Mixing
Probability Density Function sign problem when using Call Price
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Given the call price $C_t = e^-int_t^Tr(s)dsint (s-K)^+phi_S_T(T,s)ds $
we know that $$fracdCdK=-e^-int_t^Tr(s)dsint_K^infty phi_S_T(T,s)ds$$
Now when I use dirac delta function property $ int f(t)delta(t-T)dt = f(T) $when taking the second derivative with respect to $K$ I find an akward sign minus which I don't know where my calclulus is wrong :
$$fracd^2dK^2C=-e^-int_t^Tr(s)dsint frac ddK H(s-K) phi_S_T(T,s)ds$$
with $H$ the heavy side function, which gives :
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsint delta(s-K) phi_S_T(T,s)ds $$
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsphi_S_T(T,K)$$
and the right result should be without the minus sign here. Where did I go wrong? I know how to find the right result using just the indicator function, but I wanna use the delta function property here to derive it. Any help?
local-volatility calculation
asked 2 hours ago
user30614
655
add a comment |Â
up vote
2
down vote
favorite
Given the call price $C_t = e^-int_t^Tr(s)dsint (s-K)^+phi_S_T(T,s)ds $
we know that $$fracdCdK=-e^-int_t^Tr(s)dsint_K^infty phi_S_T(T,s)ds$$
Now when I use dirac delta function property $ int f(t)delta(t-T)dt = f(T) $when taking the second derivative with respect to $K$ I find an akward sign minus which I don't know where my calclulus is wrong :
$$fracd^2dK^2C=-e^-int_t^Tr(s)dsint frac ddK H(s-K) phi_S_T(T,s)ds$$
with $H$ the heavy side function, which gives :
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsint delta(s-K) phi_S_T(T,s)ds $$
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsphi_S_T(T,K)$$
and the right result should be without the minus sign here. Where did I go wrong? I know how to find the right result using just the indicator function, but I wanna use the delta function property here to derive it. Any help?
local-volatility calculation
asked 2 hours ago
user30614
655
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the call price $C_t = e^-int_t^Tr(s)dsint (s-K)^+phi_S_T(T,s)ds $
we know that $$fracdCdK=-e^-int_t^Tr(s)dsint_K^infty phi_S_T(T,s)ds$$
Now when I use dirac delta function property $ int f(t)delta(t-T)dt = f(T) $when taking the second derivative with respect to $K$ I find an akward sign minus which I don't know where my calclulus is wrong :
$$fracd^2dK^2C=-e^-int_t^Tr(s)dsint frac ddK H(s-K) phi_S_T(T,s)ds$$
with $H$ the heavy side function, which gives :
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsint delta(s-K) phi_S_T(T,s)ds $$
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsphi_S_T(T,K)$$
and the right result should be without the minus sign here. Where did I go wrong? I know how to find the right result using just the indicator function, but I wanna use the delta function property here to derive it. Any help?
local-volatility calculation
asked 2 hours ago
user30614
655
Given the call price $C_t = e^-int_t^Tr(s)dsint (s-K)^+phi_S_T(T,s)ds $
we know that $$fracdCdK=-e^-int_t^Tr(s)dsint_K^infty phi_S_T(T,s)ds$$
Now when I use dirac delta function property $ int f(t)delta(t-T)dt = f(T) $when taking the second derivative with respect to $K$ I find an akward sign minus which I don't know where my calclulus is wrong :
$$fracd^2dK^2C=-e^-int_t^Tr(s)dsint frac ddK H(s-K) phi_S_T(T,s)ds$$
with $H$ the heavy side function, which gives :
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsint delta(s-K) phi_S_T(T,s)ds $$
$$fracd^2dK^2C= -e^-int_t^Tr(s)dsphi_S_T(T,K)$$
and the right result should be without the minus sign here. Where did I go wrong? I know how to find the right result using just the indicator function, but I wanna use the delta function property here to derive it. Any help?
local-volatility calculation
local-volatility calculation
asked 2 hours ago
user30614
655
asked 2 hours ago
user30614
655
asked 2 hours ago
user30614
655
asked 2 hours ago
user30614
655
asked 2 hours ago
user30614
655
655
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Note that with $H(cdot)$ the Heaviside function
$$fracdds H(s-K) = delta(s-K)$$
but
$$fracddK H(s-K) = colorred-delta(s-K)$$
You can also use the Leibniz integral rule to write that
$$ fracddK int_K^infty phi_S_T(T,s) ds = -phi(S_T,K) $$
answered 2 hours ago
Quantuple
10.4k11141
add a comment |Â
up vote
1
down vote
Just the chain rule;
$fracddK H left (S-Kright)=delta left (S-Kright) fracddK left (S-Kright)=-delta left (S-Kright) $
answered 1 hour ago
Magic is in the chain
44915
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Note that with $H(cdot)$ the Heaviside function
$$fracdds H(s-K) = delta(s-K)$$
but
$$fracddK H(s-K) = colorred-delta(s-K)$$
You can also use the Leibniz integral rule to write that
$$ fracddK int_K^infty phi_S_T(T,s) ds = -phi(S_T,K) $$
answered 2 hours ago
Quantuple
10.4k11141
add a comment |Â
up vote
1
down vote
Note that with $H(cdot)$ the Heaviside function
$$fracdds H(s-K) = delta(s-K)$$
but
$$fracddK H(s-K) = colorred-delta(s-K)$$
You can also use the Leibniz integral rule to write that
$$ fracddK int_K^infty phi_S_T(T,s) ds = -phi(S_T,K) $$
answered 2 hours ago
Quantuple
10.4k11141
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that with $H(cdot)$ the Heaviside function
$$fracdds H(s-K) = delta(s-K)$$
but
$$fracddK H(s-K) = colorred-delta(s-K)$$
You can also use the Leibniz integral rule to write that
$$ fracddK int_K^infty phi_S_T(T,s) ds = -phi(S_T,K) $$
answered 2 hours ago
Quantuple
10.4k11141
Note that with $H(cdot)$ the Heaviside function
$$fracdds H(s-K) = delta(s-K)$$
but
$$fracddK H(s-K) = colorred-delta(s-K)$$
You can also use the Leibniz integral rule to write that
$$ fracddK int_K^infty phi_S_T(T,s) ds = -phi(S_T,K) $$
answered 2 hours ago
Quantuple
10.4k11141
answered 2 hours ago
Quantuple
10.4k11141
answered 2 hours ago
Quantuple
10.4k11141
answered 2 hours ago
Quantuple
10.4k11141
10.4k11141
add a comment |Â
add a comment |Â
up vote
1
down vote
Just the chain rule;
$fracddK H left (S-Kright)=delta left (S-Kright) fracddK left (S-Kright)=-delta left (S-Kright) $
answered 1 hour ago
Magic is in the chain
44915
add a comment |Â
up vote
1
down vote
Just the chain rule;
$fracddK H left (S-Kright)=delta left (S-Kright) fracddK left (S-Kright)=-delta left (S-Kright) $
answered 1 hour ago
Magic is in the chain
44915
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just the chain rule;
$fracddK H left (S-Kright)=delta left (S-Kright) fracddK left (S-Kright)=-delta left (S-Kright) $
answered 1 hour ago
Magic is in the chain
44915
Just the chain rule;
$fracddK H left (S-Kright)=delta left (S-Kright) fracddK left (S-Kright)=-delta left (S-Kright) $
answered 1 hour ago
Magic is in the chain
44915
answered 1 hour ago
Magic is in the chain
44915
answered 1 hour ago
Magic is in the chain
44915
answered 1 hour ago
Magic is in the chain
44915
44915
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquant.stackexchange.com%2fquestions%2f42124%2fprobability-density-function-sign-problem-when-using-call-price%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
