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How are the PDEs for f obtained from the SDE and Itô's lemma?

By substituting the ansatz S=f(W,t)S=f(W,t) into the SDE and applying Itô's lemma, the coefficients of dWdW and dtdt are matched. This yields the system of PDEs: fW=σff_W = \sigma f and ft+12fWW=μff_t + \frac{1}{2} f_{WW} = \mu f.

Conditions

  • S=f(W,t)S = f(W,t) is the assumed solution form.
  • dS=μS dt+σS dWdS = \mu S\,dt + \sigma S\,dW is the defining SDE.
  • Itô's lemma is applied to f(W,t)f(W,t).

Reasoning, step by step

  1. Write the SDE: dS=μS dt+σS dWdS = \mu S\,dt + \sigma S\,dW.
  2. Apply Itô's lemma to S=f(W,t)S=f(W,t): df(W,t)=[ft+12fWW]dt+fWdWdf(W,t) = \left[f_t + \frac{1}{2} f_{WW}\right] dt + f_W dW.
  3. Equate the dWdW coefficients: fW=σS=σff_W = \sigma S = \sigma f.
  4. Equate the dtdt coefficients: ft+12fWW=μS=μff_t + \frac{1}{2} f_{WW} = \mu S = \mu f.
  5. Conclude the PDE system.

Example

The board shows 'By Itô, fW=σS=f⋅σf_W = \sigma S = f \cdot \sigma and ft+12ff_t + \frac{1}{2} f_{WW} = μS=f⋅μ\mu S = f \cdot \mu.'

Common misconceptions

  • Failing to substitute S=f(W,t)S=f(W,t) into the right-hand side of the SDE before matching coefficients.
  • Assuming the dtdt coefficient in Itô's lemma is just ftf_t, ignoring the 12fWW\frac{1}{2} f_{WW} term.

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