# What's the variance of this Ito integral?

I am reading stochastic calculus and I have understood that the process $$X=\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t$$ has normal distribution with mean zero. How can I find the variance of $X$?

• Your question is not about Quantitative Finance. – user16651 Aug 9 '16 at 16:16
• Why?. My question belongs stochastic calculus and I think that it is so simple. – Crisis2008 Aug 9 '16 at 16:26
• Why do you think that your question is so simple? Do you know the answer!? – user16651 Aug 9 '16 at 16:29
• Check Proposition 1 of stat.uchicago.edu/~lalley/Courses/390/Lecture6.pdf – Tingiskhan Aug 9 '16 at 16:45
• "I am reading stochastic calculus"? Could you please give a source. – vonjd Aug 9 '16 at 16:51

$$\mathbb{E^P}\left[\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right]=0$$ thus $$\sigma^2=\mathbb{Var^P}\left(\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right)=\mathbb{E^P}\left[\left(\int_{0}^{1}\sqrt{\frac{\tan^{-1}t}{t}}dW_t\right)^2\right]$$ By application of Ito's isometry, we have $$\sigma^2=\mathbb{E^P}\left[\int_{0}^{1}\left(\sqrt{\frac{\tan^{-1}t}{t}}\right)^2dt\right]=\mathbb{E^P}\left[\int_{0}^{1}\frac{\tan^{-1}t}{t}dt\right]=\int_{0}^{1}\frac{\tan^{-1}t}{t}dt\tag 1$$ we know (See Maclaurin Series of $\tan^{-1}x$ in wolfram) $$\tan^{-1}t=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-1}$$ hence $$\frac{\tan^{-1}t}{t}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}$$ and $$I=\int_{0}^{1}\frac{\tan^{-1}t}{t}dt=\int_{0}^{1}\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}\right)dt=\sum_{n=1}^{\infty}\int_{0}^{1}\frac{(-1)^{n+1}}{2n-1}t^{2n-2}dt$$ therefore $$I=\sum_{n=1}^{\infty}\left[\frac{(-1)^{n+1}}{(2n-1)^2}t^{2n-1}\Big{|}_{0}^{1}\right]=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^2}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^2}=\color{red}{G}\tag 2$$ where $G$ is Catalan's constant.
$(1)$ and $(2)$ $$\sigma^2=G\simeq 0.916$$
• @Kiwiakos Yes $\operatorname{Ti}_2(1)=G$ – user16651 Aug 10 '16 at 12:43