How to calculate implied correlation via observed market price (Margrabe option)Can the Heston model be shown...
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How to calculate implied correlation via observed market price (Margrabe option)
Can 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}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
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$endgroup$
add a comment |
$begingroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$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}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Bear in mind that what you're calculating is the margrabe option implied correlation, it's not necessarily the correct correlation to use for pricing other options, it's important to be aware of that.
$endgroup$
– will
3 hours ago
add a comment |
$begingroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$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}$$
Note that $d_− = d_+ − σsqrt{T}$.
black-scholes correlation european-options implied nonlinear
New contributor
Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below:
$$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
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.
edited 5 hours ago
Daneel Olivaw
3,0431629
3,0431629
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 yesterday
TaraTara
164
164
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Tara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$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
3 hours ago
add a comment |
$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
3 hours ago
$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
3 hours ago
$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
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$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.
$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_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$tag{2}min_rholeft(M(rho)-M_{text{quote}}right)^2$$
because $(M(rho)-M_{text{quote}})^2geq0$. This 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)$, 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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$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.
$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.
$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.
$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 23 hours ago
Alex CAlex C
6,63611123
6,63611123
add a comment |
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_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$tag{2}min_rholeft(M(rho)-M_{text{quote}}right)^2$$
because $(M(rho)-M_{text{quote}})^2geq0$. This an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
$endgroup$
add a comment |
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$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_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$tag{2}min_rholeft(M(rho)-M_{text{quote}}right)^2$$
because $(M(rho)-M_{text{quote}})^2geq0$. This an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
$endgroup$
add a comment |
$begingroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$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_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$tag{2}min_rholeft(M(rho)-M_{text{quote}}right)^2$$
because $(M(rho)-M_{text{quote}})^2geq0$. This an optimization problem which can be solved through traditional techniques:
- The solution suggested by @Alex C will give you a quick, approximate answer;
- If you want arbitrary precision, you can use a simple Newton algorithm on either $(1)$ or $(2)$, this is quick to program in Excel VBA, or you can maybe even find an online tool that does it. This PDF explains the method for a vanilla call in a Black-Scholes framework to find the implied volatility, but the set-up is very similar. Another alternative is gradient descent but this would probably take longer to program and you have to do it on $(2)$;
- You can also use Excel's Solver to find a solution to $(1)$ directly. I have tried with $S_0(0)=$101$, $S_1(0)=$113.5$, $sigma_0=45%$, $sigma_1=37%$, $T=1text{ year}$ and $q_0=q_1=0$ and it has worked just fine.
$endgroup$
Let $rhotriangleqrho_{imp}$. Note that:
$$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_−)$$
and $d_pm$ as defined in your question, with $M_{text{quote}}$ the observed market price. In practice, this can be restated as:
$$tag{2}min_rholeft(M(rho)-M_{text{quote}}right)^2$$
because $(M(rho)-M_{text{quote}})^2geq0$. This 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 $(2)$, 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.
edited 5 hours ago
answered 5 hours ago
Daneel OlivawDaneel Olivaw
3,0431629
3,0431629
add a comment |
add a comment |
Tara is a new contributor. Be nice, and check out our Code of Conduct.
Tara is a new contributor. Be nice, and check out our Code of Conduct.
Tara is a new contributor. Be nice, and check out our Code of Conduct.
Tara is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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
3 hours ago