Mixing
Ito Isometry for Brownian Bridge
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I have been looking at the following expression, and I'm a bit stuck:
$$left(int_{0}^{t}frac{dW_{u}}{1-u}(B_{1} - B_{u})right)^{2}$$ where ${W_{t}}_{tin[0,1]}$ is a standard Weiner Process and ${B_{t}}_{tin[0,1]}$ is a generalized Brownian bridge (endpoints need not be the same). If I apply expectation to this, will the Ito Isometry hold? Is the resulting expression just:
$$mathbb{E}left[int_{0}^{t}frac{du}{(1-u)^{2}}(u - u^{2})^{2}right] = mathbb{E}left[int_{0}^{t}u^{2}duright]?$$ Any help or clarity is greatly appreciated.
stochastic-processes brownian-motion
asked Jan 22 at 21:33
IndexIndex
213
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add a comment |
$begingroup$
I have been looking at the following expression, and I'm a bit stuck:
$$left(int_{0}^{t}frac{dW_{u}}{1-u}(B_{1} - B_{u})right)^{2}$$ where ${W_{t}}_{tin[0,1]}$ is a standard Weiner Process and ${B_{t}}_{tin[0,1]}$ is a generalized Brownian bridge (endpoints need not be the same). If I apply expectation to this, will the Ito Isometry hold? Is the resulting expression just:
$$mathbb{E}left[int_{0}^{t}frac{du}{(1-u)^{2}}(u - u^{2})^{2}right] = mathbb{E}left[int_{0}^{t}u^{2}duright]?$$ Any help or clarity is greatly appreciated.
stochastic-processes brownian-motion
asked Jan 22 at 21:33
IndexIndex
213
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Why were you able to move the squaring inside the integral?
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– Larry B.
Jan 22 at 23:25
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@LarryB. Assuming the Ito Isometry holds.
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– Index
Jan 23 at 3:49
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How are $W$ and $B$ related?
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– John Dawkins
Jan 23 at 21:49
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@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45
add a comment |
$begingroup$
I have been looking at the following expression, and I'm a bit stuck:
$$left(int_{0}^{t}frac{dW_{u}}{1-u}(B_{1} - B_{u})right)^{2}$$ where ${W_{t}}_{tin[0,1]}$ is a standard Weiner Process and ${B_{t}}_{tin[0,1]}$ is a generalized Brownian bridge (endpoints need not be the same). If I apply expectation to this, will the Ito Isometry hold? Is the resulting expression just:
$$mathbb{E}left[int_{0}^{t}frac{du}{(1-u)^{2}}(u - u^{2})^{2}right] = mathbb{E}left[int_{0}^{t}u^{2}duright]?$$ Any help or clarity is greatly appreciated.
stochastic-processes brownian-motion
asked Jan 22 at 21:33
IndexIndex
213
$endgroup$
I have been looking at the following expression, and I'm a bit stuck:
$$left(int_{0}^{t}frac{dW_{u}}{1-u}(B_{1} - B_{u})right)^{2}$$ where ${W_{t}}_{tin[0,1]}$ is a standard Weiner Process and ${B_{t}}_{tin[0,1]}$ is a generalized Brownian bridge (endpoints need not be the same). If I apply expectation to this, will the Ito Isometry hold? Is the resulting expression just:
$$mathbb{E}left[int_{0}^{t}frac{du}{(1-u)^{2}}(u - u^{2})^{2}right] = mathbb{E}left[int_{0}^{t}u^{2}duright]?$$ Any help or clarity is greatly appreciated.
stochastic-processes brownian-motion
stochastic-processes brownian-motion
asked Jan 22 at 21:33
IndexIndex
213
asked Jan 22 at 21:33
IndexIndex
213
asked Jan 22 at 21:33
IndexIndex
213
asked Jan 22 at 21:33
IndexIndex
213
asked Jan 22 at 21:33
IndexIndex
213
213
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Why were you able to move the squaring inside the integral?
$endgroup$
– Larry B.
Jan 22 at 23:25
$begingroup$
@LarryB. Assuming the Ito Isometry holds.
$endgroup$
– Index
Jan 23 at 3:49
$begingroup$
How are $W$ and $B$ related?
$endgroup$
– John Dawkins
Jan 23 at 21:49
$begingroup$
@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45
add a comment |
$begingroup$
Why were you able to move the squaring inside the integral?
$endgroup$
– Larry B.
Jan 22 at 23:25
$begingroup$
@LarryB. Assuming the Ito Isometry holds.
$endgroup$
– Index
Jan 23 at 3:49
$begingroup$
How are $W$ and $B$ related?
$endgroup$
– John Dawkins
Jan 23 at 21:49
$begingroup$
@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45
$begingroup$
Why were you able to move the squaring inside the integral?
$endgroup$
– Larry B.
Jan 22 at 23:25
$begingroup$
Why were you able to move the squaring inside the integral?
$endgroup$
– Larry B.
Jan 22 at 23:25
$begingroup$
@LarryB. Assuming the Ito Isometry holds.
$endgroup$
– Index
Jan 23 at 3:49
$begingroup$
@LarryB. Assuming the Ito Isometry holds.
$endgroup$
– Index
Jan 23 at 3:49
$begingroup$
How are $W$ and $B$ related?
$endgroup$
– John Dawkins
Jan 23 at 21:49
$begingroup$
How are $W$ and $B$ related?
$endgroup$
– John Dawkins
Jan 23 at 21:49
$begingroup$
@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45
$begingroup$
@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45
add a comment |
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$begingroup$
Why were you able to move the squaring inside the integral?
$endgroup$
– Larry B.
Jan 22 at 23:25
$begingroup$
@LarryB. Assuming the Ito Isometry holds.
$endgroup$
– Index
Jan 23 at 3:49
$begingroup$
How are $W$ and $B$ related?
$endgroup$
– John Dawkins
Jan 23 at 21:49
$begingroup$
@John Dawkins I've been just using the fact that B is the solution to the SDE $$dB_{t} = frac{b- B_{t}}{1-t}dt + dW_{t}.$$ However, I could also use $$B_t = (1-t)a + tb + Z_{t}$$ where $Z_{t} = W_t - tW_1$ and that would be fine.
$endgroup$
– Index
Jan 24 at 13:45