Suppose that:

  • $W^*_t$ is a Wiener process under probability measure $\mathbb{P}^*$ and;
  • $\tilde{S}_t=S_0+\sigma\int_{0}^{t}S(u)dW^*_s$.

In my lecture notes, it says that $\tilde{S}_t$ is a martingale under $\mathbb{P}^*$ "due to the fact that the stochastic integral from 0 to t with respect to Brownian motion is a martingale".

Why is this quotation (in bold) indeed correct?

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    $\begingroup$ I find it odd that they call this $\bar{S}$ (with the bar) because to me they're just saying that $dS_t = \sigma S_t dW_t$. (I just used a common notation for the rest, I hope you don't mind) $\endgroup$
    – SRKX
    Dec 9, 2014 at 1:12

2 Answers 2


In the integral

$$\int_0^t S_u dW^{*}_u \, ,$$

$dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$.

So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since $\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0 \right] = 0$.

Therefore, $\mathbb{E}\left[ S_t\middle\vert \mathcal{F}_0\right] = \mathbb{E}\left[ S_0\middle\vert \mathcal{F}_0\right] + \mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \mathbb{E}\left[ S_0\middle\vert \mathcal{F}_0\right]$.

To address Gordon's comments that we need to prove that the integral is a martingale, recall that $g_t$ is a martingale if, for $s < t$, $$\mathbb{E}\left[ g_t \middle\vert \mathcal{F}_s \right] = g_s \, .$$ Now let $g_t = \int_0^t S_u dW^{*}_u$. $$\mathbb{E}\left[ g_t \middle\vert \mathcal{F}_s \right] = \mathbb{E}\left[ \int_0^s S_u dW^{*}_u\middle\vert \mathcal{F}_s\right] + \mathbb{E}\left[ \int_s^t S_u dW^{*}_u\middle\vert \mathcal{F}_s\right] = \int_0^s S_u dW^{*}_u + 0 = g_s \, . $$

  • 1
    $\begingroup$ fushsialatitude: good question. To move the expectation operator inside the integration, we need to satisfy Fubini's theorem. If $S(u)$ is "well-behavior", the theorem should be satisfied. $\endgroup$
    – wsw
    Dec 8, 2014 at 18:14
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    $\begingroup$ Wow my answer is at -1 vote. Folks please go take a Stochastic Calculus class, as this result is fairly basic. Look at slide #12 at idsc.ethz.ch/Courses/stochasticsystems/SDE.pdf; "The expectation of stochastic integrals is zero. This is what we would expect anyway." $\endgroup$
    – wsw
    Dec 8, 2014 at 18:48
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    $\begingroup$ A zero expectation does not imply the martingality. $\endgroup$
    – Gordon
    Dec 9, 2014 at 15:55
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    $\begingroup$ emcor: because of the independent increments property of $W_t$. At time $u$, $S_u$ is known but $dW^*_u \equiv W^*_{u+du}-W^*_u$ is not. $\endgroup$
    – wsw
    Dec 9, 2014 at 18:25
  • 1
    $\begingroup$ @Marco I added more to my answer to explain why that is. Hope it is clear now. $\endgroup$
    – wsw
    Jan 2, 2020 at 15:57

I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak.

The reason I am putting this answer on is twofold: first, to compliment @ William S. Wong's answer by adding greater mathematical intricacy for other users of the website, and secondly to confirm that I understand the solution. To that end, I welcome any improvements / corrections.

The reason I asked this question, is actually because I wanted to know if there was a more concise, yet still rigorous, way of approaching this problem.


Random step process

We shall call $f(t), t\geq0$ a random step process if there is a finite sequence of numbers $0=t_0<t_1<\cdots<t_n$ and square integrable random variables $\eta_0,\eta_1,\cdots,\eta_{n-1}$ such that $$f(t)=\sum_{j=0}^{n-1}\eta_j1_{[t_j,t_{j+1})}(t)$$ where $\eta_j$ is $\mathcal{F}_t$-measurable for $j=0,1,\cdots,n-1$. The set of random step processes will be denoted $M^2_{\mbox{step}}$.

The Stochastic Integral (for step processes)

The stochastic integral of a random step process $f \in M^2_{\mbox{step}}$ is defined by $$I(f)=\sum_{j=0}^{n-1}\eta_j(W(t_{j+1})-W(t_j)).$$

The stochastic integral $I(f)$ has now been defined for $M^2_{\mbox{Step}}$. We now extend this definition to a larger class of processes by approximation.

The "approximative" processes $M^2$

We denote by $M^2$ the class of stochastic processes $f(t), t\geq 0$ such that $$E\left(\int_0^\infty|f(t)|^2dt\right) < \infty$$ and there is a sequence $f_1,f_2,\cdots,\in M^2_{\mbox{step}}$ of random step processes such that $$\lim_{n\to\infty} E\left(\int_0^\infty|f(t)-f_n(t)|^2dt\right)=0.$$

In this case, we shall say that the sequence of random step processes $f_1,f_2,\cdots$ approximates $f$ in $M^2$.

The Ito Stochastic Integral (from $0$ to $\infty$)

We call $I(f) \in L^2$ the Ito Stochastic Integral (from $0$ to $\infty$) of $f\in M^2$ if $$\lim_{n\to\infty}E\left(|I(f)-I(f_n)|^2\right)=0$$ for any sequence $f_1,f_2,\cdots \in M^2_{\mbox{step}}$ that approximates $f$ in $M^2$. We shall also write $$\int_0^\infty f(t) dW(t)$$ in place of $I(f)$.

The Ito Stochastic Integral (from $0$ to $T$)

For any $T>0$ we shall denote by $M^2_{T}$ the space of all stochastic processes $f(t)$,$t \geq 0$ such that $1_{[0,T)}f \in M^2$. The Ito Stochastic Integral from $0$ to $T$ of $f \in M^2_T$ is defined by $$I_T(f) \equiv \int_0^T f(t)dW(t) = I(1_{[0,T)}f).$$

Definition of a martingale

A stochastic process $\xi(t)$ parameterized by $t \in T$ is called a Martingale with respect to filtration $\mathcal{F}_t$ if:

  • $\xi(t)$ is integrable for each $t \in T$
  • $\xi(t)$ is $\mathcal{F}_t$-measurable for each $t \in T$
  • $\xi(s) = E(\xi(t)|\mathcal{F}_s)$ for every $s, t \in T$ such that $s \leq t$.

The third bullet point is called the Martingale condition.

Lemma 1

For each random step process $f\in M_{step}^{2}$, the stochastic integral $\int_{0}^{t}f(s)dW(s)$ is a martingale.

Proof of Lemma 1

Let $0\leq s < t$ and suppose that $f \in M_{step}^{2}$ can be written of the form of our definition, whereby $$0=t_0 < t_1 < \cdots < t_k =s < t_{k+1} < \cdots < t_m = t < t_{m+1} < \cdots < t_n.$$ We shall denote the increment $W(t_{j+1}) - W(t_j)$ by $\Delta_j W$. Then $$1_{[0,t]}f = \sum_{j=0}^{m-1} \eta_j 1_{[t_j,t_{j+1}]}$$ and $$I_t(f) = I(1_{[0,t]}f) = \sum_{j=0}^{m-1} \eta_j \Delta_j W,$$ which is adapted to $\mathcal{F}_t$ and square integrable, and so integrable. It remains to compute

$$E(I_t(f) | \mathcal{F}_s) = E(\sum_{j=0}^{m-1} \eta_j \Delta_j W | \mathcal{F}_s) $$

If $j<k$, then $\eta_j$ and $\Delta_j W$ are $\mathcal{F}_s$-measurable and $$E(\eta_j \Delta_j W | \mathcal{F}_s) = \eta_j \Delta_j W. $$

This is getting to the heart of the question now. Note that: If $j\geq k$ then $\mathcal{F}_s \subset \mathcal{F}_{t_j}$ and

\begin{eqnarray*} E(\eta_{j}\Delta_{j}W|\mathcal{F}_{s}) & = & E(E(\eta_{j}\Delta_{j}W|\mathcal{F}_{t_{j}})|\mathcal{F}_{s})\mbox{ by the tower property}\\ & = & E(\eta_{j}E(\Delta_{j}W|\mathcal{F}_{t_{j}})|\mathcal{F}_{s})\mbox{ by taking out what's known}\\ & = & E(\eta_{j}|\mathcal{F}_{s})E(\Delta_{j}W)\mbox{ by independence}\\ & = & E(\eta_{j}|\mathcal{F}_{s})\times0\mbox{ by definition of Wiener process.} \end{eqnarray*} It follows that \begin{eqnarray*} E(I_{t}(f)|\mathcal{F}_{s}) & = & \sum_{j=0}^{k-1}\eta_{j}\Delta_{j}W\\ & = & I(1_{[0,s]}f)\\ & = & I_{s}(f). \end{eqnarray*}

Proof of the Martingale Property

Finally we show that for any $f\in M_{t}^{2}$ and for any $0\leq s<t$ that $$ E\left(\int_{0}^{t}f(r)dW(r)|\mathcal{F}_{s}\right)=\int_{0}^{s}f(r)dW(r). $$ We approach this by remembering that a process can be approximated by a sequence of step processes. That is, if $f$ belongs to $M_{t}^{2}$ then $1_{[0,t)}f$ belongs to $M^{2}$. Let $f_{1},f_{2},\cdots$ be a sequence of processes in $M_{\mbox{step}}^{2}$ approximating $1_{[0,t)}f$. By lemma 1, we know that $$ E\left(I(1_{[0,t)}f_{n})|\mathcal{F}_{s}\right)=I\left(1_{[0,s)}f_{n}\right) $$ for each $n$. By taking the $L^{2}$ limit of both sides of this equality as $n\to\infty$ we shall show that $$ E\left(I(1_{[0,t)}f)|\mathcal{F}_{s}\right)=I\left(1_{[0,s)}f\right) $$ which is what we need to prove.

Right hand side: observe that $1_{[0,s)}f_{1},1_{[0,s)}f_{2},\cdots$ is a sequence in $M_{\mbox{step}}^{2}$ approximating $1_{[0,s)}f$ so $$ I\left(1_{[0,s)}f_{n}\right)\to I\left(1_{[0,s)}f\right)\mbox{ in }L^{2}\mbox{ as }n\to\infty. $$ Left hand side: $1_{[0,t)}f_{1},1_{[0,t)}f_{2},\cdots$ is also a sequence in $M_{\mbox{step}}^{2}$ approximating $I(1_{[0,t)}f)$, which implies that $$ I\left(1_{[0,t)}f_{n}\right)\to I\left(1_{[0,t)}f\right)\mbox{ in }L^{2}\mbox{ as }n\to\infty. $$ It is now possible to show that, $$ E\left(I\left(1_{[0,t)}f_{n}\right)|\mathcal{F}_{s}\right)\to E\left(I\left(1_{[0,t)}f\right)|\mathcal{F}_{s}\right)\mbox{ in }L^{2}\mbox{ as }n\to\infty $$ which completes the proof. Why does this last equation hold? It's subtle... not obvious... and the answer is given Lemma 2.

Lemma 2

If $\xi$ and $\xi_{1},\xi_{2},\cdots,$ are square integrable random variables such that $\xi_{n}\to\xi$ in $L^{2}$ as $n\to\infty$ then $$ E\left(\xi_{n}|\mathcal{G}\right)\to E\left(\xi|\mathcal{G}\right)\mbox{ in }L^{2}\mbox{ as }n\to\infty $$ for any $\sigma-$field $\mathcal{G}$ on $\Omega$ contained in $\mathcal{F.}$


By Jensens inequality, \begin{eqnarray*} \left|E\left(\xi_{n}|\mathcal{G}\right)-E\left(\xi|\mathcal{G}\right)\right|^{2} & = & \left|E\left(\xi_{n}-\xi|\mathcal{G}\right)\right|^{2}\\ & \leq & E\left(\left|\xi_{n}-\xi\right|^{2}|\mathcal{G}\right), \end{eqnarray*} which implies that \begin{eqnarray*} E\left(\left|E\left(\xi_{n}|\mathcal{G}\right)-E\left(\xi|\mathcal{G}\right)\right|^{2}\right) & \leq & E\left(E\left(\left|\xi_{n}-\xi\right|^{2}|\mathcal{G}\right)\right)\\ & = & E\left(\left|\xi_{n}-\xi\right|^{2}\right)\to0 \end{eqnarray*} as $n\to\infty.$

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    $\begingroup$ While the OP answered his/her own question, I wonder if he/she just copied the answer from a book. In the OP's answer, it says that it "is possible to show" the lemma that $\int_0^t f(s) dW(s)$ is a martingale, but it "is omitted here to keep things short". What? That's exactly what we are trying to prove! $\endgroup$
    – wsw
    Dec 9, 2014 at 2:55
  • $\begingroup$ That was why I gave a -1 because this didn't answer anything. $\endgroup$
    – SmallChess
    Dec 9, 2014 at 10:16
  • $\begingroup$ @StudentT I have edited my answer. Hopefully it's clearer now. $\endgroup$
    – Kian
    Dec 9, 2014 at 16:22

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