# How to compute the expectation of integral of this random function?

Let $W_t$ be a standard wiener process and

$$Y_t=\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds$$

If $W(t_0)=\sqrt{3}$, then how can we compute $\mathbb{E}[Y(t_0)]$?

Is $\mathbb{E}[Y(t_0)]=0$?

• 1. This is not an Itô integral. 2. If $W(t_0)=\sqrt{3}$ clearly $W_t$ is not a standard Wiener process. So are you sure of your question? – Quantuple Jul 10 '16 at 21:07
• $t_0\ne 0$ yes I sure – math Jul 10 '16 at 21:34
• Okay then, my bad. – Quantuple Jul 10 '16 at 21:38
• Please pay attention to typos and grammar. – SRKX Jul 11 '16 at 2:45

1. This integral is not Ito's Integral. Indeed $Y_t$ is a random time change with time change rate $\frac{W_t}{1+W_t^2}.$ (Oksendal, Sixth edition,page 147)
Let $$f''(x)=\frac{-2x}{(1+x^2)^2}$$ then $$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$ and $$f(x)=\tan^{-1}(x)+c_1x+c_2$$ set $c_1=c_2=0$.By application of Ito's lemma we have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$ therefore $$\tan^{-1}(W_t)=\int_{0}^{t}\frac{1}{1+W_s^2}dW_s-\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds$$ in other words $$\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds=\int_{0}^{t}\frac{1}{1+W_s^2}dW_s-\tan^{-1}(W_t)$$ thus $$\mathbb{E}\left[\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds\right]=\underbrace{\mathbb{E}\left[\int_{0}^{t}\frac{1}{1+W_s^2}dW_s\right]}_{0}-\mathbb{E}[\tan^{-1}(W_t)]$$ as a result $$\mathbb{E}\left[\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds\right]=-\mathbb{E}[\tan^{-1}(W_t)]$$ Now set $t=t_0$ $$\mathbb{E}\left[\int_{0}^{t_0}\frac{W_s}{(1+W_s^2)^2}ds\right]=-\mathbb{E}[\tan^{-1}(W_{t_0})]=-\mathbb{E}[\tan^{-1}(\sqrt{3})]=-\frac{\pi}{3}$$ finally
$$\color{red}{\mathbb{E}[Y_{t_0}]=-\frac{\pi}{3}}$$