$$ I = \int_0^1W_tdt, $$

where $W_t$ is a Brownian motion. From Integral of Brownian motion w.r.t. time we have that

$$ \mathbb{E}[I]=0, $$ by Fubini's theorem. And that $$ \mathbb{V}\text{ar}[I] = \mathbb{E}[I^2] = \mathbb{E}\left[\int_0^1\int_0^1W_sW_tdsdt\right] = \int_0^1\int_0^1\mathbb{E}[W_sW_t]dsdt = \frac{1}{3}. $$ Here I do not understand how we are able to bring the expectation into the integral, Ito isometry says that $$ \mathbb{E}\left[\left(\int_0^tX_sdW_s\right)^2\right] = \int_0^t\mathbb{E}[X_s^2]ds $$ so I am unsure how it is possible to use this here?

Furthermore, I am required to find $$ \mathbb{C}\text{ov}[I, W_1], $$

I know that $$ \mathbb{C}\text{ov}[I, W_1] = \mathbb{E}[IW_1], $$ since $\mathbb{E}[I] = \mathbb{E}[W_1] = 0$, my best guess here is then that $$ \mathbb{E}[IW_1] = \min\left(\frac{1}{3},1\right) = \frac{1}{3}, $$ however, this is probably wrong.

Thank you in advance!

  • 1
    $\begingroup$ Hint: you can write $ I =\int_0^1 W_t dt = \int_0^1(1-t)dW_t$. Can you see why this is true and how it helps? For the second part you need the slighlty more general version of Ito isometry. I can make a complete answer but I'll give you some time to think about it if you like. $\endgroup$
    – R. Rayl
    May 11, 2021 at 14:35

2 Answers 2


For the first question, equality $$\mathbb{E}\left[\int_{[0,1]\times[0,1]} W_sW_tdsdt\right] = \int_{[0,1]\times[0,1]}\mathbb{E}[W_sW_t]dsdt \left(= \int_{[0,1]\times[0,1]}\min(s,t)dsdt\right) $$

is due to commuting expectation and integral (not to Ito isometry), which in turn is allowed by Fubini's theorem condition being met:

$$ \left(\int_{[0,1]\times[0,1]}\mathbb{E}\left[|W_sW_t| \right] ds dt \right)^2\leq $$ $$\int_{[0,1]\times[0,1]}\mathbb{E}\left[|W_sW_t| \right]^2 ds dt \leq \int_{[0,1]\times[0,1]} \mathbb{E}\left[W_s^2 \right] \mathbb{E}\left[W_t^2 \right] dsdt $$ $$= \int_{[0,1]\times[0,1]}st dsdt = \frac{1}{4} <\infty $$

(with first and second inequalities being applications of Jensen and Cauchy-Schwarz inequalities, respectively).

For the second question, one way is to use the definition of the time integral (Riemann pathwise) and use covariance properties:

$$ {\rm cov} \left(\int_0^1 W_tdt, W_1 \right) = {\rm cov} \left(\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} W_{\frac{k}{n}}, W_1\right) $$

$$= \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} {\rm cov} \left( W_{\frac{k}{n}}, W_1\right) = \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} \frac{k}{n} $$ $$= \lim_{n\rightarrow \infty} \frac{(n-1)n}{2n^2} = \frac{1}{2}$$


There are two questions that you are asking:

  1. How to prove $\text{Var}[I] = \frac{1}{3}$? and
  2. What is $\text{Cov}[I, W_1]$?

For Question 1, you absolutely can use Ito isometry. First, note that we can use integration by parts to obtain the formula: \begin{align} \int_0^1 W_t dt &= W_1 - \int_0^1tdW_t \\ &= \int_0^1(1-t)dW_t \end{align} So we can write $I=\int_0^1(1-t)dW_t$ (switching from Riemann integral to an Ito integral like this is often a useful trick). Now, \begin{align} \text{Var}[I] = \text{E}[I^2] &= \text{E}\bigg[\bigg(\int_0^1(1-t)dW_t\bigg)^2\bigg] \\ &= \int_0^1(1-t)^2dt \\ &= \frac{1}{3} \end{align} where we use Ito isometry (exactly as you stated) to go from the first to the second line.

For Question 2, we use Ito isometry again. But this time the slightly more general form (see the very last equation). So, \begin{align} \text{Cov}[W_1, I] = \text{E}[W_1, I] &= \text{E}\bigg[W_1 \int_0^1tdW_t\bigg] \\ &= \text{E}\bigg[\int_0^1dW_t \int_0^1tdW_t\bigg] \\ &= \int_0^1tdt \\ &= \frac{1}{2} \end{align} where we use the general Ito isometry to go from line 2 to line 3.


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