Mixing
stopping time and quadratic variation process
0
$begingroup$
Let $tau$ be a stopping time and $(M_n)_{n in mathbb{N}_0}$ be a martingale with $mathbb{E}(M_n^2)<infty$ for any $n in mathbb{N}_0$. Show that, if $langle M rangle_{tau} = 0$ (where it means the quadratic variation) a.s. then $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1.$$
I cannot understand the point, in fact $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1$$ I think it means that the stopping time is $tau=0$ but it makes no sense.
Could someone help me to resolve this exercise?
stochastic-processes martingales stopping-times
edited Jan 23 at 18:54
saz
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
$endgroup$
add a comment |
0
$begingroup$
Let $tau$ be a stopping time and $(M_n)_{n in mathbb{N}_0}$ be a martingale with $mathbb{E}(M_n^2)<infty$ for any $n in mathbb{N}_0$. Show that, if $langle M rangle_{tau} = 0$ (where it means the quadratic variation) a.s. then $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1.$$
I cannot understand the point, in fact $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1$$ I think it means that the stopping time is $tau=0$ but it makes no sense.
Could someone help me to resolve this exercise?
stochastic-processes martingales stopping-times
edited Jan 23 at 18:54
saz
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
$endgroup$
$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22
add a comment |
0
0
0
$begingroup$
Let $tau$ be a stopping time and $(M_n)_{n in mathbb{N}_0}$ be a martingale with $mathbb{E}(M_n^2)<infty$ for any $n in mathbb{N}_0$. Show that, if $langle M rangle_{tau} = 0$ (where it means the quadratic variation) a.s. then $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1.$$
I cannot understand the point, in fact $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1$$ I think it means that the stopping time is $tau=0$ but it makes no sense.
Could someone help me to resolve this exercise?
stochastic-processes martingales stopping-times
edited Jan 23 at 18:54
saz
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
$endgroup$
Let $tau$ be a stopping time and $(M_n)_{n in mathbb{N}_0}$ be a martingale with $mathbb{E}(M_n^2)<infty$ for any $n in mathbb{N}_0$. Show that, if $langle M rangle_{tau} = 0$ (where it means the quadratic variation) a.s. then $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1.$$
I cannot understand the point, in fact $$mathbb{P}(M_{tau wedge n} = M_0 , , text{for any $n in mathbb{N}_0$})=1$$ I think it means that the stopping time is $tau=0$ but it makes no sense.
Could someone help me to resolve this exercise?
stochastic-processes martingales stopping-times
stochastic-processes martingales stopping-times
edited Jan 23 at 18:54
saz
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
edited Jan 23 at 18:54
saz
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
edited Jan 23 at 18:54
saz
81.7k861128
edited Jan 23 at 18:54
saz
81.7k861128
edited Jan 23 at 18:54
saz
81.7k861128
81.7k861128
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
asked Jan 23 at 18:47
Francesca BallatoreFrancesca Ballatore
426
426
$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22
add a comment |
$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22
$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22
add a comment |
1 Answer
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$begingroup$
Hints:
- Use the optional stopping theorem to show that $$X_n := M_{n wedge tau}^2 - langle M rangle_{n wedge tau}$$ is a martingale; in particular $$mathbb{E}(M_{n wedge tau}^2) - mathbb{E}(M_0^2) = mathbb{E}(langle M rangle_{n wedge tau}). tag{1}$$
- Deduce from $langle M rangle_{tau} = 0$ that $langle M rangle_{n wedge tau} = 0$ for all $n in mathbb{N}_0$.
- Show that $$mathbb{E}((M_{n wedge tau}-M_0)^2) = mathbb{E}(M_{n wedge tau}^2)-mathbb{E}(M_0^2), qquad n in mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
- Combining the first three steps shows that $$mathbb{E}((M_{n wedge tau}-M_0)^2)=0$$ for all $n in mathbb{N}_0$. Conclude.
answered Jan 23 at 20:13
sazsaz
81.7k861128
$endgroup$
add a comment |
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1
$begingroup$
Hints:
- Use the optional stopping theorem to show that $$X_n := M_{n wedge tau}^2 - langle M rangle_{n wedge tau}$$ is a martingale; in particular $$mathbb{E}(M_{n wedge tau}^2) - mathbb{E}(M_0^2) = mathbb{E}(langle M rangle_{n wedge tau}). tag{1}$$
- Deduce from $langle M rangle_{tau} = 0$ that $langle M rangle_{n wedge tau} = 0$ for all $n in mathbb{N}_0$.
- Show that $$mathbb{E}((M_{n wedge tau}-M_0)^2) = mathbb{E}(M_{n wedge tau}^2)-mathbb{E}(M_0^2), qquad n in mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
- Combining the first three steps shows that $$mathbb{E}((M_{n wedge tau}-M_0)^2)=0$$ for all $n in mathbb{N}_0$. Conclude.
answered Jan 23 at 20:13
sazsaz
81.7k861128
$endgroup$
add a comment |
1
$begingroup$
Hints:
- Use the optional stopping theorem to show that $$X_n := M_{n wedge tau}^2 - langle M rangle_{n wedge tau}$$ is a martingale; in particular $$mathbb{E}(M_{n wedge tau}^2) - mathbb{E}(M_0^2) = mathbb{E}(langle M rangle_{n wedge tau}). tag{1}$$
- Deduce from $langle M rangle_{tau} = 0$ that $langle M rangle_{n wedge tau} = 0$ for all $n in mathbb{N}_0$.
- Show that $$mathbb{E}((M_{n wedge tau}-M_0)^2) = mathbb{E}(M_{n wedge tau}^2)-mathbb{E}(M_0^2), qquad n in mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
- Combining the first three steps shows that $$mathbb{E}((M_{n wedge tau}-M_0)^2)=0$$ for all $n in mathbb{N}_0$. Conclude.
answered Jan 23 at 20:13
sazsaz
81.7k861128
$endgroup$
add a comment |
1
1
1
$begingroup$
Hints:
- Use the optional stopping theorem to show that $$X_n := M_{n wedge tau}^2 - langle M rangle_{n wedge tau}$$ is a martingale; in particular $$mathbb{E}(M_{n wedge tau}^2) - mathbb{E}(M_0^2) = mathbb{E}(langle M rangle_{n wedge tau}). tag{1}$$
- Deduce from $langle M rangle_{tau} = 0$ that $langle M rangle_{n wedge tau} = 0$ for all $n in mathbb{N}_0$.
- Show that $$mathbb{E}((M_{n wedge tau}-M_0)^2) = mathbb{E}(M_{n wedge tau}^2)-mathbb{E}(M_0^2), qquad n in mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
- Combining the first three steps shows that $$mathbb{E}((M_{n wedge tau}-M_0)^2)=0$$ for all $n in mathbb{N}_0$. Conclude.
answered Jan 23 at 20:13
sazsaz
81.7k861128
$endgroup$
Hints:
- Use the optional stopping theorem to show that $$X_n := M_{n wedge tau}^2 - langle M rangle_{n wedge tau}$$ is a martingale; in particular $$mathbb{E}(M_{n wedge tau}^2) - mathbb{E}(M_0^2) = mathbb{E}(langle M rangle_{n wedge tau}). tag{1}$$
- Deduce from $langle M rangle_{tau} = 0$ that $langle M rangle_{n wedge tau} = 0$ for all $n in mathbb{N}_0$.
- Show that $$mathbb{E}((M_{n wedge tau}-M_0)^2) = mathbb{E}(M_{n wedge tau}^2)-mathbb{E}(M_0^2), qquad n in mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
- Combining the first three steps shows that $$mathbb{E}((M_{n wedge tau}-M_0)^2)=0$$ for all $n in mathbb{N}_0$. Conclude.
answered Jan 23 at 20:13
sazsaz
81.7k861128
edited Jan 24 at 9:10
answered Jan 23 at 20:13
sazsaz
81.7k861128
answered Jan 23 at 20:13
sazsaz
81.7k861128
answered Jan 23 at 20:13
sazsaz
81.7k861128
81.7k861128
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$begingroup$
It means that the martingale is constant (for fixed $omega in Omega$) up to time $tau$, i.e. $M_0(omega) = M_1(omega) = ldots = M_n(omega)$ for all $n leq tau(omega)$.
$endgroup$
– saz
Jan 23 at 18:57
$begingroup$
Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n?
$endgroup$
– Francesca Ballatore
Jan 23 at 19:22