Proof that $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X? The 2019 Stack Overflow Developer Survey Results Are InWhy is $ N^tau ( M - M^tau ) $ a continuous local martingale if $ M $ and $ N $ are?Almost sure convergence of stochastic processfractional Brownian motion is not a semimartingale. How to apply Ergodic theorem in the proof of this theorem?Prove that the stochastic process can not have continuous paths.Why Are Semimartingales the Largest Possible Class of Stochastic Integrators?Is the solution to an SDE a continuous semimartingale/diffusion process?Proof that poisson process has right continuous modificationOptional sampling theorem for bounded stopping time for a right continuous-submartingaleDefinition of semimartingale in Protter's bookProve limiting distribution goes to stationary distribution: $lim_ttoinfty pi_j(t) = overlinepi_j$
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Proof that $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X?
The 2019 Stack Overflow Developer Survey Results Are InWhy is $ N^tau ( M - M^tau ) $ a continuous local martingale if $ M $ and $ N $ are?Almost sure convergence of stochastic processfractional Brownian motion is not a semimartingale. How to apply Ergodic theorem in the proof of this theorem?Prove that the stochastic process can not have continuous paths.Why Are Semimartingales the Largest Possible Class of Stochastic Integrators?Is the solution to an SDE a continuous semimartingale/diffusion process?Proof that poisson process has right continuous modificationOptional sampling theorem for bounded stopping time for a right continuous-submartingaleDefinition of semimartingale in Protter's bookProve limiting distribution goes to stationary distribution: $lim_ttoinfty pi_j(t) = overlinepi_j$
$begingroup$
An identity, namely: $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X, is used rather frequently. I understand that $Y_t=e^X_t-X_0$ is the solution for initial value $X_0$ but cannot see what can be done from here to prove it directly.
probability-theory stochastic-processes stochastic-calculus
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
$endgroup$
add a comment |
$begingroup$
An identity, namely: $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X, is used rather frequently. I understand that $Y_t=e^X_t-X_0$ is the solution for initial value $X_0$ but cannot see what can be done from here to prove it directly.
probability-theory stochastic-processes stochastic-calculus
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
$endgroup$
$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41
add a comment |
$begingroup$
An identity, namely: $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X, is used rather frequently. I understand that $Y_t=e^X_t-X_0$ is the solution for initial value $X_0$ but cannot see what can be done from here to prove it directly.
probability-theory stochastic-processes stochastic-calculus
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
$endgroup$
An identity, namely: $Y_tdX_t=dY_t Rightarrow d[Y,Y]_t=Y^2d[X,X]_t$ for continuous semimartingale X, is used rather frequently. I understand that $Y_t=e^X_t-X_0$ is the solution for initial value $X_0$ but cannot see what can be done from here to prove it directly.
probability-theory stochastic-processes stochastic-calculus
probability-theory stochastic-processes stochastic-calculus
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
edited Apr 7 at 4:06
BayesIsBae
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
asked Apr 6 at 22:41
BayesIsBaeBayesIsBae
858
858
$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41
add a comment |
$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41
$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41
add a comment |
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$begingroup$
$d[Y,Y]_t=(dY_t)^2$
$endgroup$
– QFi
Apr 7 at 4:33
$begingroup$
@Zizou23 Thank you so much! I just realized this is even alluded to in the wikipedia article en.wikipedia.org/wiki/Quadratic_variation#Semimartingales. I presume the proof for this identity ($d[Y,Y]_t=(dY_t)^2$) is similar to i.imgur.com/6t8MAr6.png?
$endgroup$
– BayesIsBae
Apr 7 at 20:07
$begingroup$
see p.143 of Stochastic Calculus for Finance II: Continuous-Time Models
$endgroup$
– QFi
Apr 7 at 20:20
$begingroup$
@QFi That implies for$X(t)=X(0)+int_0^t Delta(u) d W(u)+int_0^t Theta(u) d u$, it is s.t. $d[X,X]_t=Delta^2dt$ for some adapted $Delta$. Is $Delta=fracdXsqrtdt$ necessarily?
$endgroup$
– BayesIsBae
Apr 7 at 22:41