Incorrect use of the scaling relation for Brownian motion?

Question

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?

Answer 1

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 $t\ne1$, $\mathbb E(B_tB_1)=t\ne\sqrt{t}=\mathbb E(\sqrt{t}B_1^2)$.

Answer 2

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.