# Derive the price of log contract

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.

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 $$r = 0$$.

• @noob2 But it suffices to replace $r = 0$ in the formula of (Neuberger, 1999), right?
– NN2
Dec 3, 2021 at 14:12
• @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 $r$? Dec 4, 2021 at 16:34
• @Gunner_ZZ the right hand side of the formula you wrote is already the price, so I think the left hand side must be $$E^Q_t(e^{-r(T-t)}\ln(F_t))$$
– NN2
Dec 4, 2021 at 17:49
• @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 $t$? Dec 4, 2021 at 19:09
• @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 $\ln(S_T)$) $$E^Q_t[e^{-r(T-t}\ln(S_T)] = e^{-r(T-t)}[\ln(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)]$$
– NN2
Dec 4, 2021 at 19:52