Replicating a Log-Price Integral Payoff with Stock and Variance

# Hedging strategy for payoff $\int_0^T\log S_u\mathrm{d}u$


# Hedging strategy for payoff $\int_0^T\log S_u\mathrm{d}u$












What would a hedging strategy look like for a payoff $\int_0^T\log S_u\mathrm{d}u$? I have determined under Black-Scholes stock dynamics,

$$\int_0^T\log S_u\mathrm{d}u=\int_0^t\log S_u\mathrm{d}u+\int_t^T\left[\log S_t+\left(r-\frac{\sigma^2}2\right)(u-t)+\sigma(W_u-W_t)\right],$$

so conditioned on $\mathcal{F}_t$ we have

$$\int_0^T\log S_u\mathrm{d}u\Big|\mathcal{F}_t\sim\mathcal{N}\left(\int_0^t\log S_u\mathrm{d}u+(T-t)\log S_t+\frac{T^2-t^2}2\left(r-\frac{\sigma^2}2\right),\frac{\sigma^2}3\frac{(T-t)^3}{T^2}\right),$$

and therefore the time-$t$ price of this option would be

$$V(S_t,t)=\mathrm{e}^{-r(T-t)}\left[\int_0^t\log S_u\mathrm{d}u+(T-t)\log S_t+\frac{T^2-t^2}2\left(r-\frac{\sigma^2}2\right)\right].$$

How do I use this information to develop a hedging strategy for such a claim? If I can perfectly delta hedge, in terms of the underlying, is delta really just as simple as

$$\frac{\partial}{\partial S_t}\mathrm{e}^{-r(T-t)}\left[\int_0^t\log S_u\mathrm{d}u+(T-t)\log S_t+\frac{T^2-t^2}2\left(r-\frac{\sigma^2}2\right)\right],$$

or is there more to it that I am missing?

## Answer by Soumirai (score 6, accepted)

https://quant.stackexchange.com/a/60207

I assume you want to price a derivative product that pays $\int_0^T\ln S_tdt$ at maturity time $T$, from time $t=0$. I'll ignore generalization to time $t$ because it is trivial (split the integral in two, before and after $t$ as you did).

The first trick it to do an integration by part on $\ln S_t dt$:

$d(t\ln S_t) = \ln S_t dt + \frac{t}{S_t}dS_t + \frac{t}{2S_t^2} \sigma_t^2 S_t^2 dt$

Now let's integrate both sides between $0$ and $T$ and rearrange:

$\int_0^T \ln S_t dt = \int_0^T d(t\ln S_t) - \int_0^T \frac{t}{S_t}dS_t - \int_0^T \frac{t}{2} \sigma_t^2 dt = T \ln S_T - \int_0^T \frac{t}{S_t}dS_t - \int_0^T \frac{t}{2} \sigma_t^2 dt$

So your payoff $\int_0^T \ln S_t dt$ equals three terms:

- $T \ln S_T$: $T$-times the log-contract, which is statically replicable.

- $\int_0^T -\frac{t}{S_t}dS_t$: a dynamic delta-hedging strategy (to short a quantity $\frac{t}{S_t}$ of stock at any time $t$).

- $\int_0^T -\frac{t}{2} \sigma_t^2 dt$ a third dynamic term that is a function of $S_t$'s total variance.

We price the derivative that pays $\int_0^T \ln S_t dt$ at time $T$, so you know by heart that its price is the discounted expectation under the risk-neutral measure:

$Price_0 = e^{-rT}\mathbb{E}^\mathbb{Q}( \int_0^T \ln S_t dt ) = e^{-rT}\mathbb{E}^\mathbb{Q}( T \ln S_T ) - e^{-rT}\mathbb{E}^\mathbb{Q}( \int_0^T \frac{t}{S_t}dS_t ) - e^{-rT}\mathbb{E}^\mathbb{Q}( \int_0^T \frac{t}{2} \sigma_t^2 dt )$

The two first terms are straightforward.

- $e^{-rT}\mathbb{E}^\mathbb{Q}( T \ln S_T )$ is the price off $T$ times log-contract.

- $e^{-rT}\mathbb{E}^\mathbb{Q}( \int_0^T -\frac{t}{S_t}dS_t )$ is the present value of the expected financing cost of the dynamic delta-hedging. Note that here, delta is negative (you short the stock) so financing is a benefit: you are paid interest to be short the stock.

The tricky part here is the third term: how are you going to hedge this integral of $\sigma_t^2$?

- In basic Black-Scholes framework, you have $\sigma_t = \sigma$, so

$\int_0^T \frac{-t}{2} \sigma_t^2 dt = -\frac{\sigma^2}{2} \int_0^T tdt = -\frac{T^2\sigma^2}{4}$

Nothing to hedge dynamically; it just makes the derivative cheaper by $e^{-rT}\frac{T^2\sigma^2}{4}$ (it is just some deterministic cash amount you remove from the rest of the replication strategy).

- If $\sigma_t = \sigma (t,S_t)$, i.e. $\sigma$ is a local volatility function, it becomes tricky. You would need to dynamically hedge that variance integral $\int_0^T -\frac{t}{2} \sigma_t^2 dt$, because it has some delta risk as well (it depends on $S_t$!). Maths become nasty because you have some recursivity there (hedging the variance integral will require a dynamic hedging strategy, likely to involve a variance integral as well, that would need to be delta-hedged..). In practice, all it means is that you would have to adjust your delta hedging $\int_0^T \frac{-t}{S_t}dS_t$ to account for the fact that volatility is not constant with time and spot! (different model, different delta)

- If volatility is stochastic and has its own source of risk, then simple delta-hedging is not enough. You need something else to hedge that variance risk (the variance integral). Note that the integral ($\int_0^T \frac{-t}{2} \sigma_t^2 dt$) is similar to the payoff of a variance swap ($\int_0^T \sigma_t^2 dt$), but not exactly, as it has some time weighting. So there is probably some dynamic hedging through options to be done there.