I am trying to follow some logic in a proof. This is not for a class, I am looking at a proof in a paper. I watered down the part I am stuck on.
I have a non-random even function $k:[0,1] \times [0,1] \rightarrow [0, c]$. I am unfamiliar with Brownian motions and stochastic calculus but I am trying to learn. My understanding is that if I integrate with respect to a Brownian motion without a drift, then I am integrating with respect to a martingale and I can use standard stochastic integral properties. I have primarily been looking at Introduction to Stochastic Integration by Chung and Williams, among a few other books and sources.
Since my function $k(r, s)$ is non-random and bounded, then it is a function with bounded integration and I can use a simple Riemann-Stieltjes integral to obtain my desired integral. That is,
$$\int_0^1 k(r,s) dB(r) = \sum_{i=1}^{n} k(r_i,s)[B(r_i) - B(r_{i-1})].$$
My end goal is to understand the following results I saw in the paper:
$E \left (\int^1_0 \int^1_0 k(r,s) dB(r)dB(s)\right ) = \int^1_0 k(r,r) dr $
$E \left (\int^1_0 \int^1_0 k(r^{(1)}, s^{(1)}) dB(r^{(1)}) dB(s^{(1)}) \int^1_0 \int^1_0 k(r^{(2)}, s^{(2)}) dB(r^{(2)}) dB(s^{(2)} ) \right ) = \left (\int^1_0 k(r,r)dr \right)^2 + 2 \int^1_0 \int^1_0k^2(r,s) dr ds$
Now, when I look at the Reimann-Stieltjes above, I am definately not getting these results. My guess is I am missing a theorem somewhere or I am just not understanding the set up. Some direction would be very helpful. Thanks in advance!
UPDATE:
Wouldn't $E(a[B(r_i) - B(r_{i-1})])$ always be $0$ because $[B(r_i) - B(r_{i-1})]\sim Normal(0, r_i - r_{i-1})?$
I am not fully confident that I have solved $(1)$ but this outline might help you. With that said I want to remark that Brownian motion is not of bounded variation so you cannot use Riemann-Stieltjes. The definition of Ito integral is however quite similar: $$\int_{0}^{1}k(r,s) \ dB(r) = \lim_{n\to \infty}\sum_{i=1}^{n}k(r_{i-1},s)[B(r_{i})-B(r_{i-1})].$$
Anyways, I have used this definition below but have excluded the limits, I think the calculations are correct with suitable placements of limits.
You obtain number $1$ by appealing to two facts, first the variance of Brownian motion and secondly that Brownian motion has independent increments. That is if $r_{i}\ne r_{j}$ then $$E([B(r_{i})-B(r_{i-1})][B(r_{j})-B(r_{j-1})])=0$$ however $$E[(B(r_{i})-B(r_{i-1}))^{2}]=r_{i}-r_{i-1}.$$
Therefore we can do the following in $(1)$ by choosing the same partition:
\begin{align*}(1) &= E\left(\sum_{i=1}^{n}\sum_{j=1}^{n}k(r_{i-1},r_{j-1})[B(r_{i})-B(r_{i-1})][B(r_{j})-B(r_{j-1})]\right)\\ &=\sum_{i=1}^{n}k(r_{i-1},r_{i-1})E([B(r_{i})-B(r_{i-1})]^{2})\\ &= \sum_{i=1}^{n}k(r_{i-1},r_{i-1})(r_{i}-r_{i-1})\\ &=\int_{0}^{1}k(r,r)\ dr.\end{align*}