Having a payoff of log-contract defined as $$ \Pi_T = \ln \left(\frac{S_T}{S_0} \right) $$

How would you express the MC-estimator for the price of this contract?

The stock price dynamics here is given as $$\frac{dS_t}{St}=r dt + \sigma dW^{\mathbb {Q}}_t \tag{2}$$


1 Answer 1


By definition, the payoff of a log-contract of maturity $T$ writes $$ \phi(S_T) = \ln\left(\frac{S_T}{S_0}\right) $$ Let $\Pi_t$ denote the $t$-value of such a contingent claim. We are interested in the price at $t=0$, best known as the option premium. Theory tells us that the latter premium can be computed as $$ \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ \phi(S_T) \right] $$ that is, a (discounted and risk-neutral) expectation of the future payout, the payout here being a simple function of the random variable $S_T$.

Assuming we managed to draw $M$ i.i.d.(independent and identically distributed) samples of the random variable $S_T$ - which we shall denote by$(S_T^{(m)})_{m=1,...,M}$ in what follows -, we could easily infer $M$ i.i.d. samples of the random payout $\phi(S_T)=\ln(S_T/S_0)$, then take their average and multiply by the discount factor to get the option price. This is precisely the idea behind Monte Carlo. Now how to simulate i.i.d. samples of $S_T$?

From the Black-Scholes SDE expressed under the risk-neutral measure $\mathbb{Q}$, applying Itô's lemma gets you: $$ S_T = S_0 \exp\left( (r-\frac{1}{2}\sigma^2)T + \sigma \sqrt{T} Z \right)$$ with $Z \sim \mathcal{N}(0,1)$ and hence $S_T$ is lognormally distributed.

This shows that, in order to obtain i.i.d. samples $S_T^{(m)}$, one just needs i.i.d. samples $Z^{(m)}$ out of a standard normal distribution (there should exist libraries to do that in any decent programming language) $$ S_T^{(m)} = S_0 \exp\left( (r-\frac{1}{2}\sigma^2)T + \sigma \sqrt{T} Z^{(m)} \right),\ \ \forall m=1,...,M $$

The i.i.d. sample option payouts are then $$ \phi\left(S_T^{(m)}\right) =\ln\left(\frac{S_T^{(m)}}{S_0} \right),\ \ \forall m=1,...,M $$

The Monte Carlo estimator of the option price is then given by $$ S_T^{(m)} = S_0 \exp\left( (r-\frac{1}{2}\sigma^2)T + \sigma \sqrt{T} Z^{(m)} \right),\ \ \forall m=1,...,M $$ $$ \hat{\Pi}_0 = e^{-rT} \left( \frac{1}{M} \sum_{m=1}^M \phi\left(S_T^{(m)}\right) \right) $$ with $(Z^{(m)})_{m=1,...,M}$ representing $M$ i.i.d. samples out of a standard normal distribution.

$\hat{\Pi}_0$ is an unbiased estimator of the true premium $\Pi_0$ whose variance decreases proportionally to $M^{-1/2}$ (which is a direct consequence of the CLT).

[Remark 1] This result holds for any instrument whose payoff is a function of the terminal value of the asset only, i.e. $\phi(.)$ of the form $\phi(S_T)$. This is known as a European option. The situation of a log-contract with $\phi(S_T)=\ln(S_T/S_0)$ is just a particular case.

[Remark 2] A log-contract can be priced in semi-closed form (see work of Carr-Madan in that area) $$ \Pi_0 = e^{-rT}\left( rT - \int_0^{F(0,T)} \frac{\tilde{P}(K,T)}{K^2} dK + \int_{F(0,T)}^{\infty} \frac{\tilde{C}(K,T)}{K^2} dK \right) $$ where $\tilde{P}(K,T)$ and $\tilde{C}(K,T)$ figure undiscounted European put/call option prices. This gives you a good reference to compare your MC simulations to (or assess its convergence).

  • $\begingroup$ Is it practical to add Milstein correction to the discretized GBM SDE? $\endgroup$
    – nz_
    May 22, 2016 at 7:08
  • $\begingroup$ Here, you are not trying to simulate full paths (ie realisations) but just terminal values. There is no discretisation, this is an exact solution (I'm not talking about the premium but about $S_T$). Anyway, if you use an Euler discretisation of log-returns (and not directly prices)... $\endgroup$
    – Quantuple
    May 22, 2016 at 8:24
  • $\begingroup$ Milstein does not add anything under GBM assumptions. See here for instance google.be/url?sa=t&source=web&rct=j&url=http://… $\endgroup$
    – Quantuple
    May 22, 2016 at 8:28
  • $\begingroup$ @Michal if this answer helped you please mark this question as "answered". If not please state what you do not understand? $\endgroup$
    – Quantuple
    Oct 5, 2016 at 13:56

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