracts are entered sequentially rolling over the positions for time T. so if initial forward contract has value S-K*e^-r than after T=1 the value of S becomes S*e^r and value of K becomes K*e^r so now the new forward contract has value S*e^r-K*e^r*e^-r=e^r*(S-Ke^-r). Noe after T=2 , S*e^r becomes S*e^2r and K*e^2r so that forward contract has now value S*e^2r-K*e^2r*e^-r so this continues till time T when the contract value becomes S*e^Tr-K*e^Tr*e^-r. S that delta changes from e^r to e^rT for a future contract with maturity. So in general for a future contract at time t the delta is e^rt.

thanks

#### ShaktiRathore

##### Well-Known Member
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Now lets discuss theta,
Theta is the rate of change of the derivative w.r.t. the passage of time. The theta of a call or a put is usually negative that is the value of the long option decreases with the passage of time with price and volatility of the underlying asset remaining the same.
The intuition behind is that as the time to maturity approaches the chances of the option price moving favorably decreases. As total volatility increases with time to maturity and as time decays the time to maturity of the option decreases so overall volatility of underlying decreases thereby decreasing the chances of option gaining more or becoming in the money.
value of long option is positively related to overall volatility and maturity.
or value of long option is positively related to vol*root(T-t) and T-t. (where T-t is the remaining time to maturity at time t.)
or value of long option is positively related to root(T-t) and T-t.
or value of long option is positively related to (-t)
or value of long option is negatively related to (t). hence its proved that long option value decreases with increasing t or with the passage of time.
An exception is deep in the money put. Suppose that at time t put has exercise price X and stock price is near to zero so that the option is deep in the money now tendency for stock price to further move in favorable position is very less but its chances of going in unfavorable position are more so that there is now net time value of option is negative and value option increases with the passage of time till maturity. As option shall now at deep money shall try to reach the maturity as fast as it can so that it does not loose enough money so as more time passes the option value remains same or slightly increases.
thanks

#### RiskNoob

##### Active Member
Hi ShaktiRathore, David, BT folks,

I do have a question for deriviation of delta for futures.

Let’s start with a definition of delta (for option) delta=dC/dS = N(d1), where C, the (present) value of call option, was derived from the solution of BSM PDE.

So we can apply the same idea for deriving delta for forwards and futures.
The (present) value of forward of non-dividend paying stock S, risk-free rate r is
Value_forward f = S – Ke^(-rT)

Take the derivative with respect to S, df/dS = 1 – 0 = 1, so delta of 1 is obtained for forwards.

From my understanding, due to daily settlement, the above (present) valuation cannot be used for deriving delta for futures. Hull instead used the price of future (F = S*e^(rT)) to derive the delta (dF/dS = e^(rT)) for futures. I am not so convinced about the derivation above.Wonder if any of us could help me out?

RiskNoob
Thanks ShaktiRathore for the explanation RiskNoob

#### ShaktiRathore

##### Well-Known Member
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Today lets discuss about the greek called Gamma,
Gamma is nothing but rate of change of delta w.r.t. the price of the underlying asset. Gamma is highest for the options that are close to At the money.
As option price does not change linearly but non linearly w.r.t the underlying asset price the delta hedging does not take into account the curvature in the price of option, in order to take this curvature into account for hedging gamma comes into effect.
for A delta neutral portfolio change in portfolio value is,
change in portfolio value= theta*(change in t)+.5*gamma*(change in S)^2

relationship between delta , gamma and theta is given by,
theta+ rS*(delta)+.5sigma^2*S^2*gamma=r*portfolioValue

delta=exp(-qt)*N(d1) where q is dividend yield.
derivation of gamma from delta,
where N(d1)=(1/sqrt(2pi))((-inf to d1) integral of exp(-y^2/2)dy))
N'(d1)=(1/sqrt(2pi))*(exp(-d1^2/2)
differentiating delta w.r.t S,
d(delta)/dS=d/dS(exp(-qt)*N(d1))
d(delta)/dS=exp(-qt)*d/dS(N(d1))
d(delta)/dS=exp(-qt)*(d(d1)/dS*d(N(d1)/dd1)
d(delta)/dS=exp(-qt)*((1/S*sigma*root(T))*d(N(d1)/dd1)
d(delta)/dS=exp(-qt)*((1/S*sigma*root(T))* (1/sqrt(2pi))*(exp(-d1^2/2) )
d(delta)/dS=exp(-qt)*((1/S*sigma*root(T))*N'(d1))
d(delta)/dS=exp(-qt)*N'(d1)/S*sigma*root(T)
so gamma of option is exp(-qt)*N'(d1)/S*sigma*root(T).
thanks

#### ShaktiRathore

##### Well-Known Member
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Those of you wondering where the delta came from in the above derivation,
call price,
c=Sexp(-qT)*N(d1)-X*exp(-rT)*N(d2)
c=Sexp(-qT)*N(d1)-X* exp(-rT)*N(d2)
dc/dS= d/dS[Sexp(-qT)*N(d1)-X exp(-rT)*N(d2)]
dc/dS= d/dS[Sexp(-qT)N(d1)]- exp(-rT)*d/dS[XN(d2)]
dc/dS= exp(-qT) [N(d1)+S dN(d1)/dS]-X exp(-rT)d/dS[N(d2)]
dc/dS= exp(-qT) [N(d1)+S dN(d1)/dd1*dd1/dS]-X exp(-rT) [dN(d2)/d2*dd2/dS)]
dc/dS= exp(-qT) [N(d1)+S *(N’(d1))*(1/S*sigma*root(T))]-X exp(-rT) [(1/S*sigma*root(T))*N’(d2))]….1
Assuming delta to be constant w.r.t S for very small changes in S=>
N(d1)=const
Differentiating w.r.t S both sides,
N’(d1)=0……2
Also d2= d1-sigma*root(T)=>(1/root(2pi))*exp(-.5*d2^2)=(1/root(2pi))*exp(-.5*d1^2)-sigma*root(T)
integrating both sides w.r.t x,
N(d1)=N(d2)-sigma*root(T)*x
Differentiating w.r.t S both sides,
N’(d2)= N’(d1)…..3
From 1,2 and 3,
dc/dS= exp(-qT) *[N(d1)+S *(0)*(1/S*sigma*root(T))]-X[(1/S*sigma*root(T))*0))]
dc/dS= exp(-qT) *N(d1)
or delta of option is exp(-qT) *N(d1)

thanks

#### ShaktiRathore

##### Well-Known Member
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Hi today is turn of vega,
Vega is nothing but rate of change of derivatives portfolio w.r.t. volatility of the underlying asset. Vega tends to be greatest for the option that are close to the money. derivation of vega,
c=Sexp(-qT)*N(d1)-X* exp(-rT)*N(d2)
dc/dσ= d/dσ [Sexp(-qT)*N(d1)-X exp(-rT)*N(d2)]
dc/dσ = d/dσ [Sexp(-qT)N(d1)]- exp(-rT)*d/dσ [XN(d2)]
dc/dσ = exp(-qT) [S dN(d1)/dσ]-X exp(-rT)d/dσ [N(d2)]
dc/dσ = exp(-qT) [S dN(d1)/dd1*dd1/dσ]-X exp(-rT) [dN(d2)/d2*dd2/dσ)]
dc/dσ = exp(-qT) [S N’(d1)* dd1/dσ]-X exp(-rT) [N’(d1- σ√T)* d(d1- σ√T)/dσ]
dc/dσ = exp(-qT) [S N’(d1)* dd1/dσ]-X exp(-rT) [N’(d1- σ√T)* [d(d1)/ dσ - √T]]
assume that d1- σ√T=0 so that dd1/dσ=√T
=>
dc/dσ= exp(-qT) [S N’(d1)* √T]=vega

thanks

#### ShaktiRathore

##### Well-Known Member
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Derivation of theta:

c=Sexp(-q(T-t)*N(d1)-X* exp(-r(T-t))*N(d2)
c=T1-T2
dc/dt=d(T1)/dt-d(T2)/dt
d(T1)/dt=d/dt(Sexp(-q(T-t)*N(d1))
d(T1)/dt=(qSexp(-q(T-t)*N(d1)+Sexp(-q(T-t)*dN(d1)/dt)
d(T1)/dt=(qSexp(-q(T-t)*N(d1)+Sexp(-q(T-t)*dN(d1)/dd1*dd1/dt)
d1=[ln(S/X)+(T-t)(r+.5σ²)]/σ√T-t
dd1/dt=√T-t* [-(r+.5σ²)]+[ln(S/X)+(T-t)(r+.5σ²)]*.5*(T-t)^-.5/σ(T-t)
dd1/dt=[ln(S/X)-(T-t)(r+.5σ²)]/2σ(T-t)^1.5
let ln(S/X)-(T-t)r=0 or X=S*e^-r(T-t)
=> dd1/dt=-.5σ²(T-t)]/2σ(T-t)^1.5
dd1/dt=-.5σ/2(T-t)^.5
d(T1)/dt=(qSexp(-q(T-t)*N(d1)-Sexp(-q(T-t)*N’(d1)*(.5σ/2(T-t)^.5)

d(T2)/dt=d/dt(X* exp(-r(T-t))*N(d2))
d(T2)/dt=Xd/dt(exp(-r(T-t))*N(d2))
d(T2)/dt=(rXexp(-r(T-t)*N(d2)+Xexp(-r(T-t)*dN(d2)/dt)
d(T2)/dt=(rXexp(-r(T-t)*N(d2)+Xexp(-r(T-t)*dN(d2)/dd2*dd2/dt)
dd2/dt=√T-t* [-(r-.5σ²)]+[ln(S/X)+(T-t)(r-.5σ²)]*.5*(T-t)^-.5/σ(T-t)
dd2/dt=[ln(S/X)-(T-t)(r-.5σ²)]/2σ(T-t)^1.5
dd2/dt=[(T-t)(.5σ²)]/2σ(T-t)^1.5
dd2/dt=[(.5σ)]/2(T-t)^.5
d(T2)/dt=(rXexp(-r(T-t)*N(d2)+Xexp(-r(T-t)*N’(d2)*(.5σ)]/2(T-t)^.5)
dc/dt=
(qSexp(-q(T-t)*N(d1)-Sexp(-q(T-t)*N’(d1)*(.5σ/2(T-t)^.5)-[(rXexp(-r(T-t)*N(d2)+Xexp(-r(T-t)*N’(d2)*(.5σ)]/2(T-t)^.5)]
dc/dt= [qSexp(-q(T-t)*N(d1)-rXexp(-r(T-t)*N(d2)]-(.5σ/2(T-t)^.5)[ Sexp(-q(T-t)*N’(d1)+ S*N’(d2)]
Assume N’(d1)*exp(-q(T-t))=N’(d2)
=> dc/dt= [qSexp(-q(T-t)*N(d1)-rXexp(-r(T-t)*N(d2)]-(.5σ/2(T-t)^.5)*exp(-q(T-t))*[ 2S*N’(d1)]
=> dc/dt= [qSexp(-q(T-t)*N(d1)-rXexp(-r(T-t)*N(d2)]-(σ/2(T-t)^.5)*exp(-q(T-t))*[S*N’(d1)]
dc/dt=theta
[qSexp(-q(T-t)*N(d1)-rXexp(-r(T-t)*N(d2)]- exp(-q(T-t)) (S*N’(d1)σ/2(T-t)^.5)

thanks

#### ShaktiRathore

##### Well-Known Member
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Lets discuss rho,
Rho is the rate of change of value of a derivative w.r.t the interest rate
Derivation of rho,
c=Sexp(-q(T-t)*N(d1)-X* exp(-r(T-t))*N(d2)
dc/dr=d/dr[Sexp(-q(T-t)*N(d1)-X* exp(-r(T-t))*N(d2)]
dc/dr=d/dr[Sexp(-q(T-t)*N(d1)]-d/dr[X* exp(-r(T-t))*N(d2)]
dc/dr=[Sexp(-q(T-t)*d/drN(d1)]-[X* exp(-r(T-t))*d/drN(d2)+XN(d2)d/dr(exp(-r(T-t))]
dc/dr=[Sexp(-q(T-t)*d/dd1N(d1)*dd1/dr]-[X* exp(-r(T-t))*d/dd2N(d2)*dd2/dr+XN(d2)(-(T-t)exp(-r(T-t))]
dc/dr=[Sexp(-q(T-t)*N('d1)*dd1/dr]-[X* exp(-r(T-t))*N'(d2)*dd2/dr+XN(d2)(-(T-t)exp(-r(T-t))]
d1=[ln(S/X)+(T-t)(r+.5σ²)]/σ√T-t
dd1/dr=√T-t/σ
dd2/dr=√T-t/σ
dc/dr=[Sexp(-q(T-t)*N('d1)* √T-t/σ ]-[X* exp(-r(T-t))*N'(d2)* √T-t/σ +XN(d2)(-(T-t)exp(-r(T-t))]
dc/dr=√(T-t/σ)[Sexp(-q(T-t)*N('d1)-X* exp(-r(T-t))*N'(d2)]+XN(d2)(-(T-t)exp(-r(T-t))]
Sexp(-q(T-t)*N('d1)-X* exp(-r(T-t))*N'(d2)=0
dc/dr=XN(d2)(-(T-t)exp(-r(T-t))
thanks

#### ShaktiRathore

##### Well-Known Member
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Lets discuss rho,
Rho is the rate of change of value of a derivative w.r.t the interest rate
Derivation of rho,
c=Sexp(-q(T-t)*N(d1)-X* exp(-r(T-t))*N(d2)
dc/dr=d/dr[Sexp(-q(T-t)*N(d1)-X* exp(-r(T-t))*N(d2)]
dc/dr=d/dr[Sexp(-q(T-t)*N(d1)]-d/dr[X* exp(-r(T-t))*N(d2)]
dc/dr=[Sexp(-q(T-t)*d/drN(d1)]-[X* exp(-r(T-t))*d/drN(d2)+XN(d2)d/dr(exp(-r(T-t))]

dc/dr=[Sexp(-q(T-t)*d/dd1N(d1)*dd1/dr]-[X* exp(-r(T-t))*d/dd2N(d2)*dd2/dr+XN(d2)(-(T-t)exp(-r(T-t))]

dc/dr=[Sexp(-q(T-t)*N('d1)*dd1/dr]-[X* exp(-r(T-t))*N'(d2)*dd2/dr+XN(d2)(-(T-t)exp(-r(T-t))]
d1=[ln(S/X)+(T-t)(r+.5σ²)]/σ√T-t
dd1/dr=√T-t/σ
dd2/dr=√T-t/σ

dc/dr=[Sexp(-q(T-t)*N('d1)* √T-t/σ ]-[X* exp(-r(T-t))*N'(d2)* √T-t/σ +XN(d2)(-(T-t)exp(-r(T-t))]

dc/dr=√(T-t/σ)[Sexp(-q(T-t)*N('d1)-X* exp(-r(T-t))*N'(d2)]+XN(d2)(-(T-t)exp(-r(T-t))]
Sexp(-q(T-t)*N('d1)-X* exp(-r(T-t))*N'(d2)=0
dc/dr=XN(d2)(-(T-t)exp(-r(T-t))
thanks

#### ShaktiRathore

##### Well-Known Member
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Applying put call parity to call we calculate the put prices:
p+Sexp(-qT)=c+X*exp(-rT).......PCP
1) delta of put:
differentiating both sides of PCP w.r.t S,
dp/dS+ exp(-qT)=dc/dS
dp/dS=dc/dS- exp(-qT)=delta of call- exp(-qT)
dp/dS=exp(-qT) *N(d1)- exp(-qT)
dp/dS=exp(-qT) *[N(d1)- 1]
dp/dS=-exp(-qT) *[N(-d1)]
2) gamma of put:
differentiating both sides of PCP twice w.r.t S,
d2p/dS2=d2c/dS2
gamma of put=gamma of call
3) theta of put:
differentiating both sides of PCP w.r.t t,
dp/dt+Sd/dt*exp(-q(T-t))=dc/dt+rX*exp(-r(T-t))
dp/dt+qS*exp(-q(T-t))=dc/dt+rX*exp(-r(T-t))
dp/dt= dc/dt+rX*exp(-r(T-t))-qS*exp(-q(T-t))
dp/dt= [qSexp(-q(T-t)*N(d1)-rXexp(-r(T-t)*N(d2)]- exp(-q(T-t)) (S*N’(d1)σ/2(T-t)^.5)+rX*exp(-r(T-t))-qS*exp(-q(T-t))
dp/dt= qSexp(-q(T-t))*(N(d1)-1)-rXexp(-r(T-t)*(N(d2)-1)- exp(-q(T-t)) (S*N’(d1)σ/2(T-t)^.5)
dp/dt= -qSexp(-q(T-t))*(N(-d1))+rXexp(-r(T-t))*(N(-d2))- exp(-q(T-t)) (S*N’(d1)σ/2(T-t)^.5)
3) vega of put:
differentiating both sides of PCP w.r.t σ,
dp/dσ=dc/dσ
ðVega of put=vega of call
4)rho of put:
differentiating both sides of PCP w.r.t r,
dp/dr=dc/dr-TX*exp(-rT)
dp/dr= XN(d2)((T) exp(-rT)-TX*exp(-rT)
dp/dr= TX(exp(-r(T))(N(d2)-1)
dp/dr= -TX exp(-r(T)N(-d2)
Thanks

#### ShaktiRathore

##### Well-Known Member
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This completes greeks theory from my side. any queries on how above results came and any mistake are wellcomed....

thanks

#### caramel

##### Member
hey Shakti , could it be possible to do stuff when gamma is largest , and theta . I mean applications of greeks. The above is very helpful

#### ShaktiRathore

##### Well-Known Member
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Hi Caramel,
yeah that is definitely possible that we find points of extrema for gamma,theta etc. But i think its already very long exercise but we can always find it. yeah if you insist i can do it for you.But these are the basic greek formulas that i derived from Black Scholes model with few assumptions to reach the final result.

thanks

#### ShaktiRathore

##### Well-Known Member
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Caramel we can always find the derivatives of above greeks and equalise them to zero to get extrema. However its also possible to find derivatives of greeks with some other variable .I would illustrate this with following example.

thanks

#### ShaktiRathore

##### Well-Known Member
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dd1/dσ=d/dσ[ln(S/X)+T(r+.5σ^2)/σ*root(T)]
dd1/dσ=[ σ*σT-(ln(S/X)+T(r+.5σ^2))/σ^2*root(T)]
dd1/dσ=[-(ln(S/X)+T(r-.5σ^2))/σ^2*root(T)]
dd1/dσ=[-d2/σ]
dc/dσ= exp(-qT) [S N’(d1)* √T]=vega
d(vega)/dσ= exp(-qT)S*√T*d/dσ(exp(-.5d1^2))
d(vega)/dσ=exp(-qT)S*√T*d/dd1(exp(-.5d1^2))dd1/dσ
d(vega)/dσ=exp(-qT)S*√T*(-d1)d/dd1(exp(-.5d1^2))dd1/dσ
d(vega)/dσ=exp(-qT)S*√T*(-d1)N'(d1)dd1/dσ
d(vega)/dσ=exp(-qT)S*√T*(-d1)N'(d1)[-d2/σ]

d(vega)/dσ=exp(-qT)S*√T*(d1d2)N'(d1)[1/σ]
d(vega)/dσ=vega*(d1d2)*[1/σ]
This is rate of change of vega w.r.t the sigma. This is also called vomma. Vomma measures the sensitivity of vega w.r.t volatility.If we equate above with zero than point of extrema occurs at d1=0 or d2=0 or vega=0. Thus vega is maximum/minimum when any of the above conditions is satisified.
Similarly other second derivatives occurs for delta, gamma and theta.Vanna is rate of change of delta w.r.t volatility. Charm, is the rate of change of delta w.r.t. time or the rate of change of theta w.r.t the underlying asset price.Vera is the rate of change of rho w.r.t volatility. Dvega time is the rate of change of vega with time

thanks

#### ShaktiRathore

##### Well-Known Member
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Vanna =d(delta)/dσ
=d/dσ[exp(-qt)*N(d1)]
=exp(-qt)d/dσ[N(d1)]
=exp(-qt)*d/dd1[N(d1)]*dd1/dσ
we have seen that dd1/dσ=-(d2/σ)
so Vanna==exp(-qt)*d/dd1[N(d1)]*-(d2/σ)
Vanna==-exp(-qt)*[N'(d1)]*(d2/σ).

Vera=d(rho)/dσ
Vera=d//dσ[XN(d2)(-(T-t)exp(-r(T-t))]
Vera=[X(-(T-t)exp(-r(T-t))]*d//dσ(N(d2))
d//dσ(N(d2))=d(N(d2)/dd2*dd2/dσ
d//dσ(N(d2))=(N'(d2)*dd2/dσ
dd2/dσ=(-d1/σ)
d//dσ(N(d2))=(N'(d2)*(-d1/σ)
Vera=[X(-(T-t)exp(-r(T-t))]*(N'(d2)*(-d1/σ)

thanks

#### ShaktiRathore

##### Well-Known Member
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similarly we can find the derivative of option vamma w.r.t the volatility called ultima which is nothing but the double derivative of call w.r.t the volatility.

vamma=d(vega)/dσ=vega*(d1d2)*[1/σ]
ultima=d(vamma)/dσ=d/dσ[vega*(d1d2)*[1/σ]]
ultima=vega*(d1d2)*d/dσ[[1/σ]]+[1/σ]*(d/dσ)*[vega*(d1d2)]
(d/dσ)*[vega*(d1d2)]=vega*(d/dσ)(d1d2)+d1d2*(dvega/dσ)
(d/dσ)*[vega*(d1d2)]=vega*[d2(d/dσ)(d1)+d1(d/dσ)(d2)]+d1d2*( vega*(d1d2)*[1/σ] )
(d/dσ)*[vega*(d1d2)]=vega*[d2(-d2/σ)+d1(-d1/σ)]+d1d2*( vega*(d1d2)*[1/σ] )
(d/dσ)*[vega*(d1d2)]=-vega*[d2^2/σ)+d1^2/σ]+d1d2*( vega*(d1d2)*[1/σ] )
(d/dσ)*[vega*(d1d2)]=-(vega/σ)*[d2^2+d1^2-d1d2*(d1d2) )
ultima=vega*(d1d2)*[-[1/σ^2]]-(vega/σ^2)*[d2^2+d1^2-d1d2*(d1d2) )
ultima=-(vega/σ^2)*[d1d2+d2^2+d1^2-d1d2*(d1d2) )
ultima=-(vega/σ^2)*[d1d2(1-d1d2)+d2^2+d1^2 )
This is ultima. thanks