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Good evening, how can I solve this expected value? $$ E \Bigl[ B_1 \int_0^{x} B_u du\ \Bigr] $$ where $B_t$ is a standard Brownian Motion and x > 0.

dubidub
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  • $$ E \Bigl[ B_1 \int_0^{x} B_u du\Bigr] = \int_0^{x} E \Bigl[ B_1 B_u\Bigr]\ du = \int_0^{x} \min(1,u)\ du $$ – Did Feb 09 '15 at 08:35

2 Answers2

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Integration by parts justifies

$$ \int_{0}^{x}B_u \mathrm du = xB_x - \int_{0}^{x}u \mathrm dB_u\,. $$

Therefore,

$$ \begin{array}{rcl} \displaystyle \mathbb{E}\left[B_1 \int_{0}^{x}B_u \mathrm du\right] &{}={}&\displaystyle \mathbb{E}\Big[xB_1 B_x \Big] - \mathbb{E}\left[B_1 \int_{0}^{x}u ~\mathrm dB_u \right] \\ &{}={}&\displaystyle x\mathbb{E}\Big[B_1 B_x \Big] - \mathbb{E}\left[ \int_{0}^{1} 1 ~\mathrm dB_u \int_{0}^{x}u ~\mathrm dB_u \right] \\&{}={}&\displaystyle x\mathbb{E}\Big[B^2_{1\wedge x} \Big] - \mathbb{E}\left[ \int_{0}^{1\wedge x} 1 ~\mathrm dB_u \int_{0}^{1\wedge x}u ~\mathrm dB_u \right] \\ &{}={}&\displaystyle x\cdot\left(1\wedge x\right) - \int^{1\wedge x}_{0}u~\mathrm du \\ &{}={}&\displaystyle \left(x - \frac{1}{2}\left(1\wedge x\right)\right)\cdot\left(1\wedge x\right)\,. \end{array} $$

Above, use was made of the "independent increments", "variance" and "zero expectation" properties of Brownian motion. Also, Ito's Isometry (see this answer) justifies

$$ \mathbb{E}\left[ \int_{0}^{1\wedge x} 1 ~\mathrm dB_u \int_{0}^{1\wedge x}u ~\mathrm dB_u \right] = \int^{1\wedge x}_{0}u~\mathrm du\,. $$

We have just shown,

$$ \mathbb{E}\left[B_1 \int_{0}^{x}B_u \mathrm du\right] = \left(x - \frac{1}{2}\left(1\wedge x\right)\right)\cdot\left(1\wedge x\right)\,. $$
Community
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ki3i
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  • +1, the only thing i dont fully understand is why $$ \mathbb{E}\left[ \int_{0}^{1} 1 ~\mathrm dB_u \int_{0}^{x}u ~\mathrm dB_u \right] = \mathbb{E}\left[ \int_{0}^{1\wedge x} 1 ~\mathrm dB_u \int_{0}^{1\wedge x}u ~\mathrm dB_u \right] $$ – JKnecht Oct 06 '19 at 20:23
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This article might be useful. You evaluate the integral as follows: $\int_0^x B_u du = (B_0 + B_h + B_{2h} + ... + B_x)h$ with a stepwidth $h$. Then you can use the fact for Standard Brownian motion $E(B_i B_j) = \delta_{ij}$ for some numbers $i,j$.

kryomaxim
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