Mixing
SDE for option value
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Given an SDE for an underlying:
$$dS(t) = mu(S,t)dt+sigma(S,t)dW(t)$$
the SDE for the value of the option $V=V(S,t)$ is given via Ito's lemma as:
$$dV = V_tdt+V_Smu(S,t)dt+frac12V_SSsigma^2(S,t)dt+V_Ssigma(S,t)dW(t)$$
It seems that this would results in an SDE containing $S(t)$.
How does one then obtain an SDE for the option value so that it can be simulated directly without simulating the underlyings, i.e. something like
$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$
option-pricing monte-carlo sde
edited 5 hours ago
Daneel Olivaw
2,5431528
asked 5 hours ago
Confounded
836
add a comment |Â
up vote
2
down vote
favorite
Given an SDE for an underlying:
$$dS(t) = mu(S,t)dt+sigma(S,t)dW(t)$$
the SDE for the value of the option $V=V(S,t)$ is given via Ito's lemma as:
$$dV = V_tdt+V_Smu(S,t)dt+frac12V_SSsigma^2(S,t)dt+V_Ssigma(S,t)dW(t)$$
It seems that this would results in an SDE containing $S(t)$.
How does one then obtain an SDE for the option value so that it can be simulated directly without simulating the underlyings, i.e. something like
$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$
option-pricing monte-carlo sde
edited 5 hours ago
Daneel Olivaw
2,5431528
asked 5 hours ago
Confounded
836
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given an SDE for an underlying:
$$dS(t) = mu(S,t)dt+sigma(S,t)dW(t)$$
the SDE for the value of the option $V=V(S,t)$ is given via Ito's lemma as:
$$dV = V_tdt+V_Smu(S,t)dt+frac12V_SSsigma^2(S,t)dt+V_Ssigma(S,t)dW(t)$$
It seems that this would results in an SDE containing $S(t)$.
How does one then obtain an SDE for the option value so that it can be simulated directly without simulating the underlyings, i.e. something like
$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$
option-pricing monte-carlo sde
edited 5 hours ago
Daneel Olivaw
2,5431528
asked 5 hours ago
Confounded
836
Given an SDE for an underlying:
$$dS(t) = mu(S,t)dt+sigma(S,t)dW(t)$$
the SDE for the value of the option $V=V(S,t)$ is given via Ito's lemma as:
$$dV = V_tdt+V_Smu(S,t)dt+frac12V_SSsigma^2(S,t)dt+V_Ssigma(S,t)dW(t)$$
It seems that this would results in an SDE containing $S(t)$.
How does one then obtain an SDE for the option value so that it can be simulated directly without simulating the underlyings, i.e. something like
$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$
option-pricing monte-carlo sde
option-pricing monte-carlo sde
edited 5 hours ago
Daneel Olivaw
2,5431528
asked 5 hours ago
Confounded
836
edited 5 hours ago
Daneel Olivaw
2,5431528
asked 5 hours ago
Confounded
836
edited 5 hours ago
Daneel Olivaw
2,5431528
edited 5 hours ago
Daneel Olivaw
2,5431528
edited 5 hours ago
Daneel Olivaw
2,5431528
2,5431528
asked 5 hours ago
Confounded
836
asked 5 hours ago
Confounded
836
asked 5 hours ago
Confounded
836
836
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
That's impossible. The value of your option is a function of time and the value of the underlying:
$$ V(t) triangleq V(t,S_t) $$
How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.
answered 4 hours ago
Daneel Olivaw
2,5431528
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
add a comment |Â
up vote
1
down vote
This has been indirectly discussed in this question. We assume that $mathcalF_t, , tge 0$ is the natural filtration generated by the Brownian motion $W_t,, t ge 0$. Moreover, let
$B_t= e^int_0^t r_sds$ be the money market account value at time $tge0$. Let $V_T$ be the option payoff at maturity $T$. Then
beginalign*
V_t = B_tE_Qleft(fracV_TB_Tmid mathcalF_t right),
endalign*
where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $V_t/B_t, t ge 0$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $kappa_t, 0le t le T$ such that
beginalign*
fracV_tB_t &= V_0 + int_0^t kappa_u dW_u.
endalign*
Then
beginalign*
dV_t &= dleft(B_t fracV_tB_t right)\
&=r_tV_t dt + B_t kappa_t dW_t\
&=V_tBig(r_t dt + s(V,t) dW_tBig).
endalign*
where $s(V,t)=fracB_t kappa_tV_t$.
answered 4 hours ago
Gordon
14k11455
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
That's impossible. The value of your option is a function of time and the value of the underlying:
$$ V(t) triangleq V(t,S_t) $$
How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.
answered 4 hours ago
Daneel Olivaw
2,5431528
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
add a comment |Â
up vote
2
down vote
That's impossible. The value of your option is a function of time and the value of the underlying:
$$ V(t) triangleq V(t,S_t) $$
How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.
answered 4 hours ago
Daneel Olivaw
2,5431528
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
That's impossible. The value of your option is a function of time and the value of the underlying:
$$ V(t) triangleq V(t,S_t) $$
How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.
answered 4 hours ago
Daneel Olivaw
2,5431528
That's impossible. The value of your option is a function of time and the value of the underlying:
$$ V(t) triangleq V(t,S_t) $$
How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.
answered 4 hours ago
Daneel Olivaw
2,5431528
answered 4 hours ago
Daneel Olivaw
2,5431528
answered 4 hours ago
Daneel Olivaw
2,5431528
answered 4 hours ago
Daneel Olivaw
2,5431528
2,5431528
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
add a comment |Â
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^-1(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$?
– Confounded
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision.
– Daneel Olivaw
4 hours ago
add a comment |Â
up vote
1
down vote
This has been indirectly discussed in this question. We assume that $mathcalF_t, , tge 0$ is the natural filtration generated by the Brownian motion $W_t,, t ge 0$. Moreover, let
$B_t= e^int_0^t r_sds$ be the money market account value at time $tge0$. Let $V_T$ be the option payoff at maturity $T$. Then
beginalign*
V_t = B_tE_Qleft(fracV_TB_Tmid mathcalF_t right),
endalign*
where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $V_t/B_t, t ge 0$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $kappa_t, 0le t le T$ such that
beginalign*
fracV_tB_t &= V_0 + int_0^t kappa_u dW_u.
endalign*
Then
beginalign*
dV_t &= dleft(B_t fracV_tB_t right)\
&=r_tV_t dt + B_t kappa_t dW_t\
&=V_tBig(r_t dt + s(V,t) dW_tBig).
endalign*
where $s(V,t)=fracB_t kappa_tV_t$.
answered 4 hours ago
Gordon
14k11455
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
add a comment |Â
up vote
1
down vote
This has been indirectly discussed in this question. We assume that $mathcalF_t, , tge 0$ is the natural filtration generated by the Brownian motion $W_t,, t ge 0$. Moreover, let
$B_t= e^int_0^t r_sds$ be the money market account value at time $tge0$. Let $V_T$ be the option payoff at maturity $T$. Then
beginalign*
V_t = B_tE_Qleft(fracV_TB_Tmid mathcalF_t right),
endalign*
where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $V_t/B_t, t ge 0$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $kappa_t, 0le t le T$ such that
beginalign*
fracV_tB_t &= V_0 + int_0^t kappa_u dW_u.
endalign*
Then
beginalign*
dV_t &= dleft(B_t fracV_tB_t right)\
&=r_tV_t dt + B_t kappa_t dW_t\
&=V_tBig(r_t dt + s(V,t) dW_tBig).
endalign*
where $s(V,t)=fracB_t kappa_tV_t$.
answered 4 hours ago
Gordon
14k11455
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This has been indirectly discussed in this question. We assume that $mathcalF_t, , tge 0$ is the natural filtration generated by the Brownian motion $W_t,, t ge 0$. Moreover, let
$B_t= e^int_0^t r_sds$ be the money market account value at time $tge0$. Let $V_T$ be the option payoff at maturity $T$. Then
beginalign*
V_t = B_tE_Qleft(fracV_TB_Tmid mathcalF_t right),
endalign*
where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $V_t/B_t, t ge 0$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $kappa_t, 0le t le T$ such that
beginalign*
fracV_tB_t &= V_0 + int_0^t kappa_u dW_u.
endalign*
Then
beginalign*
dV_t &= dleft(B_t fracV_tB_t right)\
&=r_tV_t dt + B_t kappa_t dW_t\
&=V_tBig(r_t dt + s(V,t) dW_tBig).
endalign*
where $s(V,t)=fracB_t kappa_tV_t$.
answered 4 hours ago
Gordon
14k11455
This has been indirectly discussed in this question. We assume that $mathcalF_t, , tge 0$ is the natural filtration generated by the Brownian motion $W_t,, t ge 0$. Moreover, let
$B_t= e^int_0^t r_sds$ be the money market account value at time $tge0$. Let $V_T$ be the option payoff at maturity $T$. Then
beginalign*
V_t = B_tE_Qleft(fracV_TB_Tmid mathcalF_t right),
endalign*
where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $V_t/B_t, t ge 0$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $kappa_t, 0le t le T$ such that
beginalign*
fracV_tB_t &= V_0 + int_0^t kappa_u dW_u.
endalign*
Then
beginalign*
dV_t &= dleft(B_t fracV_tB_t right)\
&=r_tV_t dt + B_t kappa_t dW_t\
&=V_tBig(r_t dt + s(V,t) dW_tBig).
endalign*
where $s(V,t)=fracB_t kappa_tV_t$.
answered 4 hours ago
Gordon
14k11455
answered 4 hours ago
Gordon
14k11455
answered 4 hours ago
Gordon
14k11455
answered 4 hours ago
Gordon
14k11455
14k11455
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
add a comment |Â
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct?
– Confounded
3 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used.
– Gordon
2 hours ago
add a comment |Â
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