We have a financial derivative that pays $S_T \ln\left(S_T\right)$ at maturity $t=T$

We assume a Black-Scholes world:

  • No arbitrage opportunities.
  • No dividend payments from the stock $S_t$.
  • Existence of a riskless asset yielding the risk free rate
  • Possibility to borrow and lend infinitely at the risk-free rate.
  • Possibility to buy and sell infinitely the stock $-$ even fractional amounts.
  • No transaction costs.

We also assume that the stock is tradable and that the derivative is attainable $-$ we basically assume we are in the standard pricing setting.

What's the present value of this financial derivative at $t=0$ ?

My understanding is that using risk neutral measure to calculate PV of this payoff is rather difficult. We need to change the measure to simplify the calculation.


1 Answer 1


I assume this is a homework question. You do need to change the measure and price under $Q_s$ rather than $Q$.

$E_{Q}(S_T.\ln(S_T)) = S_t.E_{Q_s}(\ln(S_T))$

Note the SDEs for $dS_t$ and hence $d\ln(S_t)$ now change by virtue of the Girsanov theorem.

In particular $d\ln(S_t) = (r+\frac{\sigma^2}{2})dt + \sigma dW^{Q_s}_t$

Taking the integral then expectation, you should find eventually that

$PV(t) = S_t(\ln(St) + (r+\frac{\sigma^2}{2})(T-t))$

  • $\begingroup$ "now change by virtue of the Girsanov theorem" how and why? $\endgroup$
    – Trajan
    Jul 3, 2020 at 17:16
  • $\begingroup$ “The SDEs ... now change” (“now that the measure is changed”). If you change the measure, the SDE $dS_t$ (and any other that depends on it) changes under the new measure. This is the Girsanov theorem. See en.wikipedia.org/wiki/Girsanov_theorem for example. $\endgroup$
    – Ivan
    Jul 4, 2020 at 18:54
  • $\begingroup$ how did you know it would be that form though? $\endgroup$
    – Trajan
    Jul 5, 2020 at 6:16
  • $\begingroup$ If you change the measure from $Q$ to $Q_s$ and apply the Girsanov theorem, you get that form for $dln(S_t)$. It is different from that under $Q$. In the first place you use Ito to get $dln(S_t)$. $\endgroup$
    – Ivan
    Jul 5, 2020 at 10:59

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