Mixing
Where is the flaw in my reasoning when computing the expectation of a Brownian motion dependent on a rate one...
$begingroup$
We have that ${W_t:tgeq 0}$ is a Brownian motion, and ${N_t:tgeq 0}$ is a rate one poisson process which is independent of the Brownian motion.
We must show that $$mathbb{E}[W_{N_t}^2]=t$$
I figured there are two ways to do this, but they yield different answers. Why is the one of them (or both) incorrect?
Method 1: Using the tower property we have that $$mathbb{E}[W_{N_t}^2]=mathbb{E}[mathbb{E}[W_{N_t}^2|mathscr{F}_{N_t}]]=mathbb{E}[N_t]=t$$
Where $mathscr{F}_{N_t}=sigma{W_s,0leq sleq N_t}$
Method 2: We can use the fact that $$W_{N_t}sim N(0,N_t)$$
Which would mean that we have $$mathbb{E}[W_{N_t}^2]=Var(W_{N_t})=N_t$$
Obviously there is something going wrong in method 2, but what is it? Didn't we use pretty much the same trick in method 1, except there we conditioned on the filtration? Why does this ''matter'' here? Does $W_{N_t}$ not have this distribution?
Any help is appreciated.
probability probability-theory brownian-motion conditional-expectation
asked Jan 24 at 14:01
S. CrimS. Crim
404212
$endgroup$
add a comment |
$begingroup$
We have that ${W_t:tgeq 0}$ is a Brownian motion, and ${N_t:tgeq 0}$ is a rate one poisson process which is independent of the Brownian motion.
We must show that $$mathbb{E}[W_{N_t}^2]=t$$
I figured there are two ways to do this, but they yield different answers. Why is the one of them (or both) incorrect?
Method 1: Using the tower property we have that $$mathbb{E}[W_{N_t}^2]=mathbb{E}[mathbb{E}[W_{N_t}^2|mathscr{F}_{N_t}]]=mathbb{E}[N_t]=t$$
Where $mathscr{F}_{N_t}=sigma{W_s,0leq sleq N_t}$
Method 2: We can use the fact that $$W_{N_t}sim N(0,N_t)$$
Which would mean that we have $$mathbb{E}[W_{N_t}^2]=Var(W_{N_t})=N_t$$
Obviously there is something going wrong in method 2, but what is it? Didn't we use pretty much the same trick in method 1, except there we conditioned on the filtration? Why does this ''matter'' here? Does $W_{N_t}$ not have this distribution?
Any help is appreciated.
probability probability-theory brownian-motion conditional-expectation
asked Jan 24 at 14:01
S. CrimS. Crim
404212
$endgroup$
$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
1
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
1
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16
add a comment |
$begingroup$
We have that ${W_t:tgeq 0}$ is a Brownian motion, and ${N_t:tgeq 0}$ is a rate one poisson process which is independent of the Brownian motion.
We must show that $$mathbb{E}[W_{N_t}^2]=t$$
I figured there are two ways to do this, but they yield different answers. Why is the one of them (or both) incorrect?
Method 1: Using the tower property we have that $$mathbb{E}[W_{N_t}^2]=mathbb{E}[mathbb{E}[W_{N_t}^2|mathscr{F}_{N_t}]]=mathbb{E}[N_t]=t$$
Where $mathscr{F}_{N_t}=sigma{W_s,0leq sleq N_t}$
Method 2: We can use the fact that $$W_{N_t}sim N(0,N_t)$$
Which would mean that we have $$mathbb{E}[W_{N_t}^2]=Var(W_{N_t})=N_t$$
Obviously there is something going wrong in method 2, but what is it? Didn't we use pretty much the same trick in method 1, except there we conditioned on the filtration? Why does this ''matter'' here? Does $W_{N_t}$ not have this distribution?
Any help is appreciated.
probability probability-theory brownian-motion conditional-expectation
asked Jan 24 at 14:01
S. CrimS. Crim
404212
$endgroup$
We have that ${W_t:tgeq 0}$ is a Brownian motion, and ${N_t:tgeq 0}$ is a rate one poisson process which is independent of the Brownian motion.
We must show that $$mathbb{E}[W_{N_t}^2]=t$$
I figured there are two ways to do this, but they yield different answers. Why is the one of them (or both) incorrect?
Method 1: Using the tower property we have that $$mathbb{E}[W_{N_t}^2]=mathbb{E}[mathbb{E}[W_{N_t}^2|mathscr{F}_{N_t}]]=mathbb{E}[N_t]=t$$
Where $mathscr{F}_{N_t}=sigma{W_s,0leq sleq N_t}$
Method 2: We can use the fact that $$W_{N_t}sim N(0,N_t)$$
Which would mean that we have $$mathbb{E}[W_{N_t}^2]=Var(W_{N_t})=N_t$$
Obviously there is something going wrong in method 2, but what is it? Didn't we use pretty much the same trick in method 1, except there we conditioned on the filtration? Why does this ''matter'' here? Does $W_{N_t}$ not have this distribution?
Any help is appreciated.
probability probability-theory brownian-motion conditional-expectation
probability probability-theory brownian-motion conditional-expectation
asked Jan 24 at 14:01
S. CrimS. Crim
404212
asked Jan 24 at 14:01
S. CrimS. Crim
404212
asked Jan 24 at 14:01
S. CrimS. Crim
404212
asked Jan 24 at 14:01
S. CrimS. Crim
404212
asked Jan 24 at 14:01
S. CrimS. Crim
404212
404212
$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
1
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
1
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16
add a comment |
$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
1
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
1
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16
$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
1
1
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
1
1
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16
add a comment |
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$begingroup$
@AddSup Why is that wrong when, for a normal Brownian motion, we have that $W_tsim N(0,t)$? Or is that also not the case? And in the first method I condition on the filtration right, so there I used it correctly then, I assume.
$endgroup$
– S. Crim
Jan 24 at 14:15
1
$begingroup$
Method 2 is wrong... what do you possibly mean by $N(0,X)$ for a random variable $X$...? Variances are always deterministic
$endgroup$
– saz
Jan 24 at 15:38
1
$begingroup$
Also, in Method 1, I think you should condition on $N_t$, not your $mathcal F_{N_t}$. I don't know the exact definition of $mathcal F_{N_t}$, but if it can be defined and means what it intends to mean, then $W_{N_t}inmathcal F_{N_t}$. So the inner conditional expectation does nothing: $E(W_{N_t}^2|mathcal F_{N_t})=W_{N_t}^2$. (I deleted my earlier comment, which is why S. Crim's precedes mine. I thought I deleted it immediately...)
$endgroup$
– AddSup
Jan 25 at 14:36
$begingroup$
I agree with AddSup, just use $N_t$ instead of the $sigma$-algebra in Method $1$. Method $2$ is really an incomplete Method 1: you have conditioned on $N_t$ to get $E(W^2_{N_t}) = N_t$ and now you take the expectation over $N_t$ to get $t$. I must disagree with saz, the variance of a normal distribution can be treated as a random variable, e.g. in Bayesian statistics (we would use a prior for the variance if there was uncertainty over its value).
$endgroup$
– Alex
Jan 26 at 14:16