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suppose I have a family of i.i.d standard normal random variables $Y_{n,k}$ and I define $X^N_t:=\sum_{n=0}^N\sum_{k=1}^{2^n}Y_{n,k}\phi_{n,k}(t)$ for $t\in [0,1]$ where $\phi_{n,k}$ are the Schauder functions.Furthermore, I know that $(X_t^N)$ is a martingale bounded in $L^2$ and therefore converges a.s. and in $L^2$ to a random variable $X$.

Why am I allowed to interchange expectation and the sum in the following expression

$E[X_sX_t]=\sum\sum E[Y_{n,k}Y_{l,m}]\phi_{n,k}\phi_{l,m}$

Note: The two sums are running both over two variables. The first over $n,m$ and the second over $k,l$.

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    @ procastinator: General, I have to show \int_{X\times Y}|F(x,y)|d(x,y) < \infty. If this is true, then $\int_{X\times Y}F(x,y)d(x,y) = \int_X\int_YF(x,y)dydx$. The expression on the RHS is $\int_X\int_YF(x,y)dydx$ and not $\int_{X\times Y}|F(x,y)|d(x,y) < \infty$. Or am I wrong? Actually I'm not quite sure how this product measure should look like.2012-07-05

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By the Cauchy-Schwarz inequality, the mapping \[ L^2(P)\times L^2(P) \ni (X,Y) \mapsto E(XY) \] is continuous, which can be seen as follows: Let $(X_0, Y_0) \in L^2(P)^2$ and $\epsilon > 0$. Let $\delta = \min\{\epsilon, \frac{\epsilon}{\max\{\|X_0\|_2, \|Y_0\|_2+\epsilon\}}\} $, then for $(X,Y) \in L^2(P)^2$ with $\|X-X_0\|_2 + \|Y-Y_0\|_2 < \delta$ it holds \begin{align*} \left|E(X_0Y_0) - E(XY)\right| &\le \left|E[X_0(Y_0 - Y)]\right| + \left|E[Y(X_0-X)]\right|\\ &\le \|X_0\|_2 \|Y_0 - Y\|_2 + \|Y\|_2 \|X-X_0\|_2\\ &\le \max\{\|X_0\|_2, \|Y_0\|_2 + \epsilon\} \bigl(\|X-X_0\|_2 + \|Y-Y_0\|_2\bigr)\\ &< \epsilon. \end{align*} Now we can continue: As $X^N_t \to X_t$ and $X^M_s \to X_s$ in $L^2(P)$, we have $E(X_sX_t) = \lim_{N,M} E(X^N_s Y^M_s)$. As expectation is linear, the last term equals $\lim_{N,M} \sum_{n\le N} \sum_{m\le M} \sum_{k=1}^{2^n}\sum_{l=1}^{2^m} E(Y_{n,k}Y_{m,l})\phi_{n,k}(t)\phi_{m,l}(s)$.

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    Sry ... my mistake, we can choose \delta < \epsilon, then \|Y_0 - Y\| \le \delta < \epsilon. Will correct it above ...2012-07-09