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Is Ito's isometry true for conditional expectations too?

I mean, is it true that:$$\mathbb{E}\left[\left(\int_0^tX_sdB_s\right)^2\ |\ \mathcal{F}_t^B\right]=\mathbb{E}\left[\int_0^tX^2_sds\ |\ \mathcal{F}_t^B\right]$$ where $B_t$ is a Brownian motion and $\mathcal{F}_t^B$ the natural filtration?

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1 Answer 1

This equality cannot hold true as the following counterexample shows:

For fixed $t>0$ set $X_s(\omega) := 1_{[0,t]}(s)$. Then the left-hand side is equal to

$$\mathbb{E} \bigg( \underbrace{\left(\int_0^t X_s \, dB_s \right)^2}_{B_t^2} | \mathcal{F}_t^B \bigg) = \mathbb{E}(B_t^2|\mathcal{F}_t^B)=B_t^2$$

whereas the right-hand side equals

$$\mathbb{E} \left(\int_0^t X_s^2 \, ds |\mathcal{F}_t^B \right) = \mathbb{E}(t|\mathcal{F}_t^B)=t$$

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