What would be integration of Brownian Motion with $t$ and is there a general formula for $f(t)$ , i.e.
What is $E[\int_{0}^{t} t W_t dt]$? Is it $0$?
Any general rule or result for $E[\int_{0}^{t} f(t) W_t dt]$?
Asked
Active
Viewed 70 times
1
uday
- 300
-
for example, if $f$ is continuous, denote by $F$ its anti-derivative. Then apply Ito formula to $F(t)W_t$, we can get $$F(t)W_t - F(0)W_0 - \int_0^t F(s) dW_s = \int_0^t f(s) W_s ds $$ is this enough for your application ? p.s. the integral in your first point is not zero. – Chival Jul 11 '15 at 20:47
-
You are right! I forgot to add the expectation – uday Jul 13 '15 at 17:37
-
Ok, then you can see from the above formula, the expectation that you want is equal to $$\mathbb{E}\left[ ,, -\int_0^t F_s , dW_s ,, \right] . $$To answer your first question: yes. To answer your second question: if $f$ is continuous, then $F$ is differentiable (continuous), therefore $$t\geq 0 \longmapsto \int_0^t F_s , dW_s $$ defines a square-integrable martingale ==> Zero expectation. Of course you can weaken the assumption on $f$ a little bit. – Chival Jul 13 '15 at 20:09