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How to show that the process $X_t=tW_t - \int_0^t W_s ds $ is a martingale?

I guess I have to use the definition of martingale and properties of Wiener process, but I stack with this integral. Please, help me. What I've done:

$E(X_t|F_r)=E(tW_t- \int_0^t W_s ds|F_r)=tE(W_t-W_r|F_r)+tE(W_r|F_r)-E(\int_0^t W_s ds|F_r)=tE(W_t-W_r)+tW_r-E(\int_0^t W_s ds|F_r)=tW_r-E(\int_0^t W_s ds|F_r)$

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We want to show that $X_t = tW_t - \int_0^t W_s \mathrm{d}s$ is a martingale. Let $r\leq t$ then $$ \mathbb{E}[X_t\ | \ \mathcal{F}_r]= \mathbb{E}[tW_t\ |\ \mathcal{F}_r] - \mathbb{E}\left[\int_0^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right]. $$

As you noticed $$\mathbb{E}[tW_t\ |\ \mathcal{F}_r] = \mathbb{E}[t(W_t-W_r) + tW_r\ |\ \mathcal{F}_r] = t \underbrace{\mathbb{E}[W_t - W_r \ | \mathcal{F}_r]}_{=\mathbb{E}[W_t-W_r] = 0} + t \underbrace{\mathbb{E}[W_r\ |\ \mathcal{F}_r]}_{=W_r} = tW_r,$$ now we take the second part, that is, $$\begin{align*}\mathbb{E}\left[\int_0^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right] =& \mathbb{E}\left[\int_0^r W_s \mathrm{d}s + \int_r^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right] \\=& \mathbb{E}\left[\int_0^r W_s \mathrm{d}s \ | \ \mathcal{F}_r\right]+ \mathbb{E}\left[\int_r^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right] \\ =& \int_0^rW_s \mathrm{d} s + \mathbb{E}\left[\int_r^t W_r + (W_s-W_r) \mathrm{d}s \ | \ \mathcal{F}_r\right] \\ =& \int_0^r W_s \mathrm{d}s + \mathbb{E}\left[\int_r^t W_r \mathrm{d}s \ | \ \mathcal{F}_r \right] + \mathbb{E}\left[\int_r^t (W_s-W_r) \mathrm{d}s \ | \ \mathcal{F}_r \right] \\ =& \int_0^r W_s \mathrm{d}s + \mathbb{E}\left[(t-r)W_r \ | \ \mathcal{F}_r \right] + \mathbb{E}\left[\int_r^t (W_s-W_r) \mathrm{d}s \ | \ \mathcal{F}_r \right] \\ =& \int_0^r W_s \mathrm{d}s + (t-r)W_r + \int_r^t \underbrace{\mathbb{E}[W_s - W_r | \mathcal{F}_r]}_{=0} \mathrm{d}s = \\ =& (t-r)W_r + \int_0^r W_s \mathrm{d}s .\end{align*}$$ I have used the fact that $ \mathbb{E}\left[\int_r^t (W_s-W_r) \mathrm{d}s \ | \ \mathcal{F}_r\right]=\int_r^t \mathbb{E}[W_s-W_r | \mathcal{F}_r] \mathrm{d}s$ via Fubini's theorem for more detail on this you may want to see this link.

Hence, $$\mathbb{E}[X_t\ | \ \mathcal{F}_r] = tW_r - (t-r)W_r - \int_0^r W_s \mathrm{d}s =rW_r - \int_0^r W_s \mathrm{d}s = X_r. $$

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  • $\begingroup$ thanks !!!!! You're the best $\endgroup$ – wiwnes691 May 11 '16 at 17:47
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    $\begingroup$ No problem. For your second question I think that according to the rules of this portal, you will have to either create a new topic with this question or edit your main post and input it there. Admins here don't like if you post a question as an answer to the topic. You can also vote up or down the answers and accept the one which you prefer. I can write you the solution for your second question, but I will wait for your edits first. Cheers. $\endgroup$ – m_gnacik May 11 '16 at 18:24
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Using Integration by Part formula we have : $$t.W_t -\int_0^t W_s ds = \int_0^t s.dW_s+ 0$$ (the $0$-term comes from quadratic covariation between $t$ and $W_t$ which is null)

So $X_t=\int_0^t s.dW_s$, to show that this is indeed a martingale is straightforward using the facts that : $W_t$ is a square integrable martingale and $s$ is a bounded predictable process.

You get the martingale property for $X$ from lemma 3 here.

Best regards

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