Standard Brownian motion and stopping time
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Let be $B$ standard Brownian motion and let $S leq T$ two stopping times with $E(T) < infty $ and $E(S) < infty$. Prove that hold $$ E[(B_T - B_S)^2] = E[B_T^2 - B_S^2] = E(T-S).$$
Please help me solve this.
stochastic-processes stochastic-calculus brownian-motion stopping-times
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add a comment |
$begingroup$
Let be $B$ standard Brownian motion and let $S leq T$ two stopping times with $E(T) < infty $ and $E(S) < infty$. Prove that hold $$ E[(B_T - B_S)^2] = E[B_T^2 - B_S^2] = E(T-S).$$
Please help me solve this.
stochastic-processes stochastic-calculus brownian-motion stopping-times
$endgroup$
add a comment |
$begingroup$
Let be $B$ standard Brownian motion and let $S leq T$ two stopping times with $E(T) < infty $ and $E(S) < infty$. Prove that hold $$ E[(B_T - B_S)^2] = E[B_T^2 - B_S^2] = E(T-S).$$
Please help me solve this.
stochastic-processes stochastic-calculus brownian-motion stopping-times
$endgroup$
Let be $B$ standard Brownian motion and let $S leq T$ two stopping times with $E(T) < infty $ and $E(S) < infty$. Prove that hold $$ E[(B_T - B_S)^2] = E[B_T^2 - B_S^2] = E(T-S).$$
Please help me solve this.
stochastic-processes stochastic-calculus brownian-motion stopping-times
stochastic-processes stochastic-calculus brownian-motion stopping-times
asked Jan 11 at 20:34
user631885
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1 Answer
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$begingroup$
For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.
By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T mid S]] = E[B_S^2]$.
For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.
$E[B_T^2] = E[E[B_T^2 mid T]] = E[T]$.
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
$endgroup$
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
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1 Answer
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1 Answer
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$begingroup$
For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.
By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T mid S]] = E[B_S^2]$.
For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.
$E[B_T^2] = E[E[B_T^2 mid T]] = E[T]$.
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
$endgroup$
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
add a comment |
$begingroup$
For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.
By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T mid S]] = E[B_S^2]$.
For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.
$E[B_T^2] = E[E[B_T^2 mid T]] = E[T]$.
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
$endgroup$
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
add a comment |
$begingroup$
For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.
By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T mid S]] = E[B_S^2]$.
For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.
$E[B_T^2] = E[E[B_T^2 mid T]] = E[T]$.
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
$endgroup$
For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.
By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T mid S]] = E[B_S^2]$.
For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.
$E[B_T^2] = E[E[B_T^2 mid T]] = E[T]$.
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
answered Jan 11 at 20:43
angryavianangryavian
42.6k23481
42.6k23481
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
add a comment |
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
$begingroup$
Thank you, i didn't think that way.
$endgroup$
– user631885
Jan 11 at 21:18
add a comment |
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