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Claim: $V_t = W^2 _t - t$ is a $\mathbb{P}$-martingale.

I have shown via Ito's formula, that $dV_t = 2 W_t \, dW_t$.

For reference, I will list this "Proposition": If $X$ is a stochastic process with volatility $\sigma _t$ (that is, $dX_t = \sigma _t \, dW_t + \mu _t \, dt$) which satisfies the technical condition $\mathbb{E}[( \int_0^T \sigma^2 _s \, ds)^{\frac{1}{2}}] < \infty$, then: $X$ is a martingale $\iff$ $X$ is driftless ($\mu _t \equiv 0$).

Now, my book states the following: $V_t$ is a proper martingale; if we let $X$ be $(\int_{0}^{T} W^2 _t \, dt)^{\frac{1}{2}}$, then it is enough to show (by the above proposition) that $\mathbb{E}[X] < \infty$. In fact,

$$(\mathbb{E}[X])^2 \leq \mathbb{E}[X^2] = \int_0^T \mathbb{E}(W^2 _t) \, dt = \frac{1}{2} T^2.$$

My attempt: Now, obviously $V$ is driftless since $dV_t$ has no drift term. But according to our "Proposition" shouldn't we be checking that $\mathbb{E}[( \int_0^T 4 W^2 _t \, dt)^{\frac{1}{2}}] < \infty$ ???

Furthermore, is $(\mathbb{E}[X])^2$ $\leq$ $\mathbb{E}[X^2]$ an intrinsic property of the expectation? I don't see where they get that from...

I do see that $\mathbb{E}[X^2] = \mathbb{E}[((\int_0^T W^2 _t \, dt)^{\frac{1}{2}})^2] = \mathbb{E}[(\int_{0}^{T} W^2 _t \, dt)] = \int_0^T \mathbb{E}[W^2 _t] \, dt = \frac{1}{2} T^2$. This is since $Var(W_t) = \mathbb{E}[W^2 _t] - (\mathbb{E}[W_t])^2 = \mathbb{E}[W^2 _t] + 0 = t$ $\implies \mathbb{E}[W^2 _t] = t.$

So: If $\mathbb{E}[X]^2 \leq \mathbb{E}[X^2]$, then we have that $\mathbb{E}[X] \leq \pm \frac{1}{\sqrt{2}} T$. So the expectation is bounded. Then by our proposition, we have that $V_t$ is a $\mathbb{P}$-martingale.

I am just confused why we want to check $\mathbb{E}[(\int_{0}^{T} W^2 _t \, dt)^\frac{1}{2}] < \infty$? Is it because our $\mathcal{F}$-previsible process is $(2W_t)$? So we just ignore the constant multiple, square $W_t$, and integrate over $W^2 _t$???

Thanks. Any and all help is appreciated. I am so confused about this technical condition.

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    $\begingroup$ I would try something like: $W_t^2 - t = (W_t - W_s )^2+ W_s^2 + 2(W_t - W_s) W_s - t$. Now what happens if you condition on $\mathcal{F_s}$? $\endgroup$
    – Kore-N
    May 26, 2016 at 20:05
  • $\begingroup$ If my math is correct then conditioning on $\mathcal{F}$ gives $\mathbb{E}[V_t | \mathcal{F_s}] = W_s^2 - s$. Which proves $V_t$ is a $\mathbb{P}$-martingale. Thanks so much! $\endgroup$
    – Javier
    May 27, 2016 at 0:11

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