# Covariance between Wiener process and stochastic integral

I want to calculate $cov($$w_t , \int_0^{t}s^{n}dw_s) . I have tried integration by parts:$$\int_0^{t}s^{n}dw_s = t^{n}w_{t} - n\int_{0}^{t}s^{n-1}w_sds$$Further, I think to use this formula n times, but I can't do it on this step. My idea was to get the equation at the end that will contain next components (maybe with some coefficients): \int_0^{t}sdw_s, \int_0^tw_sds, w_t and the component that if I count a mathematical expactation from it I will get the cov($$w_t$, $\int_0^{t}s^{n}dw_s)$ (the mathemetical expectation for the first three components I know). And from this equation I would count the covariance, but I can't understand how I can continue. I'll be happy for any idea.

• Use directly that $$E\left(w_t\cdot\int_0^ts^ndw_s\right)=E\left(\int_0^t1dw_s\cdot\int_0^ts^ndw_s\right)=\int_0^t1\cdot s^n\,ds$$ – Did Apr 29 '18 at 13:28
• @Did You are awesome! Thank you! – Sergey Kravchenko Apr 29 '18 at 14:13

## 3 Answers

$$cov(w_t , \int_0^{t}s^{n}dw_s) = E[w_t\int_0^{t}s^{n}dw_s] = E[\int_0^{t} dw_s\int_0^{t}s^{n}dw_s] = E[\int_0^{t}s^{n}ds] = \int_0^{t}s^{n}ds$$

(*) Ito isometry:

$$E[(\int_0^{t}x_sdw_s)^2] = E[\int_0^{t}x_s^2ds]$$

Generalisation:

$$E[\int_0^{t}x_sdw_s\int_0^{t}y_sdw_s] = E[\int_0^{t}x_sy_sds]$$

• Not sure how your answer adds much new on the topic. It's exactly what @Did wrote (somewhat more compactly) in his comment. – saz May 4 '18 at 13:06
• @saz I guess my answer gives intuition that what did wrote is based on a generalisation of ito isometry? – BCLC Nov 17 '20 at 22:24

The most clever way is, as @Did suggested, to use Itô's isometry. However, you can also use the integration by parts formula:

$$\int_0^t s^n \, dW_s = t^n W_t - n \int_0^t s^{n-1} W_s \, ds.$$

Multiplying both sides by $W_t$ and taking the expectation we get

\begin{align*} \mathbb{E} \left( W_t \cdot \int_0^t s^n \, dW_s \right) &= t^n \underbrace{\mathbb{E}(W_t^2)}_{t} - n \int_0^t s^{n-1} \underbrace{\mathbb{E}(W_s W_t)}_{\min\{s,t\}=s}\,d s \\ &= t^{n+1} - n \int_0^t s^n \, ds =\frac{1}{n+1} t^{n+1}.\end{align*}

The solution is $$cov(w_t, \int_0^ts^ndw_s)=E(w_t - Ew_t)(\int_0^ts^ndw_s- E\int_0^ts^ndw_s)=E(w_t\int_0^ts^ndw_s-w_tE\int_0^ts^ndw_s)=\int_0^ts^nds-Ew_tE\int_0^ts^ndw_s=\int_0^ts^nds=\frac{t^{n+1}}{n+1}$$