Problem of expectation of product of Brownian motion

Question

$B(t),\:t\ge0$ is a standard Brownian motion. $0<s,t<\infty$.

Is it correct to write $$E\left[sB(\frac{1}{s})tB(\frac{1}{t})\right]=min(s,t)$$

I've used the property $E[B(t)B(s)]=min(s,t)$.

Answer 1

Yes, it's right. The verification is easy:

$\mathbb{E}(sB(\frac{1}{s})tB(\frac{1}{s}))=st\mathbb{E}(B(\frac{1}{s})B(\frac{1}{t}))=st\min\{\frac{1}{s},\frac{1}{t}\}=\min\{\frac{st}{s},\frac{st}{t}\}=\min\{s,t\}$