Problem with proving Brownian Motion 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 $&lt;Y&gt;_t=t$.
We have that $dM_t=dX_t+X_tdt.   $ Hence $$ d&lt;Y&gt;_t=dY_t * dY_t = \frac{1}{2}dM_t*dM_t = \frac{1}{2} (dX_t+X_tdt)(dX_t+X_tdt)=(1).$$
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 $&lt;Y&gt;_t \neq t.$ 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.
Definition of Standard BM:
$(i) B(0):=0$
$(ii)$ The increments $B(t) - B(s)$ are independent for all $0\leq s &lt; t$
$(iii)$ 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).
The quadratic variation of $ B_{e^{2t}} $ is $ e^{2t} $. If we let $ Z_t = B_{e^{2t}} $, then $ d\langle Z \rangle_t = 2e^{2t} dt $, which leads to $ d\langle X\rangle_t = 2dt $.