# How to show that $E\left[ \int_0^t \sigma(s) e^{iuX(s)} dW(s)\right] = 0$?

Let $\sigma(t)$ be a given deterministic function of time and define the process $X_t$ by $$X(t) = \int_0^t \sigma(s)dW(s)$$ I want to show $$E\left[ \int_0^t \sigma(s) e^{iuX(s)} dW(s)\right] = 0$$

It is part of exercise 4.3 in Bjork "Arbitrage Theory in continuous time", and you can see the solution here: The solution does not explain it.

• Where did you get that solution from? You should cite your source.
– SRKX
Dec 28 '16 at 16:10
• If you were not looking for a proof, then you can use Proposotion 4.4 in Bjork, Arbitrage Theory in Continous time. Nov 30 '17 at 19:18

Some basic details

$\quad$ The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $\sigma_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Itô integral of $\sigma_t$ with respect to $W_t$ up to time t is a random variable $$X_t=\int_{0}^{t}\sigma(s)dW_s=\lim_{n\to\infty}\,\sum_{i=1}^{n}\sigma(t_{i-1})(W(t_i)-W(t_{i-1}))$$ since $\sigma(s)$ is deterministic process and $W({t_i})-W(t_{i-1})\sim\mathcal{N}(0,t_i-t_{i-1})$ , thus $X_t$ is normally distributed such that \begin{align*} \mathbb{E}[X_t]&=\mathbb{E}\left[\lim_{n\to\infty}\sum_{i=1}^{n}\sigma(t_{i-1})(W_{t_i}-W_{t_{i-1}})\right]\\ &=\lim_{n\to\infty}\mathbb{E}\left[\sum_{i=1}^{n}\sigma(t_{i-1})(W_{t_i}-W_{t_{i-1}})\right]\\ &=\lim_{n\to\infty}\sum_{i=1}^{n}\mathbb{E}\left[\sigma(t_{i-1})(W_{t_i}-W_{t_{i-1}})\right]\\ &=\lim_{n\to\infty}\sum_{i=1}^{n}\sigma(t_{i-1})\mathbb{E}\left[W_{t_i}-W_{t_{i-1}}\right]\\ &=0 . \end{align*} By application of Ito's isometry formula, we have $$\text{Var}(X_t)=\text{Var}\left(\int_{0}^{t}\sigma(s)dW_s\right)=\mathbb{E}\left[\left(\int_{0}^{t}\sigma(s)dW_s\right)^2\right]=\mathbb{E}\left[\int_{0}^{t}\sigma^2_sds\right]=\int_{0}^{t}\sigma^2_sds$$

Proof

If $Y(t)$ be a regular adapted process such that $\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds < \infty$ then $$\mathbb{E}\left[\int_{0}^{t}Y(s)dW_s\right]=0$$ Set $Y(s)=\sigma(s) e^{iuX(s)}$, we have $$\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds=\int_{0}^{t}\mathbb{E}\left[ \sigma^2(s) e^{2iuX(s)}\right]ds=\int_{0}^{t}\sigma^2(s)\mathbb{E} \left[ e^{2iuX(s)}\right]ds\tag 1$$ $X(t)$ is normally distributed with zero mean and a variance given by $$\text{Var}(X_s)=\int_{0}^{s}\sigma^2(v)dv$$ thus $$\mathbb{E} \left[ e^{2iuX(s)}\right]=\exp\left(2\text{i}\,u\,\mathbb{E}[X_s]-2u^2\text{Var}(X_s)\right)\tag 2$$

As a result $$\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds=\int_{0}^{t}\sigma^2(s)\exp\left(-2u^2\int_{0}^{s}\sigma^2(v)dv\right)ds<\infty$$

• Let $\sigma(t)$ has good properties ;)
– user16651
Dec 26 '16 at 16:03
• What a terrible answer! In order to show that $\int_0^t E[Y^2(S)]ds < \infty$, you use that $X$ is normally distributed, with mean $0$ and variance $\int_0^s \sigma^2(v)dv$. Yet, you give NO argument for these claims. Dec 26 '16 at 16:15
• @Jin5: This is a perfectly valid answer that uses the basic property that an Ito integral of a deterministic integrand is normally distributed. A better comment would have been to ask for clarification if this isn't clear to you. See e.g. Theorem 4.4.9 in Shreve's "Stochastic Calculus for Finance II". Dec 26 '16 at 17:49
• Please keep it civil. We have a 'be nice'-policy and that policy is enforced. Dec 26 '16 at 22:42
• @Jin5: Take it easy. The question asks to show why the expectation of the given Ito integral is zero. Behrouz Maleki states the condition for which an Ito integral is a martingale and then shows that this is indeed satisfied here. From the way I understand Maaniya's question, this link between square-integrability and the martingale property of the Ito integral is exactly the step that was missing in the provided sample solution. Dec 26 '16 at 23:34