Identity for exponential of Brownian motion using scaling relation

Question

Let $B$ be a Brownian motion and $s\wedge t := \min\{s,t\}$, $s\vee t := \max\{s,t\}$.

I stumbled over the following identity:

$$ \mathbb E[\exp(B(s\wedge t) + B(s\vee t))] \\=\mathbb E[\exp(B(s\wedge t) + B(s\wedge t) + [B(s\vee t) - B(s\wedge t)])] \\=\mathbb E[\exp(2\sqrt{(s\wedge t)} B(1))] \cdot \mathbb E[\exp(\sqrt{\vert t-s \vert}B(1))] $$

Evidently, this result can be obtained by using the scaling relation $B(t)=_d \sqrt{t}B(1)$, but if I try to apply this to the equation I get a different result.

How can one explain this identity?

Answer 1

$B(s \vee t)-B(s \wedge t)$ and $2B(s \wedge t)$ are indepedent since the Brownian motion has independent increments. Thus $$\mathbb{E}e(2B(s \wedge t) + B(s \vee t)-B(s \wedge t)) = \mathbb{E}e(2B(s \wedge t)) \cdot \mathbb{E}(B(s \vee t)-B(s \wedge t))$$

Moreover, obviously, $$B(s \wedge t) + B(s \wedge t) = 2 B(s \wedge t) \stackrel{d}{=} 2 \sqrt{s \wedge t} \cdot B(1)$$ by scaling property. Similarly, $$B(s \vee t)-B(s \wedge t) \stackrel{d}{=} B(s \vee t - s \wedge t) = B(|t-s|) \stackrel{d}{=} \sqrt{|t-s|} \cdot B(1)$$ where we used the stationarity of the increments (and again the scaling relation, for the last step).

Answer 2

Obviously, $B(s\wedge t)+B(s\vee t)=2B(u)+B'(v)$ with $u=s\wedge t$, $v=s\vee t-s\wedge t=|t-s|$ and $B'(v)=B(v+u)-B(u)=_dB(v)$ while being independent of $B(u)$. The rest follows.