Standard Brownian Conditional expectation

Question

(Given the process $(B(t))t≥0$ of Brownian motion, define the random variables

$$Y=\int_0^{1}B(s)\,ds $$ $$X=B(1) $$

Determine the quantities $E(Y|X)$, $Var(Y−E(Y|X))$ and the conditioned density $f_{Y|X}$

I already compute mean and variance using Fubini's theorem $E(Y) = \int_0^{1}E(B(s))\,ds =0$ and $Var(Y)= E(Y^2)=\dfrac{1}{3}$

I don't know if it's help me, I have this formula aswell $E(Y|X) = E(Y)+Cov(Y,X) + Cov(X,X)^{-1} + (X-E(X))$

Thanks for your help,

As I can't edit my comment anymore,

Thanks for the hint but I don't catch why it helps me. My problem is to deal with the quantities $\int_0^{1}E(B(s))\,ds$|X). I tried to use the formula : $Cov(Y,X) = E[(Y(t)-E(X)(X-E(X))=E[Y(t)B(1)]$ for $t<1$ $B(1)= Y(t)+(B1-Y(t))$ then we have $B(1)Y(t) = Y(t)^2+Y(t)(B(1)-Y(t))$ then $E[B(1)Y(t)]= E[Y^2] + E[Y(t)]E[B(1)-Y(t)]$ as increments are independent we find that $E[B(1)Y(t)]= \dfrac{1}{3}$ Then if I do the same for $t>1$ I find $E[B(1)Y(t)] = 1$.

Answer 1

Hint: you can write $$ B(s) = sX + T(s) $$ where $T$ is a Brownian bridge.

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you get $$ E\left[Y|X\right] = E\left[\int_0^1 sXds\right] + E\left[\int_0^1 T(s)ds\right] = \frac X2 $$because $T$ has a symetric law.

and then $$ var(Y - E[Y|X]) = var\left(\int_0^1 T(s) ds\right) $$

Answer 2

2

$$Y=\int_0^{1}B(s)\,ds $$ $$X=B(1) $$

We use fubini theorem for continuous process $X(t)$, $t\in [0,T]$ ,under appropriate assumptions it holds that : $E(\int_0^{T}X(t)\,dt) = \int_0^{T}E(X(t))\,dt$ and $E(\int_0^{T}X(t)\,dt)^2 = \int_0^{T} \int_0^{T}E(X(t)X(s))\,dtds=\int_0^{T} (\int_0^{T}E(X(t)X(s))\,dt)ds$

Thus we have $E(Y)= \int_0^{1}E(B(t))\,dt=0$ and we find $Var(Y) = E(Y^2)= E[\int_0^{1}B(t)\,dt\int_0^{1}B(s)\,ds]=\int_0^{1} \int_0^{1}E(B(t)B(s))\,dtds=$$\int_0^{1} \int_0^{1}min(t,s)\,dtds=\int_0^{1} (\int_0^{t}s\,ds+\int_t^{1}t\,ds)\,dt=\dfrac{1}{3}$

A stochastic process $X(t)\in D$ is called a Gaussian process if for each choice {t1,...,tn} $c$ D,$n \in N$ of indices , the vector $ (X(t1),...,X(tn)) $ follow a multivariate normal distribution.

By definition, a brownian motion is a gaussian process.

Theorem: Let random vectors $Y=(Y1,...,Ym) $ and $X=(X1,...,Xn)^{T}$ have a joint Gaussian distribution Then :

1) if $Cov(Y,X)=0$ then $X,Y$ are independent ( $Hint$ : show that characteristic function factorizes)

2) $G:= Y-E(Y)-Cov(Y,X)Cov(X,X)^{-1} (X-E(X))$ and X are independent. $Cov(G,X)=E[G(X-E(X))^{T}]=Cov(Y,X)-Cov(Y,X)=0$ as G and X are normally distributed their uncorrelations implies independence.

3) Because of independence of G and X we have $E(G|X)=E(G)=0$ by properties of conditionnal expectation we have :

$0=E(G|X)=E(Y|X)-E(Y)-Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$ $E(Y|X)=E(Y)+Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$

4) $E[(Y-E(Y|X))^2]=Cov(Y,Y)-Cov(Y,X)[Cov(X,X)]^{-1}Cov(Y,X))$

So for our problem : $E(Y|X)=E(Y)+Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$ $=0+Cov(Y,X)*1*X$

$Cov(Y,X)=E[(Y-E(Y))(X-E(X))]=E[YX]=E[B(1)\int_0^{1}B(s)]\,ds=E[\int_0^{1}B(1)B(s)\,ds]$

$1<s$ $B(1)=B(s)+(B(1)-B(s))$

$B(1)B(s)= B(s)^2+ B(s)(B(1)-B(s))$

$E[\int_0^{1}B(1)B(s)]=E[\int_0^{1}B(s)^2+ B(s)(B(1)-B(s))\,ds]=E[\int_0^{1}B(s)^2\,ds+\int_0^{1} B(s)(B(1)-B(s))\,ds] =\int_0^{1}E[B(s)^2]\,ds+\int_0^{1} E[B(s)(B(1)-B(s))]\,ds]$

With the independence of the increments $E[B(s)(B(1)-B(s))]=E[B(s)]E[(B(1)-B(s))]=0$

Thus we have $Cov(Y,X)=E[XY]=E[\int_0^{1}B(1)B(s)]= \int_0^{1}s\,ds=\dfrac{1}{2}$

and then $ E(Y|X) = \dfrac{1}{2}X$

from $4)$ we have $E[(Y-E(Y|X))^2]=\dfrac{1}{3}-\dfrac{1}{2}X[1*\dfrac{1}{2}X]$

$=\dfrac{1}{3}-\dfrac{1}{4}X^2$

Finally, $Var[Y-E(Y|X)]$

$=E[(Y-E(Y|X))^2]-E[Y-E(Y|X)]^2$

$= \dfrac{1}{3}-\dfrac{1}{4}X^2-\dfrac{1}{4}X^2$

$=\dfrac{1}{3}-\dfrac{1}{2}X^2$

I think I did it correctly but I am not sure for the part with the independence of the increments as if I found the same than mookid ( merci d'ailleurs). feel free to tell me if there is an error , I am going to compute density now.