I tried to calculate this integral use Ito's lemma, $W_{t}$ is the Wiener Process. $$I_{T}=\int_{0}^{T}\sqrt{|W_{t}|}dW_{t}$$

We have $d f\left(W_{t}\right)=f^{\prime}\left(W_{t}\right) d W_{t}+\frac{1}{2} f^{\prime \prime}\left(W_{t}\right) d t$, by letting $f(W_{t})=\frac{2}{3}|W_{t}|^{\frac{3}{2}}$, we have $$d(\frac{2}{3}|W_{t}|^{\frac{3}{2}})=\sqrt{|W_{t}|}dW_{t}+\frac{1}{4}\frac{1}{\sqrt{|W_{t}|}}dt$$ Then we can write $$I_{T}=\frac{2}{3}|W_{T}|^{\frac{3}{2}}-\int_{0}^{T}\frac{1}{4}\frac{1}{\sqrt{|W_{t}|}}dt$$ I don't know whether this is correct and I'm new to stochastic integral. If above is correct, how to calculate $\mathbb{E}[I_{T}]$ and $Var(I_{T})$.

  • $\begingroup$ You’re trying to differentiate the function $f(x)=|x|$ which is tricky and not very rigorous in this context. You do not need Ito’s Lemma in this case, see my answer below. $\endgroup$ – Daneel Olivaw Nov 22 '19 at 21:25

The integral $I_T$ is an Itô stochastic integral therefore its expectation is $0$. This is because $I_T$ is a martingale (see e.g. Theorem 4.3.1 in Shreve), hence: $$\mathbb{E}[I_T]=I_0=0$$ You can also see this by considering the definition of a stochastic integral, which involves the sum of terms of the form $f(W_{t_i})(W_{t_{i+1}}-W_{t_i})$, and using the independence of Brownian increments $W_{t_i}-W_0$ and $W_{t_{i+1}}-W_{t_i}$.

From the above, we get: $$\mathbb{V}[I_T]=\mathbb{E}[I_T^2]$$ Given $I_T$ is an Itô integral and that the process $Z_t\triangleq \sqrt{|W_t|}$ is adapted to the filtration generated by $W_t$, by Itô's Isometry: $$\begin{align} \mathbb{E}[I_T^2]&=\mathbb{E}\left[\int_0^TZ_t^2\text{d}t\right] \\[3pt] &=\int_0^T\mathbb{E}[|W_t|]\text{d}t \end{align}$$ $W_t$ is normally distributed. By symmetry of the Normal distribution, the expectation of $|W_t|$ is equal to twice the expectation of $1_{\{W_t\geq0\}}W_t$, namely: $$\begin{align} \mathbb{E}[1_{\{W_t\geq0\}}W_t]&=\int_0^\infty w\frac{1}{\sqrt{2\pi t}}e^{-\frac{w^2}{2t}}\text{d}w \\[3pt] &=\int_0^\infty v\frac{1}{\sqrt{2\pi}}e^{-\frac{v^2}{2}}\sqrt{t}\text{d}v \\[3pt] &=\sqrt{\frac{t}{2\pi}}\int_0^\infty ve^{-\frac{v^2}{2}}\text{d}v \\[8pt] &=\sqrt{\frac{t}{2\pi}} \end{align}$$ where we've made the change of variables $v=w/\sqrt{t}$. Thus $\mathbb{E}[|W_t|]=\sqrt{2t/\pi}$, from which it comes: $$\begin{align} \mathbb{V}[I_T]&=\sqrt{\frac{2}{\pi}}\int_0^T\sqrt{t}\text{d}t \\[6pt] &=\sqrt{\frac{8}{9\pi}}T^{3/2} \end{align}$$


Shreve, S. (2004). Stochastic Calculus for Finance II, Springer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.