# How to calculate the mean and variance of this Ito integral?

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 $$Var(I_{T})$$.

• 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. Nov 22, 2019 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}