Stochastic integration by parts The 2019 Stack Overflow Developer Survey Results Are InIntegration of Wiener process: $int_t_1^t_2 dB(s)$Condition for existence of a stochastic differential equationBrownian motion and stochastic integrationBrownian motion motivation of constructionstochastic integration with respect to quadratic variationBrownian Motion and stochastic integration on the complete real lineDifferentiating Stochastic IntegralWhich inequalities are there with stochastic integration?Stochastic Integration of $B^2dB$?Calculate a multiple Ito integral
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Stochastic integration by parts
The 2019 Stack Overflow Developer Survey Results Are InIntegration of Wiener process: $int_t_1^t_2 dB(s)$Condition for existence of a stochastic differential equationBrownian motion and stochastic integrationBrownian motion motivation of constructionstochastic integration with respect to quadratic variationBrownian Motion and stochastic integration on the complete real lineDifferentiating Stochastic IntegralWhich inequalities are there with stochastic integration?Stochastic Integration of $B^2dB$?Calculate a multiple Ito integral
$begingroup$
My professor asserts
However, Oksendal asserts in his textbook:
$X_t Y_t-X_0 Y_0=int_0^t X_s d Y_s+int_0^t Y_s d X_s+int_0^tdXdY$
These are not equivalent - consider $e^t/2W_t$ for standard brownian motion $W_t$ - the "$g_tdW_t$" term for $e^t/2$ is such that $g_t=0$.
Is my professor's version incorrect?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
$endgroup$
add a comment |
$begingroup$
My professor asserts
However, Oksendal asserts in his textbook:
$X_t Y_t-X_0 Y_0=int_0^t X_s d Y_s+int_0^t Y_s d X_s+int_0^tdXdY$
These are not equivalent - consider $e^t/2W_t$ for standard brownian motion $W_t$ - the "$g_tdW_t$" term for $e^t/2$ is such that $g_t=0$.
Is my professor's version incorrect?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
$endgroup$
$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago
add a comment |
$begingroup$
My professor asserts
However, Oksendal asserts in his textbook:
$X_t Y_t-X_0 Y_0=int_0^t X_s d Y_s+int_0^t Y_s d X_s+int_0^tdXdY$
These are not equivalent - consider $e^t/2W_t$ for standard brownian motion $W_t$ - the "$g_tdW_t$" term for $e^t/2$ is such that $g_t=0$.
Is my professor's version incorrect?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
$endgroup$
My professor asserts
However, Oksendal asserts in his textbook:
$X_t Y_t-X_0 Y_0=int_0^t X_s d Y_s+int_0^t Y_s d X_s+int_0^tdXdY$
These are not equivalent - consider $e^t/2W_t$ for standard brownian motion $W_t$ - the "$g_tdW_t$" term for $e^t/2$ is such that $g_t=0$.
Is my professor's version incorrect?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
asked Apr 8 at 1:22
BayesIsBaeBayesIsBae
858
858
$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago
add a comment |
$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago
$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago
add a comment |
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$begingroup$
If one thinks of $, d X ,d Y$ to mean $d langle X, Yrangle$ then all the integration by parts formulae seem to be consistent with each other: $$,d langle X, Y rangle_t = b_tg_t ,d langle W_1, W_2rangle_t= begincases 0 , text if W_1 text and W_2 text are independent \ b_tg_t ,d t , text if W_1 = W_2 =Wendcases.$$ Your counterexample is unclear to me.
$endgroup$
– Sayantan
Apr 8 at 2:58
$begingroup$
that is, $de^t/2=(1/2)e^t/2dt + 0dW_t$. Further, $dXdY$ is apparently literally just the product of the two differentials.
$endgroup$
– BayesIsBae
Apr 8 at 4:22
$begingroup$
$,d X ,d Y $ can be interpreted as a product of differentials, but these products must also follow the formal rules set down in Oksendal's book, cf. theorem 4.1.2 (p. 44). According to these rules $, dX_t , d Y_t = b_tg_t ,d t$ when $W_1=W_2 = W$. In your example, the second formula (3.3.28), gives you $$ e^t/2 W_t - e^0/2W_0 = int_0^t W_s ,d (e^s/2) + int_0^t e^s/2 ,d W_s$$ because $b_s =0, g_s=1$, while the third formula also gives you the same thing, because $,d s cdot ,d W_s =0$ according to the rules mentioned above. So there is no contradiction.
$endgroup$
– Sayantan
Apr 8 at 7:40
$begingroup$
@Sayantan I see, thank you for pointing this out
$endgroup$
– BayesIsBae
2 days ago