The SDE of the reciprocal

Let BtB_t be a standard Brownian motion, and suppose XtX_t is a process whose reciprocal satisfies d ⁣(1Xt)=1Xt(2dtdBt)d\!\left(\tfrac{1}{X_t}\right) = \tfrac{1}{X_t}\big(2\,dt - dB_t\big). The process XtX_t then satisfies an SDE of the form dXt=Xt(adt+bdWt)dX_t = X_t\big(a\,dt + b\,dW_t\big) for integers a,ba, b. Find aba\cdot b.

Show hints (2)+
  1. Let Y=1/XY=1/X; then X=1/YX=1/Y. Apply Itô to g(y)=y1g(y)=y^{-1}, keeping 12g(dY)2\tfrac12 g''(dY)^2.
  2. (dY)2=Y2dt(dY)^2=Y^2dt; you get dX=X(dt+dB)dX=X(-dt+dB), so a=1,b=1a=-1,b=1.

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Asked at: Jane Street, Citadel

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