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Consider the ON-sequence $\{\varphi_{k}\}_{k}\in L^{2}(\mathbb{R})$ and let $$I_{k}\in \mathrm{Bernoulli}(\lambda_{k}),\;\sum_{k=1}^\infty \lambda_{k}<\infty$$ $ \{I_k\} $ are all independent. Also let $A$ be a bounded intervall in $\mathbb{R}$ Is it allowed to change order of integration in the following manner? $$\mathbb{E}\left[\intop_A \intop_{A^c}\sum_{k=1}^\infty \sum_{j=1}^\infty I_k I_j \varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y) \, dx \, dy\right]=$$ $$\intop_{A}\intop_{A^{c}}\sum\limits _{k=1}^{\infty}\sum\limits _{j=1}^{\infty}\mathbb{E}\left[I_{k}I_{j}\right]\varphi_{k}(x)\overline{\varphi_{j}(x)}\overline{\varphi_{k}(y)}\varphi_{j}(y)dxdy$$

This is how far i have come on my own: I would like to invoke Borell Cantelli lemma, which says that the sum in the integral is finite a.s.. However for each $\omega\in\Omega$ outside a zero-set the number of terms will depend on $\omega$ so there is no single function that bounds the integrand for almost all $\omega$ and there for we may not move the expectation inside the integral.

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    How about Tonnelli's theorem? It seems that the integrand is non-negative.2012-08-19
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    One basic problem: It's not always true that $\mathbf{E}(I_k I_j) = \lambda_k \lambda_j$.2012-08-19
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    sorry, they are independent as well. Should have mentioned it.2012-08-19
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    @srsly That doesn't matter. The problem is the case $k=j$.2012-08-19
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    That is true. So the left-hand side is incorrect. We would have to split up the sums. But could we still move the expectation inside?2012-08-19
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    Yes. By Fubini and Did's answer.2012-08-19
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    If you mean $I_k$ is a Bernoulli-distributed random variable, then $I_k\sim\mathrm{Bernoulli}(\lambda_k)$ would be the standard way to express that; not $I_k\in\mathrm{Bernoulli}(\lambda_k)$.2012-08-19
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    He did not motivate the exchange of order however. Could someone elaborate? The integrand is only finite if one integrates with respect to the probability measure first so how can you use fubini?2012-08-19
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    The standard way to check whether using Fubini is permissible, for $f(x,y)$, say, is to attempt to integrate $|f(x,y)|$. Since $|f(x,y)|$ is nonnegative, changing the order of integration is *always* permissible (this is what's sometimes called Tonelli's theorem). If the result of integrating $|f(x,y)|$ is finite, then exchanging the order of integration is permissible in the original integral of $f(x,y)$. In his answer, did checked exactly this condition. (Also note that $\mathbf{E}$, $\sum$, and $\int$ are all different ways of saying $\int$ with respect to measures of particular types.)2012-08-19
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    I was not aware of this, i thought you had to dominate the integrand. Thank you for clarifying2012-08-19

2 Answers 2

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Consider $$S=\mathbb{E}\left[\intop_{A}\intop_{A^{c}}\sum\limits _{k=1}^{\infty}\sum\limits _{j=1}^{\infty}\left|I_{k}I_{j}\varphi_{k}(x)\overline{\varphi_{j}(x)}\overline{\varphi_{k}(y)}\varphi_{j}(y)\right|dxdy\right].$$ The sum $S$ of this series of nonnegative terms always exists in $\mathbb R_+\cup\{+\infty\}$. If $S$ is finite, Fubini theorem applied to the product measure of $\mathbb P$, two Lebesgue measures and two counting measures, ensures that every possible integral exists.

To show that $S$ is indeed finite, note that $\mathbb{E}(I_{k}I_{j})=\lambda_k\lambda_j$ for every $k\ne j$ and $\mathbb{E}(I_{k}^2)=\lambda_k$ for every $k$. Hence, for every $A$, $$S\leqslant\sum_{k=1}^{\infty}\lambda_k\left(\int_{\mathbb R}\left|\varphi_{k}(x)\right|^2dx\right)^2+\sum_{k\ne j}\lambda_k\lambda_j\left(\int_{\mathbb R}\left|\varphi_{k}(x)\varphi_{j}(x)\right|dx\right)^2, $$ and, for every $(k,j)$, $$ \left(\int_{\mathbb R}\left|\varphi_{k}(x)\varphi_{j}(x)\right|dx\right)^2\leqslant\int_{\mathbb R}\left|\varphi_{k}(x)\right|^2dx\cdot\int_{\mathbb R}\left|\varphi_{j}(x)\right|^2dx=1, $$ hence $$ S\leqslant\sum_{k=1}^{\infty}\lambda_k+\sum_{k\ne j}\lambda_k\lambda_j=\sum_{k=1}^{\infty}\lambda_k(1-\lambda_k)+\left(\sum_{k=1}^{\infty}\lambda_k\right)^2\leqslant\sum_{k=1}^{\infty}\lambda_k++\left(\sum_{k=1}^{\infty}\lambda_k\right)^2, $$ which is finite by hypothesis, QED.

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    How did you reach the first conclusion? How do you motivate moving the expectation inside the intergral?2012-08-19
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    If $k=j$ we have $\mathbf{E}(I_k I_j) = \lambda_k$, not $\lambda_k^2$, so you should end with something like $S\leq (\sum\lambda_k)^2 + \sum\lambda_k < \infty$, shouldn't you?2012-08-19
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    srsly: See edited version.2012-08-19
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    @SeanEberhard Indeed. Thanks.2012-08-19
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(Edit: this answer is no longer relevant to the question, following OP's edits.)

Nevermind the expectation. You have

$$\int\int\sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j\varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y)\,dx\,dy = \sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j \delta_{kj}^2 = \sum_{j=1}^\infty I_j.$$

Now taking expectation gives

$$\mathbf{E}\int\int\sum_{k=1}^\infty\sum_{j=1}^\infty I_k I_j\varphi_k(x)\overline{\varphi_j(x)}\overline{\varphi_k(y)}\varphi_j(y)\,dx\,dy = \sum_{j=1}^\infty \lambda_j.$$

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    my bad, the integrand should not be positive.2012-08-19
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    Then first take absolute value to the integrand, apply Tonelli's theorem and show that it is indeed integrable. Then by Fubini's theorem you can interchange everything!2012-08-19
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    Had to change it once more... :/2012-08-19
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    Hmmm I think mayby if I split the integral in positive and negative parts and use intersections with $B_{n}=\{\omega:I_{k}=0,k\geq n\}$ in conjunction with monotone convergence it could work?2012-08-19