How to calculate implied correlation via observed market price (Margrabe option) The 2019...
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How to calculate implied correlation via observed market price (Margrabe option)
The 2019 Stack Overflow Developer Survey Results Are InCan the Heston model be shown to reduce to the original Black Scholes model if appropriate parameters are chosen?Calculate volatility from call option priceImplied Correlation using market quotesImplied Vol vs. Calibrated VolInterpretation of CorrelationPricing of Black-Scholes with dividendHow do they calculate stocks implied volatility?Implied correlationEuropean option Vega with respect to expiry and implied volatilityIs American option price lower than European option price?
$begingroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$M_{quote} = e^{−q_0T}times S_0(0)times N(d_+)−e^{−q_1T}times S_1(0)times N(d_−)$$
where:
$$begin{align}
& d_pm = frac{logfrac{S_0(0)}{S_1(0)}+(q_1 − q_0 ±σ^2/2)T}{sigmasqrt{T}}
\[4pt]
& sigma = sqrt{sigma^2_0 + sigma^2_1 − 2rho_{imp}sigma_0 sigma_1}
end{align}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
edited yesterday
Daneel Olivaw
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$M_{quote} = e^{−q_0T}times S_0(0)times N(d_+)−e^{−q_1T}times S_1(0)times N(d_−)$$
where:
$$begin{align}
& d_pm = frac{logfrac{S_0(0)}{S_1(0)}+(q_1 − q_0 ±σ^2/2)T}{sigmasqrt{T}}
\[4pt]
& sigma = sqrt{sigma^2_0 + sigma^2_1 − 2rho_{imp}sigma_0 sigma_1}
end{align}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
edited yesterday
Daneel Olivaw
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday
add a comment |
$begingroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$M_{quote} = e^{−q_0T}times S_0(0)times N(d_+)−e^{−q_1T}times S_1(0)times N(d_−)$$
where:
$$begin{align}
& d_pm = frac{logfrac{S_0(0)}{S_1(0)}+(q_1 − q_0 ±σ^2/2)T}{sigmasqrt{T}}
\[4pt]
& sigma = sqrt{sigma^2_0 + sigma^2_1 − 2rho_{imp}sigma_0 sigma_1}
end{align}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
edited yesterday
Daneel Olivaw
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$M_{quote} = e^{−q_0T}times S_0(0)times N(d_+)−e^{−q_1T}times S_1(0)times N(d_−)$$
where:
$$begin{align}
& d_pm = frac{logfrac{S_0(0)}{S_1(0)}+(q_1 − q_0 ±σ^2/2)T}{sigmasqrt{T}}
\[4pt]
& sigma = sqrt{sigma^2_0 + sigma^2_1 − 2rho_{imp}sigma_0 sigma_1}
end{align}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
black-scholes correlation european-options implied nonlinear
edited yesterday
Daneel Olivaw
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Daneel Olivaw
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Daneel Olivaw
3,0981629
edited yesterday
Daneel Olivaw
3,0981629
edited yesterday
Daneel Olivaw
3,0981629
3,0981629
asked 2 days ago
TaraTara
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
TaraTara
186
asked 2 days ago
TaraTara
186
186
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday
add a comment |
1
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday
1
1
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$frac{partial sigma}{partial rho}(rho)=-frac{sigma_0sigma_1}{sigma(rho)}<0$$
Therefore $sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(cdot)$ is also monotonic in volatility $sigma$ thus you can find an unique solution to the equation:
$$tag{1}M_{text{quote}}=M(rho)$$
where:
$$M(rho)=e^{−q_0T}S_0(0)N(d_+)−e^{−q_1T}S_1(0)N(d_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$begin{align}
&min_rholeft(M(rho)-M_{text{quote}}right)^2tag{2}
\
& text{s.t. } rho in [-1,1]
end{align}$$
because $(M(rho)-M_{text{quote}})^2geq0$. This is an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$ with root value $rho=0$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
$endgroup$
add a comment |
$begingroup$
We know that $-1lerho_{imp}le 1$ so perhaps the simplest approach is to try the possible values $rho_{imp}={-1,-0.9,-0.8,cdots,0.8,0.9,+1}$, to calculate resulting $sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed correlations to pin it down more precisely. It is a manual but relatively simple method.
answered 2 days ago
Alex CAlex C
6,65611123
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$frac{partial sigma}{partial rho}(rho)=-frac{sigma_0sigma_1}{sigma(rho)}<0$$
Therefore $sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(cdot)$ is also monotonic in volatility $sigma$ thus you can find an unique solution to the equation:
$$tag{1}M_{text{quote}}=M(rho)$$
where:
$$M(rho)=e^{−q_0T}S_0(0)N(d_+)−e^{−q_1T}S_1(0)N(d_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$begin{align}
&min_rholeft(M(rho)-M_{text{quote}}right)^2tag{2}
\
& text{s.t. } rho in [-1,1]
end{align}$$
because $(M(rho)-M_{text{quote}})^2geq0$. This is an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$ with root value $rho=0$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
$endgroup$
add a comment |
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$frac{partial sigma}{partial rho}(rho)=-frac{sigma_0sigma_1}{sigma(rho)}<0$$
Therefore $sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(cdot)$ is also monotonic in volatility $sigma$ thus you can find an unique solution to the equation:
$$tag{1}M_{text{quote}}=M(rho)$$
where:
$$M(rho)=e^{−q_0T}S_0(0)N(d_+)−e^{−q_1T}S_1(0)N(d_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$begin{align}
&min_rholeft(M(rho)-M_{text{quote}}right)^2tag{2}
\
& text{s.t. } rho in [-1,1]
end{align}$$
because $(M(rho)-M_{text{quote}})^2geq0$. This is an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$ with root value $rho=0$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
$endgroup$
add a comment |
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$frac{partial sigma}{partial rho}(rho)=-frac{sigma_0sigma_1}{sigma(rho)}<0$$
Therefore $sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(cdot)$ is also monotonic in volatility $sigma$ thus you can find an unique solution to the equation:
$$tag{1}M_{text{quote}}=M(rho)$$
where:
$$M(rho)=e^{−q_0T}S_0(0)N(d_+)−e^{−q_1T}S_1(0)N(d_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$begin{align}
&min_rholeft(M(rho)-M_{text{quote}}right)^2tag{2}
\
& text{s.t. } rho in [-1,1]
end{align}$$
because $(M(rho)-M_{text{quote}})^2geq0$. This is an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$ with root value $rho=0$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
$endgroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$frac{partial sigma}{partial rho}(rho)=-frac{sigma_0sigma_1}{sigma(rho)}<0$$
Therefore $sigma$ is monotonic in implied correlation. In addition, the Margrabe pricing function $M(cdot)$ is also monotonic in volatility $sigma$ thus you can find an unique solution to the equation:
$$tag{1}M_{text{quote}}=M(rho)$$
where:
$$M(rho)=e^{−q_0T}S_0(0)N(d_+)−e^{−q_1T}S_1(0)N(d_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$begin{align}
&min_rholeft(M(rho)-M_{text{quote}}right)^2tag{2}
\
& text{s.t. } rho in [-1,1]
end{align}$$
because $(M(rho)-M_{text{quote}})^2geq0$. This is an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$ with root value $rho=0$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
edited 8 hours ago
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
answered yesterday
Daneel OlivawDaneel Olivaw
3,0981629
3,0981629
add a comment |
add a comment |
$begingroup$
We know that $-1lerho_{imp}le 1$ so perhaps the simplest approach is to try the possible values $rho_{imp}={-1,-0.9,-0.8,cdots,0.8,0.9,+1}$, to calculate resulting $sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed correlations to pin it down more precisely. It is a manual but relatively simple method.
answered 2 days ago
Alex CAlex C
6,65611123
$endgroup$
add a comment |
$begingroup$
We know that $-1lerho_{imp}le 1$ so perhaps the simplest approach is to try the possible values $rho_{imp}={-1,-0.9,-0.8,cdots,0.8,0.9,+1}$, to calculate resulting $sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed correlations to pin it down more precisely. It is a manual but relatively simple method.
answered 2 days ago
Alex CAlex C
6,65611123
$endgroup$
add a comment |
$begingroup$
We know that $-1lerho_{imp}le 1$ so perhaps the simplest approach is to try the possible values $rho_{imp}={-1,-0.9,-0.8,cdots,0.8,0.9,+1}$, to calculate resulting $sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed correlations to pin it down more precisely. It is a manual but relatively simple method.
answered 2 days ago
Alex CAlex C
6,65611123
$endgroup$
We know that $-1lerho_{imp}le 1$ so perhaps the simplest approach is to try the possible values $rho_{imp}={-1,-0.9,-0.8,cdots,0.8,0.9,+1}$, to calculate resulting $sigma$ values, d± values, and $M_{quote}$ values, then see which of these is closest to the observed market price. If desired you can then search a finer grid between two adjacent assumed correlations to pin it down more precisely. It is a manual but relatively simple method.
answered 2 days ago
Alex CAlex C
6,65611123
answered 2 days ago
Alex CAlex C
6,65611123
answered 2 days ago
Alex CAlex C
6,65611123
answered 2 days ago
Alex CAlex C
6,65611123
6,65611123
add a comment |
add a comment |
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1
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
yesterday