How to verify sticky delta property on a stochastic volatility modelSticky Delta Property - Heston ModelFor pricing, what types of Exotic Options are suitable using Local Volatility Model or a Stochastic Volatility Model?Black Scholes - how to calculate delta with a vol skewCalibrating stochastic volatility model from price history (not option prices)Standard Stochastic Volatility Models VS Moving Average Stochastic Volatility ModelHull White Stochastic Volatility Model in MatlabHow to use a stochastic volatility model to price a quanto optionEuler discretisation error for stochastic volatility modelDetecting stochastic volatilityHedging error in a stochastic volatility modelSSR definition in Bergomi in relation to sticky strike and sticky delta
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How to verify sticky delta property on a stochastic volatility model
Sticky Delta Property - Heston ModelFor pricing, what types of Exotic Options are suitable using Local Volatility Model or a Stochastic Volatility Model?Black Scholes - how to calculate delta with a vol skewCalibrating stochastic volatility model from price history (not option prices)Standard Stochastic Volatility Models VS Moving Average Stochastic Volatility ModelHull White Stochastic Volatility Model in MatlabHow to use a stochastic volatility model to price a quanto optionEuler discretisation error for stochastic volatility modelDetecting stochastic volatilityHedging error in a stochastic volatility modelSSR definition in Bergomi in relation to sticky strike and sticky delta
$begingroup$
Given a stochastic model for the evolution of St, with a given SDE for its volatility, how can you tell if the given model satisfy the sticky delta (or the sticky strike) property? Is it possible to prove analytically this property? Or the only way is to actually compute the prices?
stochastic-volatility
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Given a stochastic model for the evolution of St, with a given SDE for its volatility, how can you tell if the given model satisfy the sticky delta (or the sticky strike) property? Is it possible to prove analytically this property? Or the only way is to actually compute the prices?
stochastic-volatility
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03
add a comment |
$begingroup$
Given a stochastic model for the evolution of St, with a given SDE for its volatility, how can you tell if the given model satisfy the sticky delta (or the sticky strike) property? Is it possible to prove analytically this property? Or the only way is to actually compute the prices?
stochastic-volatility
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given a stochastic model for the evolution of St, with a given SDE for its volatility, how can you tell if the given model satisfy the sticky delta (or the sticky strike) property? Is it possible to prove analytically this property? Or the only way is to actually compute the prices?
stochastic-volatility
stochastic-volatility
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
asked May 27 at 9:24
Giacomo GiannoniGiacomo Giannoni
62
62
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Giacomo Giannoni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03
add a comment |
$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03
$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.
That being said, if you look at the following quantity
$$ Delta = left. fracpartial Vpartial S_0 rightvert_Theta $$
that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $Theta = (v_0,theta,kappa,rho,xi)$ in Heston), then you can say that:
For a (log-) space homogeneous diffusion model, $Delta = left. fracpartial Vpartial S_0 rightvert_Theta$ will be a sticky-moneyness Delta.
A (log-) space homogeneous model is simply one where
$$ fracdS_tS_t = mu_t dt + sigma_t dW_t $$
where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T geq t$.
As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$
$$ C(xi S_0, xi K, T; Theta) = xi C(S_0, K, T; Theta), ,,forall xi > 0 $$
such that (Euler's theorem, or just deriving the above wrt $xi$ and setting $xi = 1$
$$ C = Delta S_0 + fracpartial Cpartial K K tag1 $$
Now, if you assume the model generates a volatility surface $Sigma(S_0;K,T,Theta)$ where $Sigma$ is the function defined through
$$ C(S_0,K,T;Theta) := C_BS(S_0, K, T; Sigma(S_0,K,T;Theta)) $$
then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that
$$ fracpartial Sigmapartial S_0(S_0,K,T;Theta) = -fracKS_0 fracpartial Sigmapartial K(S_0,K,T;Theta) tag2 $$
which is indeed the definition of the sticky-moneyness rule.
Indeed, sticky moneyness suggests that
$$ Sigma(S_0+delta S_0, K, T) = Sigma(S_0, K^*, T) $$
provided, as the name indicates, that $$fracK^*S_0 = fracKS_0+delta S_0 iff K^* = K(1 + delta S_0/S_0)^-1$$
Under such circumstances,
beginalign
fracpartial Sigmapartial S_0(S_0, K, T) &= lim_delta S_0 to 0 fracSigma(S_0+delta S_0, K, T) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 + delta S_0/S_0)^-1, Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 - delta S_0/S_0), Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta K to 0 fracSigmaleft(S_0, K-delta K, Tright) - Sigma(S_0, K, T)fracS_0Kdelta K nonumber\
&= -fracKS_0 fracpartial Sigmapartial K(S_0, K, T)
endalign
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
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1 Answer
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1 Answer
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$begingroup$
I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.
That being said, if you look at the following quantity
$$ Delta = left. fracpartial Vpartial S_0 rightvert_Theta $$
that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $Theta = (v_0,theta,kappa,rho,xi)$ in Heston), then you can say that:
For a (log-) space homogeneous diffusion model, $Delta = left. fracpartial Vpartial S_0 rightvert_Theta$ will be a sticky-moneyness Delta.
A (log-) space homogeneous model is simply one where
$$ fracdS_tS_t = mu_t dt + sigma_t dW_t $$
where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T geq t$.
As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$
$$ C(xi S_0, xi K, T; Theta) = xi C(S_0, K, T; Theta), ,,forall xi > 0 $$
such that (Euler's theorem, or just deriving the above wrt $xi$ and setting $xi = 1$
$$ C = Delta S_0 + fracpartial Cpartial K K tag1 $$
Now, if you assume the model generates a volatility surface $Sigma(S_0;K,T,Theta)$ where $Sigma$ is the function defined through
$$ C(S_0,K,T;Theta) := C_BS(S_0, K, T; Sigma(S_0,K,T;Theta)) $$
then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that
$$ fracpartial Sigmapartial S_0(S_0,K,T;Theta) = -fracKS_0 fracpartial Sigmapartial K(S_0,K,T;Theta) tag2 $$
which is indeed the definition of the sticky-moneyness rule.
Indeed, sticky moneyness suggests that
$$ Sigma(S_0+delta S_0, K, T) = Sigma(S_0, K^*, T) $$
provided, as the name indicates, that $$fracK^*S_0 = fracKS_0+delta S_0 iff K^* = K(1 + delta S_0/S_0)^-1$$
Under such circumstances,
beginalign
fracpartial Sigmapartial S_0(S_0, K, T) &= lim_delta S_0 to 0 fracSigma(S_0+delta S_0, K, T) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 + delta S_0/S_0)^-1, Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 - delta S_0/S_0), Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta K to 0 fracSigmaleft(S_0, K-delta K, Tright) - Sigma(S_0, K, T)fracS_0Kdelta K nonumber\
&= -fracKS_0 fracpartial Sigmapartial K(S_0, K, T)
endalign
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
$endgroup$
add a comment |
$begingroup$
I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.
That being said, if you look at the following quantity
$$ Delta = left. fracpartial Vpartial S_0 rightvert_Theta $$
that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $Theta = (v_0,theta,kappa,rho,xi)$ in Heston), then you can say that:
For a (log-) space homogeneous diffusion model, $Delta = left. fracpartial Vpartial S_0 rightvert_Theta$ will be a sticky-moneyness Delta.
A (log-) space homogeneous model is simply one where
$$ fracdS_tS_t = mu_t dt + sigma_t dW_t $$
where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T geq t$.
As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$
$$ C(xi S_0, xi K, T; Theta) = xi C(S_0, K, T; Theta), ,,forall xi > 0 $$
such that (Euler's theorem, or just deriving the above wrt $xi$ and setting $xi = 1$
$$ C = Delta S_0 + fracpartial Cpartial K K tag1 $$
Now, if you assume the model generates a volatility surface $Sigma(S_0;K,T,Theta)$ where $Sigma$ is the function defined through
$$ C(S_0,K,T;Theta) := C_BS(S_0, K, T; Sigma(S_0,K,T;Theta)) $$
then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that
$$ fracpartial Sigmapartial S_0(S_0,K,T;Theta) = -fracKS_0 fracpartial Sigmapartial K(S_0,K,T;Theta) tag2 $$
which is indeed the definition of the sticky-moneyness rule.
Indeed, sticky moneyness suggests that
$$ Sigma(S_0+delta S_0, K, T) = Sigma(S_0, K^*, T) $$
provided, as the name indicates, that $$fracK^*S_0 = fracKS_0+delta S_0 iff K^* = K(1 + delta S_0/S_0)^-1$$
Under such circumstances,
beginalign
fracpartial Sigmapartial S_0(S_0, K, T) &= lim_delta S_0 to 0 fracSigma(S_0+delta S_0, K, T) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 + delta S_0/S_0)^-1, Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 - delta S_0/S_0), Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta K to 0 fracSigmaleft(S_0, K-delta K, Tright) - Sigma(S_0, K, T)fracS_0Kdelta K nonumber\
&= -fracKS_0 fracpartial Sigmapartial K(S_0, K, T)
endalign
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
$endgroup$
add a comment |
$begingroup$
I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.
That being said, if you look at the following quantity
$$ Delta = left. fracpartial Vpartial S_0 rightvert_Theta $$
that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $Theta = (v_0,theta,kappa,rho,xi)$ in Heston), then you can say that:
For a (log-) space homogeneous diffusion model, $Delta = left. fracpartial Vpartial S_0 rightvert_Theta$ will be a sticky-moneyness Delta.
A (log-) space homogeneous model is simply one where
$$ fracdS_tS_t = mu_t dt + sigma_t dW_t $$
where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T geq t$.
As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$
$$ C(xi S_0, xi K, T; Theta) = xi C(S_0, K, T; Theta), ,,forall xi > 0 $$
such that (Euler's theorem, or just deriving the above wrt $xi$ and setting $xi = 1$
$$ C = Delta S_0 + fracpartial Cpartial K K tag1 $$
Now, if you assume the model generates a volatility surface $Sigma(S_0;K,T,Theta)$ where $Sigma$ is the function defined through
$$ C(S_0,K,T;Theta) := C_BS(S_0, K, T; Sigma(S_0,K,T;Theta)) $$
then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that
$$ fracpartial Sigmapartial S_0(S_0,K,T;Theta) = -fracKS_0 fracpartial Sigmapartial K(S_0,K,T;Theta) tag2 $$
which is indeed the definition of the sticky-moneyness rule.
Indeed, sticky moneyness suggests that
$$ Sigma(S_0+delta S_0, K, T) = Sigma(S_0, K^*, T) $$
provided, as the name indicates, that $$fracK^*S_0 = fracKS_0+delta S_0 iff K^* = K(1 + delta S_0/S_0)^-1$$
Under such circumstances,
beginalign
fracpartial Sigmapartial S_0(S_0, K, T) &= lim_delta S_0 to 0 fracSigma(S_0+delta S_0, K, T) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 + delta S_0/S_0)^-1, Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 - delta S_0/S_0), Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta K to 0 fracSigmaleft(S_0, K-delta K, Tright) - Sigma(S_0, K, T)fracS_0Kdelta K nonumber\
&= -fracKS_0 fracpartial Sigmapartial K(S_0, K, T)
endalign
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
$endgroup$
I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.
That being said, if you look at the following quantity
$$ Delta = left. fracpartial Vpartial S_0 rightvert_Theta $$
that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $Theta = (v_0,theta,kappa,rho,xi)$ in Heston), then you can say that:
For a (log-) space homogeneous diffusion model, $Delta = left. fracpartial Vpartial S_0 rightvert_Theta$ will be a sticky-moneyness Delta.
A (log-) space homogeneous model is simply one where
$$ fracdS_tS_t = mu_t dt + sigma_t dW_t $$
where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T geq t$.
As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$
$$ C(xi S_0, xi K, T; Theta) = xi C(S_0, K, T; Theta), ,,forall xi > 0 $$
such that (Euler's theorem, or just deriving the above wrt $xi$ and setting $xi = 1$
$$ C = Delta S_0 + fracpartial Cpartial K K tag1 $$
Now, if you assume the model generates a volatility surface $Sigma(S_0;K,T,Theta)$ where $Sigma$ is the function defined through
$$ C(S_0,K,T;Theta) := C_BS(S_0, K, T; Sigma(S_0,K,T;Theta)) $$
then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that
$$ fracpartial Sigmapartial S_0(S_0,K,T;Theta) = -fracKS_0 fracpartial Sigmapartial K(S_0,K,T;Theta) tag2 $$
which is indeed the definition of the sticky-moneyness rule.
Indeed, sticky moneyness suggests that
$$ Sigma(S_0+delta S_0, K, T) = Sigma(S_0, K^*, T) $$
provided, as the name indicates, that $$fracK^*S_0 = fracKS_0+delta S_0 iff K^* = K(1 + delta S_0/S_0)^-1$$
Under such circumstances,
beginalign
fracpartial Sigmapartial S_0(S_0, K, T) &= lim_delta S_0 to 0 fracSigma(S_0+delta S_0, K, T) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 + delta S_0/S_0)^-1, Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta S_0 to 0 fracSigmaleft(S_0, K(1 - delta S_0/S_0), Tright) - Sigma(S_0, K, T)delta S_0 nonumber \
&= lim_delta K to 0 fracSigmaleft(S_0, K-delta K, Tright) - Sigma(S_0, K, T)fracS_0Kdelta K nonumber\
&= -fracKS_0 fracpartial Sigmapartial K(S_0, K, T)
endalign
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
answered May 27 at 10:28
QuantupleQuantuple
10.8k11545
10.8k11545
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Giacomo Giannoni is a new contributor. Be nice, and check out our Code of Conduct.
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Giacomo Giannoni is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots.
$endgroup$
– will
May 27 at 9:40
$begingroup$
I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question...
$endgroup$
– Giacomo Giannoni
May 27 at 10:03