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I have a process $W:= \{W_t\}_{t>0}$ and defined $Y_t = tW_{1/t}$ for all $t>0$

I want to prove that $E[Y_tY_s] = s$

Substituting in the values and extracting the constants out, I am left with

$tsE[W_{1/t}W_{1/s}]$

Now for this to be equal to $s$, I would need the expected value of the two processes to be equal to $1/t$, however I can not tell how to compute that and was wondering if I could get some pointers.

jinx
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1 Answers1

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$EW_uW_v=\min \{u,v\}$ for all $u, v >0$. So you get $tsE[W_{1/t}W_{1/s}]=ts \min ({\frac 1t, \frac 1s})$ which is nothing but $\min (t,s)$.

Proof of $EW_uW_v=\min \{u,v\}$: Suppose $u <v$ Then $EW_uW_v=E(W_u(W_v-W_u))+EW_u^{2}=E(W_u(W_v-W_u))+EW_u^{2}=0+u=u$. [$E(W_u(W_v-W_u))-(EW_u) (EW_v-W_u)=0$ by independence].

Kavi Rama Murthy
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