I am reading the Neuberger [1999] Log Contract paper and really confused on the log contract. So if the payoff is $\ln(S_T)$, then we can easily solve the price of such derivative: $$f_t^s = e^{-r(T-t)}[\ln(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)]$$ So my question is that when we have log contract whose underlying is $F_t = S_te^{r(T-t)}$, how do we derive the price of derivative with payoff $\ln(F_T)$, as indicated in the paper: $$f_t^F = \ln(F_t) -\frac{1}{2}\sigma^2(T-t)$$ It looks like $f_t^F = f_t^se^{r(T-t)}$, but why since natural log is a nonlinear tranformation.


1 Answer 1


Applying the Ito lemma, you prove easily that the dynamics of $F_t$ in risk-neutral measure $\Bbb Q$ is $$ \frac{dF_t}{F_t} = \sigma dW_t $$ (the drift is $0\cdot dt$, in stead of $r\cdot dt$ as in the dynamics of $S_t$)

Thus, it suffices to apply the formula of (Neuberger, 1999) to derive the price of the derivative with payoff $\ln (F_T)$ by replacing $r = 0$.

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    $\begingroup$ @noob2 But it suffices to replace $r = 0$ in the formula of (Neuberger, 1999), right? $\endgroup$
    – NN2
    Dec 3, 2021 at 14:12
  • $\begingroup$ @NN2 Thanks, so after we get the $E_t^Q[\ln(F_T)] = \ln(F_t)-\frac{1}{2}\sigma^2(T-t)$, we discount it back at $r=0$, instead of actual risk free rate $r$? $\endgroup$
    – Gunner_ZZ
    Dec 4, 2021 at 16:34
  • $\begingroup$ @Gunner_ZZ the right hand side of the formula you wrote is already the price, so I think the left hand side must be $$E^Q_t(e^{-r(T-t)}\ln(F_t))$$ $\endgroup$
    – NN2
    Dec 4, 2021 at 17:49
  • $\begingroup$ @NN2 so if we apply Ito lemma on $\ln F_t$, we have:$$d\ln F_t = -\frac{1}{2}\sigma^2dt + \sigma dW_t$$ Integrate both sides from $t$ to $T$, then: $$\ln F_T = \ln F_t -\frac{1}{2}\sigma^2(T-t) - \sigma(W_T-W_t) $$ Take expectations on both sides: $$E_t^Q[\ln(F_T)] = \ln(F_t) -\frac{1}{2}\sigma^2(T-t) $$ why the right hand side is already the price? Don't we need the discount the expectation back to $t$? $\endgroup$
    – Gunner_ZZ
    Dec 4, 2021 at 19:09
  • $\begingroup$ @Gunner_ZZ I didn't calculate the price of the payoff $\ln (S_T)$ or $\ln(F_T)$. but as you wrote in the question, the formula of (Neuberger, 1999) is already the price of $\ln(S_T)$: $$f_t^s = e^{-r(T-t)}[\ln(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)]\tag{1}$$ then the price of the payoff $\ln(F_T)$ must be (also as you wrote in the question) $\ln(F_t) -\frac{1}{2}\sigma^2(T-t)$ by the argument I made in the answer. $$$$ Besides, I think the left hand side of $(1)$ must be (as $(1)$ is the price of $\ln(S_T)$) $$E^Q_t[e^{-r(T-t}\ln(S_T)] = e^{-r(T-t)}[\ln(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)]$$ $\endgroup$
    – NN2
    Dec 4, 2021 at 19:52

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