Mixing
Distribution of $int_{t}^{T} W(s)ds$ where $W$ is Brownian motion
$begingroup$
What is the distribution of $int_{t}^{T} W(s)ds$ given that $W$ is a Brownian motion?
So far, I have the following,
$$int_{t}^{T} W(s)ds = (T-t)W(t) + int_{t}^{T} (T-s)dW(s)$$
Also, $$int_{0}^{T} (T-s)dW(s) sim N(0,frac{T^{3}}{3})$$
probability-theory normal-distribution brownian-motion
edited Jan 15 at 15:46
Did
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
$endgroup$
add a comment |
$begingroup$
What is the distribution of $int_{t}^{T} W(s)ds$ given that $W$ is a Brownian motion?
So far, I have the following,
$$int_{t}^{T} W(s)ds = (T-t)W(t) + int_{t}^{T} (T-s)dW(s)$$
Also, $$int_{0}^{T} (T-s)dW(s) sim N(0,frac{T^{3}}{3})$$
probability-theory normal-distribution brownian-motion
edited Jan 15 at 15:46
Did
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
$endgroup$
add a comment |
$begingroup$
What is the distribution of $int_{t}^{T} W(s)ds$ given that $W$ is a Brownian motion?
So far, I have the following,
$$int_{t}^{T} W(s)ds = (T-t)W(t) + int_{t}^{T} (T-s)dW(s)$$
Also, $$int_{0}^{T} (T-s)dW(s) sim N(0,frac{T^{3}}{3})$$
probability-theory normal-distribution brownian-motion
edited Jan 15 at 15:46
Did
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
$endgroup$
What is the distribution of $int_{t}^{T} W(s)ds$ given that $W$ is a Brownian motion?
So far, I have the following,
$$int_{t}^{T} W(s)ds = (T-t)W(t) + int_{t}^{T} (T-s)dW(s)$$
Also, $$int_{0}^{T} (T-s)dW(s) sim N(0,frac{T^{3}}{3})$$
probability-theory normal-distribution brownian-motion
probability-theory normal-distribution brownian-motion
edited Jan 15 at 15:46
Did
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
edited Jan 15 at 15:46
Did
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
edited Jan 15 at 15:46
Did
248k23224463
edited Jan 15 at 15:46
Did
248k23224463
edited Jan 15 at 15:46
Did
248k23224463
248k23224463
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
asked Jun 2 '13 at 11:35
PeAcEPeAcE
111
111
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
By definition $X=displaystyleint_t^TW(s)mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is
$$
E[X^2]=int_t^Tint_t^TE[W(s)W(r)]mathrm drmathrm ds=int_t^Tint_t^Ttext{______}=text{___}.$$
answered Jun 2 '13 at 11:40
DidDid
248k23224463
$endgroup$
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition $X=displaystyleint_t^TW(s)mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is
$$
E[X^2]=int_t^Tint_t^TE[W(s)W(r)]mathrm drmathrm ds=int_t^Tint_t^Ttext{______}=text{___}.$$
answered Jun 2 '13 at 11:40
DidDid
248k23224463
$endgroup$
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
add a comment |
$begingroup$
By definition $X=displaystyleint_t^TW(s)mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is
$$
E[X^2]=int_t^Tint_t^TE[W(s)W(r)]mathrm drmathrm ds=int_t^Tint_t^Ttext{______}=text{___}.$$
answered Jun 2 '13 at 11:40
DidDid
248k23224463
$endgroup$
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
add a comment |
$begingroup$
By definition $X=displaystyleint_t^TW(s)mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is
$$
E[X^2]=int_t^Tint_t^TE[W(s)W(r)]mathrm drmathrm ds=int_t^Tint_t^Ttext{______}=text{___}.$$
answered Jun 2 '13 at 11:40
DidDid
248k23224463
$endgroup$
By definition $X=displaystyleint_t^TW(s)mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is
$$
E[X^2]=int_t^Tint_t^TE[W(s)W(r)]mathrm drmathrm ds=int_t^Tint_t^Ttext{______}=text{___}.$$
answered Jun 2 '13 at 11:40
DidDid
248k23224463
edited Jan 15 at 15:44
answered Jun 2 '13 at 11:40
DidDid
248k23224463
answered Jun 2 '13 at 11:40
DidDid
248k23224463
answered Jun 2 '13 at 11:40
DidDid
248k23224463
248k23224463
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
add a comment |
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
Why is the fact $X=int Wds$ is Gaussian by definition? I know one can discretise it into Riemannian sum which after splitting $W$ into tiny independent increments is just a linear sum of independent normal distributions, but I don't think this thread of logic is "trivial" at first glance.
$endgroup$
– Vim
Jan 15 at 15:36
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
@Vim Anyway, it is Gaussian.
$endgroup$
– Did
Jan 15 at 15:44
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
$begingroup$
ok. I'm just wondering what "definition" do you mean here.
$endgroup$
– Vim
Jan 15 at 15:51
add a comment |
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