# Show $X_t=t\cdot W_{1/t}$ is a martingale using Ito

For $W_t$ a BM, I want to show that $X_t=t\cdot W_{1/t}$ is a martingale using Ito.

As shown in
$\quad$ https://math.stackexchange.com/questions/182107
$\quad$ https://math.stackexchange.com/questions/1358542
we know $X_t$ is a BM, and in particular a martingale.

However, I was wondering if it was possible to show the martingale feature more "directly" by showing that the resulting SDE of $X_t$ from using Ito has a zero drift term, which tells us that we have a local martingale. Ensuring appropriate integrability assumptions we arrive at a proper martingale. Similarly as in
$\quad$ https://math.stackexchange.com/questions/525390
$\quad$ https://quant.stackexchange.com/questions/2435

Having $f(t,Y_t)$ Ito tells us that

$$df(t,Y_t) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial y}dY_t + \frac12\frac{\partial^2 f}{\partial y^2}d\langle Y\rangle_t$$

But now I am unsure how to populate $Y_t$ with $W_{1/t}$. If we had $X_t=t\cdot W_t$ instead we could easily set $f(t,x) = t\cdot x$ and $Y_t=W_t$. But I am unsure how to treat $W_{1/t}$ in this context.

Any help would be appreciated.

Try $u=\frac{1}{t}$, $\hat{W}_{u}=W_{\frac{1}{t}}$, and then apply Ito lemma for $f(u,x)=\frac{1}{u}x$.
• I get $df(u,\hat{W}_u) = -(1/u^2)\cdot\hat{W}_udu + (1/u)d\hat{W}_u$ and using $du/dt =-1/t^2$ I get $df(u,\hat{W}_u) = \hat{W}_udt+(1/u) d\hat{W}_u = W_{1/t}dt + t d\hat{W}_u$. But what do I do with $d\hat{W}_u$ ? – Phil-ZXX Jan 28 '17 at 20:44