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Probably a bit trivial, but I was curious about the validity of interchanging the following integrals (where $W_t$ is Brownian Motion):

$\mathbb{E}[\int^{t}_{0} W^2_s ds] =? \int^{t}_{0} \mathbb{E}[W^2_s] ds$

In context, we know that $\mathbb{E}[\int^{t}_{0} W_s dW_s]^2 = \mathbb{E}[\int^{t}_{0} W^2_s ds]$

I just wanted to verify that its valid to make the jump that the above is equal to $\int^{t}_{0} s ds = \frac{1}{2}s^2$, since $\mathbb{E}[W^2_s] = s$ with Brownian Motion.

measure
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    Just a minor comment on notation - the variable used in the integrand shouldn't also be used in the limits. e.g. write $\displaystyle \int_0^t s,\mathsf ds$ instead. – Math1000 Jul 06 '15 at 14:15
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    Agreed, thanks! I made the edits. – measure Jul 06 '15 at 14:19
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    In your equation with the question mark, those are not stochastic integrals and Fubini's theorem applies directly. –  Jul 06 '15 at 14:31
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    @ByronSchmuland Thanks. The fact that the integrand is stochastic adds no complications? Since the Isometry shows that the double integral is finite, Fubini holds? – measure Jul 06 '15 at 14:37
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    @measure Reading the title of your question makes one think if you want to know whether $E[\int_0^tW_s, dW_s] = \int_0^tE[W_s], dW_s$ holds. – Calculon Jul 06 '15 at 15:14

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Fubini's theorem on $[0,t]\times\Omega$ gives $$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.$$