# What is the inverse of the covariance operator in Brownian motion?

Heidi
in fact the answer is given in the book (written by philippe Martin).
we have
$$(\tau_1| A^{-1} | \tau_2) = 2D \ min(\tau_1 ,\tau_2) = 2D(\tau_1 \theta (\tau_2 -\tau_1)+\tau_2 \theta (\tau_1 -\tau_2))$$
So
$$-1/2D \frac{d^2}{d\tau_1^2} (\tau_1| A^{-1} | \tau_2) = \delta( \tau_1 - \tau_2)$$

the author writes then that we see that we have the operators equality
$$-1/2D \frac{d^2}{d\tau^2} = A$$

is it obvious for you?
thanks

Edit:
does the derivative act in the same way on every members of the
covariant matrix 1/A? in this case we could consider that we have the product ot A and 1/A giving a dirac.

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