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Asymptotics of binary call option in Black–Scholes model in the large deviations sense
The Next CEO of Stack OverflowAn approximation of an integralA proof of Stirling's FormulaHow are canonical and grand canonical ensemble related in the framework of large deviations theory?Explicit Laplace approximation for tail of gaussian distributionOn the evaluation of the integral $int_-fracba^frac1-balogleft(ax+bright)expleft(-frac12x^2right)mathrmdx$.Calculation of integralHaving trouble in an exercice on Large Deviations.Evaluating an integral by setting it up as a differential equationProve that a double integral is equal to $0$Calculate limit of integral by bounded convergence or dominated convergence theorem
$begingroup$
Consider Black–Scholes model where asset log-price is given by
$$
X _ t = sigma W _ t + mu t
$$
for $W_t$ – Brownian motion.
I want to show that
$$
P left[ X _ t > k right] approx exp left( - frac k ^ 2 2 sigma ^ 2 t right) quad text as t downarrow 0
$$
in the large deviations sense, i.e.
$$
lim _tdownarrow 0 tcdot log P[X_t>k] = frac-k^22sigma^2
$$
After rescaling and getting rid of vanishing prefactor $frac1sqrt2pi$ we get
$$
beginaligned
lim _tdownarrow 0 tcdot log P[X_t>k] &= lim _tdownarrow 0 tcdot log int_frack-mu tsigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
\&=lim _tdownarrow 0 tcdot log int_fracksigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
endaligned
$$
whereas in the second equality we applied dominated convergence.
Now I was looking into the mean-value theorem, but I am not sure how to rewrite the term in order to apply it directly.
Another idea would be to reformulate the problem in order to be able to apply Cramer theorem, since $varphi ^* (z) = frac (z - mu) ^ 2 2 sigma ^ 2 $ is Legendre transform of the logarithm of the moment-generating function of the normal distribution, and Cramer theorem relates probability of the large deviations events in terms of Legendre transform of the logarithm of the mgf.
Thank you in advance!
integration large-deviation-theory
asked 2 days ago
nakajuicenakajuice
1,40311423
$endgroup$
add a comment |
$begingroup$
Consider Black–Scholes model where asset log-price is given by
$$
X _ t = sigma W _ t + mu t
$$
for $W_t$ – Brownian motion.
I want to show that
$$
P left[ X _ t > k right] approx exp left( - frac k ^ 2 2 sigma ^ 2 t right) quad text as t downarrow 0
$$
in the large deviations sense, i.e.
$$
lim _tdownarrow 0 tcdot log P[X_t>k] = frac-k^22sigma^2
$$
After rescaling and getting rid of vanishing prefactor $frac1sqrt2pi$ we get
$$
beginaligned
lim _tdownarrow 0 tcdot log P[X_t>k] &= lim _tdownarrow 0 tcdot log int_frack-mu tsigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
\&=lim _tdownarrow 0 tcdot log int_fracksigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
endaligned
$$
whereas in the second equality we applied dominated convergence.
Now I was looking into the mean-value theorem, but I am not sure how to rewrite the term in order to apply it directly.
Another idea would be to reformulate the problem in order to be able to apply Cramer theorem, since $varphi ^* (z) = frac (z - mu) ^ 2 2 sigma ^ 2 $ is Legendre transform of the logarithm of the moment-generating function of the normal distribution, and Cramer theorem relates probability of the large deviations events in terms of Legendre transform of the logarithm of the mgf.
Thank you in advance!
integration large-deviation-theory
asked 2 days ago
nakajuicenakajuice
1,40311423
$endgroup$
$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday
add a comment |
$begingroup$
Consider Black–Scholes model where asset log-price is given by
$$
X _ t = sigma W _ t + mu t
$$
for $W_t$ – Brownian motion.
I want to show that
$$
P left[ X _ t > k right] approx exp left( - frac k ^ 2 2 sigma ^ 2 t right) quad text as t downarrow 0
$$
in the large deviations sense, i.e.
$$
lim _tdownarrow 0 tcdot log P[X_t>k] = frac-k^22sigma^2
$$
After rescaling and getting rid of vanishing prefactor $frac1sqrt2pi$ we get
$$
beginaligned
lim _tdownarrow 0 tcdot log P[X_t>k] &= lim _tdownarrow 0 tcdot log int_frack-mu tsigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
\&=lim _tdownarrow 0 tcdot log int_fracksigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
endaligned
$$
whereas in the second equality we applied dominated convergence.
Now I was looking into the mean-value theorem, but I am not sure how to rewrite the term in order to apply it directly.
Another idea would be to reformulate the problem in order to be able to apply Cramer theorem, since $varphi ^* (z) = frac (z - mu) ^ 2 2 sigma ^ 2 $ is Legendre transform of the logarithm of the moment-generating function of the normal distribution, and Cramer theorem relates probability of the large deviations events in terms of Legendre transform of the logarithm of the mgf.
Thank you in advance!
integration large-deviation-theory
asked 2 days ago
nakajuicenakajuice
1,40311423
$endgroup$
Consider Black–Scholes model where asset log-price is given by
$$
X _ t = sigma W _ t + mu t
$$
for $W_t$ – Brownian motion.
I want to show that
$$
P left[ X _ t > k right] approx exp left( - frac k ^ 2 2 sigma ^ 2 t right) quad text as t downarrow 0
$$
in the large deviations sense, i.e.
$$
lim _tdownarrow 0 tcdot log P[X_t>k] = frac-k^22sigma^2
$$
After rescaling and getting rid of vanishing prefactor $frac1sqrt2pi$ we get
$$
beginaligned
lim _tdownarrow 0 tcdot log P[X_t>k] &= lim _tdownarrow 0 tcdot log int_frack-mu tsigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
\&=lim _tdownarrow 0 tcdot log int_fracksigma sqrt t^inftyexpleft(frac-x^22right)mathrm dx
endaligned
$$
whereas in the second equality we applied dominated convergence.
Now I was looking into the mean-value theorem, but I am not sure how to rewrite the term in order to apply it directly.
Another idea would be to reformulate the problem in order to be able to apply Cramer theorem, since $varphi ^* (z) = frac (z - mu) ^ 2 2 sigma ^ 2 $ is Legendre transform of the logarithm of the moment-generating function of the normal distribution, and Cramer theorem relates probability of the large deviations events in terms of Legendre transform of the logarithm of the mgf.
Thank you in advance!
integration large-deviation-theory
integration large-deviation-theory
asked 2 days ago
nakajuicenakajuice
1,40311423
asked 2 days ago
nakajuicenakajuice
1,40311423
edited 2 days ago
nakajuice
asked 2 days ago
nakajuicenakajuice
1,40311423
asked 2 days ago
nakajuicenakajuice
1,40311423
asked 2 days ago
nakajuicenakajuice
1,40311423
1,40311423
$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday
add a comment |
$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday
$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $1/t$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $k/sigmasqrtt$. This is a direct consequence of Laplace's method. First let $u=xsigmasqrtt/k$. Rewrite the integral on $[1,infty]$ with integrand $exp(-f(u)/t)$, where $f(u):=k^2u^2/2sigma^2$. Apply Laplace's method, computing $f''(u)$, to finish the proof.
answered 2 days ago
Alex R.Alex R.
25.1k12452
$endgroup$
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $1/t$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $k/sigmasqrtt$. This is a direct consequence of Laplace's method. First let $u=xsigmasqrtt/k$. Rewrite the integral on $[1,infty]$ with integrand $exp(-f(u)/t)$, where $f(u):=k^2u^2/2sigma^2$. Apply Laplace's method, computing $f''(u)$, to finish the proof.
answered 2 days ago
Alex R.Alex R.
25.1k12452
$endgroup$
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
add a comment |
$begingroup$
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $1/t$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $k/sigmasqrtt$. This is a direct consequence of Laplace's method. First let $u=xsigmasqrtt/k$. Rewrite the integral on $[1,infty]$ with integrand $exp(-f(u)/t)$, where $f(u):=k^2u^2/2sigma^2$. Apply Laplace's method, computing $f''(u)$, to finish the proof.
answered 2 days ago
Alex R.Alex R.
25.1k12452
$endgroup$
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
add a comment |
$begingroup$
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $1/t$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $k/sigmasqrtt$. This is a direct consequence of Laplace's method. First let $u=xsigmasqrtt/k$. Rewrite the integral on $[1,infty]$ with integrand $exp(-f(u)/t)$, where $f(u):=k^2u^2/2sigma^2$. Apply Laplace's method, computing $f''(u)$, to finish the proof.
answered 2 days ago
Alex R.Alex R.
25.1k12452
$endgroup$
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $1/t$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $k/sigmasqrtt$. This is a direct consequence of Laplace's method. First let $u=xsigmasqrtt/k$. Rewrite the integral on $[1,infty]$ with integrand $exp(-f(u)/t)$, where $f(u):=k^2u^2/2sigma^2$. Apply Laplace's method, computing $f''(u)$, to finish the proof.
answered 2 days ago
Alex R.Alex R.
25.1k12452
edited 2 days ago
answered 2 days ago
Alex R.Alex R.
25.1k12452
answered 2 days ago
Alex R.Alex R.
25.1k12452
answered 2 days ago
Alex R.Alex R.
25.1k12452
25.1k12452
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
add a comment |
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$.
$endgroup$
– nakajuice
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
$begingroup$
@nakajuice: Corrected, thanks!
$endgroup$
– Alex R.
2 days ago
add a comment |
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$begingroup$
The question sounds like it belongs to quant.SE due to financial context, however, the problem I am trying to tackle is mathematical, so I post this question here.
$endgroup$
– nakajuice
2 days ago
$begingroup$
The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$int_x^infty e^-t^2/2 dt = frac e^-x^2/2 x - int_x^infty frac e^-t^2/2 t^2 dt = dots ,.$$
$endgroup$
– Maxim
yesterday