Skip to main content
2 of 2
added 2 characters in body
Michael Hardy
  • 1
  • 12
  • 85
  • 126

From $\int_0^1W_t\,dW_t={1\over 2}(W_t^2-t)$. One can deduce with the change of variable $s=t-\rho$, that your integral is equal to: $$ {1\over 2}(W_1^2-W_\rho^2-W_{1-\rho}^2) $$

rjm
  • 75
  • 3