Problem with proving Brownian Motion

Question

Hey i want to prove that $Y_t= \frac{1}{\sqrt{2}}M_t $ is a B.M, while $$ M_t = X_t-X_0 + \int_{0}^{t} X_s ds $$ and $$ X_t=e^{-t} B_{e^{2t}}, $$ where $(B_t)_{t \geq 0}$ is a standart B.M..

In order to show this i wanted to show that $<Y>_t=t$.

We have that $dM_t=dX_t+X_tdt. $ Hence $$ d<Y>_t=dY_t * dY_t = \frac{1}{2}dM_t*dM_t = \frac{1}{2} (dX_t+X_tdt)(dX_t+X_tdt)=(1).$$ Since $dX_t= d(e^{-t}B_{e^{2t}}) = -e^{-t} B_{e^{2t}} dt + e^{-t} dB_{e^{2t}}, $

we get $$ (1)= \frac{1}{2} e^{-2t}dB_{e^{2t}} dB_{e^{2t}} = \frac{1}{2} e^{-2t}e^{2t} dt = \frac{1}{2} dt. $$ Hence $<Y>_t \neq t.$ So $Y_t$ is not a Brownian Motion? So what am i missing? Furthermore i am quiet sure that $Y_t$ is not a martingale. Any help would be appreciate.

Answer 1

Definition of Standard BM:

$(i) B(0):=0$

$(ii)$ The increments $B(t) - B(s)$ are independent for all $0\leq s < t$

$(iii)$ The increments $B(t) - B(s) \sim N(0, (t-s))$

$(iv)$ The function $t \rightarrow B(t)$ is almost surely continuous.

We need to show that the above hold for $Y_t$. I will start by trying to show that $Var(Y_t)=t$:

$Var(\frac{1}{\sqrt{2}}M_t)=\frac{1}{2}Var(M_t)=\\=0.5\mathbb{E}[M_t^2]-0.5\mathbb{E}[M_t]^2$

First note that $X_0=1B(1) \sim N(0,1)$

Now taking expectation of $M_t$:

$\mathbb{E}[M_t]=\mathbb{E}[X_t]-\mathbb{E}[X_0]+\int_0^t \mathbb{E}[X_s]ds=\\=0-0+0$

Now:

$M_t^2=\left(X_t-X_0+\int_{s=0}^{t}X_s ds\right)^2=\\=X_t^2-2X_tX_0+2X_t\int_{s=0}^{t}X_s ds-X_0^2+2X_0\int_{s=0}^{t}X_s ds+\int_{s=0}^{t}X_s ds\int_{h=0}^{t}X_h dh$

To work out the expectation over $M_t^2$, I will do the following:

$\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)\sim N(0,e^{2t})$

So that:

$X_t=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)$

And:

$X_t^2=e^{-2t}\frac{e^{2t}}{t}B(t)^2$

Taking the expectation:

$$\mathbb{E}[M_t^2]=\mathbb{E}[X_t^2]-2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]+2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]-\mathbb{E}[X_0^2]+2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]+\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]$$

Now, the following terms are straight forward:

$$\mathbb{E}[X_t^2]=e^{-2t}\frac{e^{2t}}{t}\mathbb{E}[B(t)^2]=e^{-2t}e^{2t}=1$$

$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]=min(1,t)2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$$

$$\mathbb{E}[X_0^2]=1$$

To workout the other three expectations, I will try to derive the process for $X_t$:

$\frac{\partial X}{\partial t}=\frac{2}{\sqrt{t}}-0.5e^{-0.5t}$

$\frac{\partial X}{\partial B_t}=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$

$\frac{\partial^2 X}{\partial B_t^2}=0$

By Ito's Lemma:

$$X(B_t,t)= X(B_0,t_0)_{=0} + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial t}+\frac{\partial X}{\partial B_h}a(B_h,h)_{=0}+\frac{1}{2}\frac{\partial^2 X}{\partial B^2}_{=0}b(B_h,h)^2_{=1}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial B}b(X_h,h)_{=1}\right)dB_h=\\=\int_{h=0}^{h=t}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=t}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h$$

So that:

$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]=\\=2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}\left[B_t\int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right\} ds\right]$$

And:

$$2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]=\\=2\mathbb{E}\left[X_0 \int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h \right\} ds \right]$$

And:

$$\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]=\\=\mathbb{E}\left[\int_{s=0}^{t}\left(\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) ds\int_{u=0}^{t}\left(\int_{h=0}^{h=u}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=u}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) du\right]$$

I will have a go at working out these later.

Ps: I am not 100% sure that $M_t$ is an Ito process (which would of course immediately imply that $Y_t$ cannot be a Brownian motion).

Answer 2

Hey i tried this again and got this:

$<Y>_t=t$, since i forgot a factor $2$. (From the answer from @Hoil Lee) Hence it remains to show that $Y_t$ is a martingale. We have $$ \mathbb{E}[M_t-M_s \vert \mathcal{F}_s]= \mathbb{E}[X_t-X_s + \int_s^t X_u du \vert \mathcal{F}_s]= \mathbb{E}[X_t-X_s \vert \mathcal{F}_s]+ \mathbb{E}[\int_s^t X_u -X_s+X_u \vert \mathcal{F}_s] = \mathbb{E}[X_t-(t-s-1)X_s \vert \mathcal{F}_s] + \int_s^t \mathbb{E}[ X_u-X_s \vert \mathcal{F}_s]=(1)$$ Since the fact from @Jan Stuller, we have that $ X_t \sim e^-t \frac{\sqrt{e^{2t}}}{\sqrt{t}}B_t $ and we know that $B_t-B_s \amalg \mathcal{F}_s $, we can conclude that $$ (1)=\mathbb{E}[X_t-(t-s-1)X_s] + \int_s^t \mathbb{E}[ X_u-X_s]=0,$$ which proofs that $Y_t$ is a Martingale and hence a Brownian Motion.