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The following result is well documented and is a result of the Stochastic Processes course I am following. Below the result, I present the standard proof presented in my course (Method 1) and my incomplete proof using an alternative approach (Method 2).

Let $W$ be a standard Brownian motion with $W_0 = 0$ adapted to some filtration $(\mathscr{F}_t)_{t≥0}$ $$X_t = W_t − tW_1 \text{ for } \space t \in [0, 1]$$ And let $$\beta_t = (t + 1)X_{\frac{t}{t+1}} \text{ for } \space t \ge 0$$ Then $(\beta_t)_{t \ge 0}$ is a standard Brownian motion in its own filtration.

I know that the process is Brownian Motion and that Method 1 (below) is the standard way of demonstrating this fact. However, I am wondering how I can get Method 2 to work. Perhaps $dW_{\frac{t}{t+1}}$ can be expressed in some way that allows for a cancellation in the drift term to allow us to conclude that the process is a martingale.

Method 1: Note that $$X_t = (1 − t)W_t − t(W_1 − W_t)$$ so that $X_t$ is the sum of two independent and Gaussian random variables. Thus, $X$ is continuous and Gaussian and $E [X_t] = 0$. This implies $β$ is also continuous and Gaussian and $E [\beta_t] = 0$. Suppose $s \leq t$. Then one can fairly straightforwardly show that $E[ \beta _t \beta_s] = s$. Thus characterising $\beta$ as a Brownian Motion.

Method 2: By Ito's Product Rule, $$d \beta _t = (t+1) dX_{\frac{t}{t+1}} + X_{\frac{t}{t+1}} dt$$ $$ = (W_{\frac{t}{t+1}} - W_1)dt + (t+1) dW_{\frac{t}{t+1}} $$ If I can show that the drift term disappears, then this is a local martingale and will allow me to invoke Levy's Characterisation of a Brownian Motion (by also showing that $[ \beta, \beta]_t = t$).

I would be grateful for any assistance.

Edit: I have written a continuation of Method 2 below, but am seeking verification or feedback so that I can be sure that it is correct. I believe I have resolved the issues above, but this is a relatively new topic to me and would like confirmation

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2 Answers 2

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This is more a very long comment:

Formally all your formulas are correct but the conclusion at the end of Method 2 is implausible to me.

From $$ \beta_t=(1+t)X_{\tfrac t{1+t}}\,, $$ we have $$ \beta_{\tfrac s{1-s}}=\frac 1{1-s} X_s\,. $$ Using integration by parts (Ito), $$\tag1 d\beta_{\tfrac s{1-s}}=\frac{dX_s}{1-s}+\frac{X_s\,ds}{(1-s)^2}\,. $$ Using, $$ s=\frac{t}{1+t}\,,\quad t=\frac{s}{1-s}\,,\quad\frac 1{1-s}=1+t\,,\quad ds=\frac{dt}{(1+t)^2}\,,\quad dt=\frac{ds}{(1-s)^2} $$ I get $$\tag2 d\beta_t=(1+t)\,dX_{\tfrac{t}{1+t}}+X_{\tfrac{t}{1+t}}\,dt $$ as you did. Using $$ dX_s=dW_s-W_1\,ds\,,\quad\quad dX_{\tfrac{t}{1+t}}=dW_{\tfrac{t}{1+t}}-\frac{W_1\,dt}{(1+t)^2} $$ I get from (2) \begin{align} d\beta_t&=(1+t)\,dW_{\tfrac{t}{1+t}}-\frac{W_1\,dt}{1+t}+\left(W_{\tfrac{t}{1+t}}-tW_1\right)\,dt\\[2mm] &=(1+t)\,dW_{\tfrac{t}{1+t}}+\left(W_{\tfrac{t}{1+t}}-W_1\right)\,dt\,\tag3 \end{align} as you did.

Now we are supposed to show that $\beta_t=\int_0^td\beta_s$ is a martingale. Clearly, the filtration of $\beta_t$ includes $W_1$ for all $t$ by definition of the Brownian bridge $X\,.$ When $t\in[0,1)$ this filtration is much larger than that of $W_{\frac{t}{1+t}}\,.$ Therefore we cannot conclude that $$ \int_0^t(1+s)\,dW_{\tfrac{s}{1+s}} $$ is a martingale under the large filtration (as required). Then we cannot conclude that the remaining task is to show that in (3) the drift $(W_{\tfrac{t}{1+t}}-W_1)\,dt$ must disappear. This cannot be expected simply by just looking at it in isolation.

  • A few more thoughts.

Imho Method 1 is the preferred way to go because it is much simpler. Having established with it that $\beta_t$ is a Brownian motion we can write (1) as $$\tag4 (1-s)\,d\beta_{\tfrac s{1-s}}=dX_s+\frac{X_s\,ds}{1-s}\,. $$ Because $$ \int_0^t(1-s)\,d\beta_{\tfrac s{1-s}} $$ is a martingale with quadratic variation $$ \int_0^t(1-s)^2\,d\Big(\frac s{1-s}\Big)=\int_0^t(1-s)^2\,\frac{ds}{(1-s)^2}=t $$ it is a Brownian motion. Let's call this BM $B_t\,.$ Then (4) becomes the well-known SDE for the Brownian bridge $$\tag5 dX_s=-\frac{X_s\,ds}{1-s}+dB_s $$ which has the well-known explicit solution $$\tag6 X_s=(1-s)\int_0^s\frac{dB_t}{1-t}\,. $$

  • The interesting conclusion from Method 1 is that the integral of the RHS of (3) is a martingale under the filtration of $\beta$ despite it having a term that looks like a drift.
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  • $\begingroup$ I have just updated and found yet another BM $B_t$ completing the zoo. Regarding your question about $dW_{t/(1+t)}\,:$ Think of the time changed BM $W_{t/(1+t)}$ as being a martingale $M_t$ with quadratic variation $t/(1+t)\,.$ Then $dW_{t/(1+t)}$ is $dM_t\,.$ This is as good as it gets. Ito's formula won't help much here. $\endgroup$
    – Kurt G.
    Commented Apr 28 at 18:20
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The following is a partial proof with one assumption needed to get it over the line. This assumption is labelled as such below.


Martingale Property

By Ito's Product Rule, $$d \beta _t = (t+1) dX_{\frac{t}{t+1}} + X_{\frac{t}{t+1}} dt$$ $$ = (W_{\frac{t}{t+1}} - W_1)dt + (t+1) dW_{\frac{t}{t+1}} $$

As mentioned in the question. Therefore integrating gives us:

$$ \beta_t = \int _0^t(W_{\frac{u}{u+1}} - W_1)du + \int _0^t (u+1) dW_{\frac{u}{u+1}} \tag1$$

Now we test $\beta$ for the martingale property by considering the conditional expectation of $\beta _t$ with respect to $\mathscr{F}_s$ where we consider $s<t$:

$$ E(\beta_t | \mathscr{F}_s) = E \Big{(}\int _0^t(W_{\frac{u}{u+1}} - W_1)du + \int _0^t (u+1) dW_{\frac{u}{u+1}} | \mathscr{F}_s \Big{)} $$

Now we can split up the integrals from $0$ to $t$ into two separate integrals from $0$ to $s$ and $s$ to $t$ using linearity:

$$ = \color{red}{\int _0^s(W_{\frac{u}{u+1}} - W_1)du + \int _0^s (u+1) dW_{\frac{u}{u+1}}} + E \Big{(}\int _s^t(W_{\frac{u}{u+1}} - W_1)du + \int _s^t (u+1) dW_{\frac{u}{u+1}} | \mathscr{F}_s \Big{)} \tag2$$

$$ = \color{red}{\beta _s} + E \Big{(}\int _s^t(W_{\frac{u}{u+1}} - W_1)du + \int _s^t (u+1) dW_{\frac{u}{u+1}} | \mathscr{F}_s \Big{)} \tag3 $$

Now for $\beta$ to be a martingale we simply need to show that this expectation is equal to zero.

First we note that these integrals are now from $s$ to $t$ but we cannot drop the conditioning on $\mathscr{F}_s$. As pointed out in the comments, since $\frac{s}{1+s} < s$ we cannot remove the conditioning by $\mathscr{F}_s$

$$E \Big{(}\int _s^t(W_{\frac{u}{u+1}} - W_1)du | \mathscr{F}_s \Big{)} + E \Big{(} \int _s^t (u+1) dW_{\frac{u}{u+1}} | \mathscr{F}_s \Big{)} \tag4 $$

Both expectations must be zero in order to conclude that $E(\beta_t | \mathscr{F}_s) = \beta _s$ to show that $\beta$ satisfies the martingale property. However, it is unclear how to show this. We know, of course, that they must be equal to zero since Method 1 proves that this is indeed a Brownian Motion.

We need these expectations to be zero. This is our key assumption. The remainder of the proof is complete.


Quadratic Variation

It remains to show that the quadratic variation is equal to $t$. Using the definition of $\beta$ above, we have that $$[\beta, \beta]_t = \int_0^t (s+1)^2 (dW_{s / (s+1)})^2$$ And we know that $$ [W,W]_t = t$$ Therefore, we know that the quadratic variation under the new index is: $$[W,W]_{t/(t+1)} = \frac{t}{t+1}$$ Which is equivalent to $$(dW_{s / (s+1)})^2 = \frac{1}{(s+1)^2} dt$$

And so by substitution into our expression for the quadratic variation of $\beta$ we find that

$$[\beta, \beta]_t = \int_0^t (s+1)^2 \frac{1}{(s+1)^2} dt = t \tag5$$

And so the proof is complete.

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  • $\begingroup$ Since I could not add the tag brownian-bridge (to many) I recommend to mention that in the title. I also recommend that you number your equations to make a discussion easier. The quadratic variation part should be correct. The martingale part is almost correct but at one crucial point it fails: if your reasoning worked then we could also say that $$\int_s^tdW_{\frac{u}{1+u}}=W_{\frac{t}{1+t}}-W_{\frac{s}{1+s}}$$ is independent of ${\cal F}_s\,.$ But this is not true since $\frac{s}{1+s}<s\,.$ Even worse: ${\cal F}_s$ contains $W_1$ as it must when it is the filtration of $\beta\,.$ $\endgroup$
    – Kurt G.
    Commented Apr 29 at 4:16
  • $\begingroup$ I have updated the title and numbered the equations. That makes sense. So that means we cannot drop the conditioning on $\mathscr{F}_s$. But the expectations in (4) must both still be equal to zero somehow as we know that $\beta$ is a martingale. Is there some other way to get there? @KurtG. $\endgroup$
    – FD_bfa
    Commented Apr 29 at 6:54
  • $\begingroup$ I learned from this very interesting question that (as I wrote) when we first use Method 1 to prove that $\beta$ is a BM we can conclude that this expression in your eq. (3) has expectation zero. You are trying the solo ascent through the Eiger north face while I took the train. Not using method one seems hard and could be a long standing open question. $\endgroup$
    – Kurt G.
    Commented Apr 29 at 7:15
  • $\begingroup$ Thank you for your help @KurtG. One of the motivations for this question was that when I first attempted the problem, I did not think to use Method 1 and my attempt was to try to show that it was a Brownian Motion using Method 2. $\endgroup$
    – FD_bfa
    Commented Apr 29 at 7:23
  • $\begingroup$ As far as I can tell the master of Brownian bridge is George Lowther. Since I rather do everything I can myself I have never read all the proofs of his Lemmas. Perhaps he has nother method but I would doubt. Let me know. $\endgroup$
    – Kurt G.
    Commented Apr 29 at 7:26

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