Deriving Expected Returns in the Gibson–Schwartz Commodity Model

# Gibson & Schwartz two factor model: mathematical derivation of the total expected return of a commodity contingent claim


# Gibson & Schwartz two factor model: mathematical derivation of the total expected return of a commodity contingent claim












I have been recently introduced to the Gibson & Schwartz two factor model (1990, link). According to the model, the dynamics of the spot commodity (oil) price ($S$) and of the convenience yield ($\delta$) can be captured by the following two drift and diffusion processes with correlated Brownian Motion increments:

$$dS = (r-\delta) S dt + \sigma_1 S dz_1 $$ $$d\delta = (k(\alpha -\delta) - \lambda \sigma_2 )dt + \sigma_2 dz_2 $$ $$ dz_1 dz_2 = \rho dt $$

In the above equation $\lambda$ represents the price per unit of convenience yield risk and $r$ the risk-free rate.

For a contingent claim (future contract) $B(S,\delta, \tau)$ , where $\tau$ is the time to maturity, in page 961 of the article (or page 4 of the pdf, see referenced link), in footnote 6 it is stated:

> The no-arbitrage condition leads to the following relationship between the total (expected) return of the claim $\mu_B$ and its risk exposure: $$\mu_B = r + \lambda ' \frac{SB_S \sigma _1}{B} +\lambda \frac{B_{\delta} \sigma _2}{B}$$

In the above equation $\lambda '$ stands for the price per unit of commodity (oil) price risk.

While I understand on high level the meaning of the no-arbitrage assumption, and its connection to the expectation under the risk-neutral probability measure $\mathbb{E^Q}[dB] = rBdt$, it is still unclear to me how the above equation of $\mu_B$ is mathematically derived. Could someone provide a detailed derivation or proof of this equation?

## Answer by Andrea (score 1)

https://quant.stackexchange.com/a/81772

Let's apply Ito to $B(S, \delta)$

$d B = (\cdots) dt + B_S dS + B_{\delta} d \delta$

Since $B$ is a price, we know its drift must be $rB$, so the above reads

$d B = r B dt + B_S dS + B_{\delta} d \delta = r B dt + B_S \sigma _1 S dz_1 + B_{\delta} \sigma_2 d z_2$

If we apply a drift change of $\lambda'$ and $\lambda$ to $z_1$ and $z_2$, we obtain

$d B = r B dt + B_S \sigma_1 S (\lambda' dt + dz_1) + B_{\delta} \sigma_2 (\lambda dt + dz_2)$

and only looking at the drift

$r B + B_S \sigma _1 S \lambda' + B_{\delta} \sigma_2 \lambda = B \left( r + \lambda' \frac{B_S \sigma_1 S }{B} + \lambda \frac{B_{\delta} \sigma_2}{B}\right )$

as claimed.