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I want to calculate

$$CoV\left(B_1,\int_0^1 B_t dt\right) = \int_0^1 CoV\left(B_t,B_1\right) dt= \int_0^1 \min(t,1)dt = 1/2$$

On the other hand, one could also use the scaling relation for Brownian motion, $B_t=_d\sqrt{t} B_1$:

$$\int_0^1 CoV\left(B_t,B_1\right) dt = \int_0^1 \mathbb E (B_t B_1) dt = \int_0^1 \mathbb E (\sqrt{t} B_1^2) dt = \int_0^1\sqrt{t} dt =2/3$$

Why am I not allowed to apply the scaling relation at this point?

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up vote 1 down vote accepted

Sure $B_t=_d\sqrt{t}B_1$ but to make your substitution legal, one would need $(B_t,B_1)=_d(\sqrt{t}B_1,B_1)$, which is not true. As a matter of fact, for $t\ne1$, $\mathbb E(B_tB_1)=t\ne\sqrt{t}=\mathbb E(\sqrt{t}B_1^2)$.

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It's not true that $B_t = \sqrt{t} B_1$.

What is true is that $B_t =_{\rm d} \sqrt{t} B_1$, i.e. they are equal in distribution, but not in value.

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Thanks, edited it. But that doesn't answer the question, does it? – madison54 Mar 1 '13 at 14:06
Yes - when you compute ${\rm E}(XY)$ it can be true that $Y=_{\rm d} X$, but that doesn't justify the substitution $Y=X$ to get ${\rm E}(X^2)$. – Chris Taylor Mar 1 '13 at 14:07

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