I have to find the mean and variance of "white noise" - Brownian motion of $$ X = \int^1_0 t dW(t)$$

So we start the mean: $$ E[X] = E[\int^1_0 tdW(t)] $$

I'm thinking this way: $$ E[X] = tW(1) - tW(0) - \int^1_0 1W(t)dt $$

Then I'm not sure where to go next as the mean of $W(t)$ is $0.$

  • $\begingroup$ Hint: Think of the integral as a sum and find the expectation. Assume the limit can be passed through the expectation. For variance, use the same trick. $\endgroup$ – EDZ Feb 21 '18 at 4:12
  • $\begingroup$ Also Integration by parts doesn't work the same way for Ito integrals. $\endgroup$ – EDZ Feb 21 '18 at 4:13
  • $\begingroup$ I'm having a hard time setting it up with the sum part. Can you show me how you would set it up? Thank you. $\endgroup$ – MelHA Feb 21 '18 at 4:25
  • $\begingroup$ Sure and is this for a measure theory class? If not, you can be fairly loose with passing the expectation through the limit. $\endgroup$ – EDZ Feb 21 '18 at 4:29
  • $\begingroup$ No, it's for a modeling class! We're learning about white noise and Brownian motion, and we were just introduced to integrating things in that regard. We did learn that $E[\int^b_a f(t) dW(t)] = 0$ but what's throwing me off is the $t$ here in my question. $\endgroup$ – MelHA Feb 21 '18 at 4:31


$$E[\int_0^1 t dW(t)] \sim E[\sum_{i=0}^{N - 1} \frac{i}{N} [W(\frac{i+1}{N}) - W(\frac{i}{N})]$$

$$E[\sum_{i=0}^{N - 1} \frac{i}{N} [W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i}{N}E[W(\frac{i+1}{N}) - W(\frac{i}{N})]$$

$$\sum_{i=0}^{N - 1} \frac{i}{N}E[W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i}{N}\cdot0 = 0$$

Assuming we can pass the expectation through the limit, we get:

$E[\int_0^1 t dW(t)] = 0$

For Variance,

$$Var(\int_0^1 t dW(t)) \sim \sum_{i=0}^{N - 1} \frac{i^2}{N^2}Var[W(\frac{i+1}{N}) - W(\frac{i}{N})]$$

By independence.

$$\sum_{i=0}^{N - 1} \frac{i^2}{N^2}Var[W(\frac{i+1}{N}) - W(\frac{i}{N})] = \sum_{i=0}^{N - 1} \frac{i^2}{N^2}\frac{1}{N}$$

The above looks a lot like the expression:

$$\int_0^1 t^2 dt$$

Which is equivalent in the limit.

Therefore: $$Var(\int_0^1 t dW(t)) = \int_0^1 t^2 dt = \frac{1}{3}$$

This result is based Ito Isometry FYI, which is the fundamental theorem of stochastic calculus.

$$E[(\int_0^T f(t)dW(t))^2] = \int_0^T f(t)^2dt$$

  • $\begingroup$ Wow! Thank you so so much! It makes sense! $\endgroup$ – MelHA Feb 21 '18 at 4:48

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