Mixing
Understanding specific step in Wald Lemma's proof
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A friend and I are having a hard time getting our head around a specific passage in proving Wald's Lemma for Brownian motion. The proof comes specifically from "Brownian Motion" by Mörters and Peres.
The theorem and proofs are as follows.
Let ${B(t):tgeq0}$ be a standard Brownian motion, and let $T$ be a stopping time such that either
(i) $E(T) < infty$
(ii) ${B(t wedge T):tgeq 0}$ is dominated by and integrable random variable.
Then we have that $E[B(T)] = 0$.
Proof:
We first show that if a stopping time satisfies $(i)$ then it also satisfies $(ii)$.
Suppose $(i)$ is true. Define
$$M_k = max_{0leq t leq 1}mid B(t+k)- B(k) mid qquad text{and} qquad M= sum_{k=1}^{lceil T rceil}M_k$$
then we have that
$$E(M) = Eleft[ sum_{k=1}^{lceil T rceil}M_kright] = sum_{k=0}^infty E[I_{{T>k-1}}M_k] = \=E(M_0)sum_{k=0}^infty P(T>k-1) = E(M_0)E(T+1) < infty$$
It is clear to us why this product is almost surely finite, and the rest of the proof is also clear. However, what we really don't understand is how $sum_{k=0}^infty P(T>k-1) = E(T+1)$. This is the only passage that is obscure to us and we would appreciate any help.
proof-explanation brownian-motion martingales
asked Jan 27 at 12:15
Easymode44Easymode44
419213
$endgroup$
add a comment |
$begingroup$
A friend and I are having a hard time getting our head around a specific passage in proving Wald's Lemma for Brownian motion. The proof comes specifically from "Brownian Motion" by Mörters and Peres.
The theorem and proofs are as follows.
Let ${B(t):tgeq0}$ be a standard Brownian motion, and let $T$ be a stopping time such that either
(i) $E(T) < infty$
(ii) ${B(t wedge T):tgeq 0}$ is dominated by and integrable random variable.
Then we have that $E[B(T)] = 0$.
Proof:
We first show that if a stopping time satisfies $(i)$ then it also satisfies $(ii)$.
Suppose $(i)$ is true. Define
$$M_k = max_{0leq t leq 1}mid B(t+k)- B(k) mid qquad text{and} qquad M= sum_{k=1}^{lceil T rceil}M_k$$
then we have that
$$E(M) = Eleft[ sum_{k=1}^{lceil T rceil}M_kright] = sum_{k=0}^infty E[I_{{T>k-1}}M_k] = \=E(M_0)sum_{k=0}^infty P(T>k-1) = E(M_0)E(T+1) < infty$$
It is clear to us why this product is almost surely finite, and the rest of the proof is also clear. However, what we really don't understand is how $sum_{k=0}^infty P(T>k-1) = E(T+1)$. This is the only passage that is obscure to us and we would appreciate any help.
proof-explanation brownian-motion martingales
asked Jan 27 at 12:15
Easymode44Easymode44
419213
$endgroup$
2
$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04
add a comment |
$begingroup$
A friend and I are having a hard time getting our head around a specific passage in proving Wald's Lemma for Brownian motion. The proof comes specifically from "Brownian Motion" by Mörters and Peres.
The theorem and proofs are as follows.
Let ${B(t):tgeq0}$ be a standard Brownian motion, and let $T$ be a stopping time such that either
(i) $E(T) < infty$
(ii) ${B(t wedge T):tgeq 0}$ is dominated by and integrable random variable.
Then we have that $E[B(T)] = 0$.
Proof:
We first show that if a stopping time satisfies $(i)$ then it also satisfies $(ii)$.
Suppose $(i)$ is true. Define
$$M_k = max_{0leq t leq 1}mid B(t+k)- B(k) mid qquad text{and} qquad M= sum_{k=1}^{lceil T rceil}M_k$$
then we have that
$$E(M) = Eleft[ sum_{k=1}^{lceil T rceil}M_kright] = sum_{k=0}^infty E[I_{{T>k-1}}M_k] = \=E(M_0)sum_{k=0}^infty P(T>k-1) = E(M_0)E(T+1) < infty$$
It is clear to us why this product is almost surely finite, and the rest of the proof is also clear. However, what we really don't understand is how $sum_{k=0}^infty P(T>k-1) = E(T+1)$. This is the only passage that is obscure to us and we would appreciate any help.
proof-explanation brownian-motion martingales
asked Jan 27 at 12:15
Easymode44Easymode44
419213
$endgroup$
A friend and I are having a hard time getting our head around a specific passage in proving Wald's Lemma for Brownian motion. The proof comes specifically from "Brownian Motion" by Mörters and Peres.
The theorem and proofs are as follows.
Let ${B(t):tgeq0}$ be a standard Brownian motion, and let $T$ be a stopping time such that either
(i) $E(T) < infty$
(ii) ${B(t wedge T):tgeq 0}$ is dominated by and integrable random variable.
Then we have that $E[B(T)] = 0$.
Proof:
We first show that if a stopping time satisfies $(i)$ then it also satisfies $(ii)$.
Suppose $(i)$ is true. Define
$$M_k = max_{0leq t leq 1}mid B(t+k)- B(k) mid qquad text{and} qquad M= sum_{k=1}^{lceil T rceil}M_k$$
then we have that
$$E(M) = Eleft[ sum_{k=1}^{lceil T rceil}M_kright] = sum_{k=0}^infty E[I_{{T>k-1}}M_k] = \=E(M_0)sum_{k=0}^infty P(T>k-1) = E(M_0)E(T+1) < infty$$
It is clear to us why this product is almost surely finite, and the rest of the proof is also clear. However, what we really don't understand is how $sum_{k=0}^infty P(T>k-1) = E(T+1)$. This is the only passage that is obscure to us and we would appreciate any help.
proof-explanation brownian-motion martingales
proof-explanation brownian-motion martingales
asked Jan 27 at 12:15
Easymode44Easymode44
419213
asked Jan 27 at 12:15
Easymode44Easymode44
419213
asked Jan 27 at 12:15
Easymode44Easymode44
419213
asked Jan 27 at 12:15
Easymode44Easymode44
419213
asked Jan 27 at 12:15
Easymode44Easymode44
419213
419213
2
$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04
add a comment |
2
$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04
2
2
$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04
add a comment |
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$begingroup$
If $S:Omega to mathbb{N}$ is a random variable, then $$mathbb{E}(S) = sum_{j geq 0} mathbb{P}(S>j).$$ You can prove this by verifying that $$S(omega) = sum_{j=0}^{infty} 1_{{j<S(omega)}}$$ and applying the monotone convergence theorem.
$endgroup$
– saz
Jan 27 at 12:39
$begingroup$
I am mind blown. How we could never come across this result is disturbing. Thank you very much @saz.
$endgroup$
– Easymode44
Jan 27 at 13:04