Commodity Prices, Convenience Yield and Inventory Behaviour

Abstract

Markets for mineral commodities serve for delivering resources extracted from the earth to consumers. Price dynamics on these markets depend on the behaviour of commodity traders holding inventories. In this chapter, we examine a commodity market model with economic agents acting on the demand side and extracting utility from holding storage. In this model, commodity price fluctuations are driven by stochastic supply shocks. The market equilibrium dynamic is described by the stochastic difference equations for the commodity price and the convenience yield, which indicates the marginal utility of inventory holding. We show how the inventory behaviour of traders contributes to the autocorrelation of prices. Analysis of the model reveals the cases of price-stabilizing and -destabilizing inventory behaviour. In the former case, inventories are adjusted to smooth price fluctuations. In the latter case, the demand for inventories reinforces a price change caused by supply shock.

Appendices

Appendices

1.1 A.1 Proposition 6.1

The solution for s_t, y_t is found as a linear approximation of Eqs. (6.21), (6.22) near the stationary point s^∗ = (b/r)^1/θ, x^∗ = 1. Consider the storage-consumption ratio (6.19):

$$ {s}_t/{y}_t={\left(b/{\delta}_t\right)}^{1/\theta }. $$

The linear approximation of the right-hand side for δ^∗ = r implies:

$$ {\left(b/{\delta}_t\right)}^{1/\theta }={s}^{\ast }{\left(r/{\delta}_t\right)}^{1/\theta}\approx {s}^{\ast}\left(1-\left({\delta}_t-r\right)/\theta r\right). $$

Hence,

$$ \mathit{\ln}\left({s}_t/{s}^{\ast}\right)-\mathit{\ln}{y}_t\approx -\frac{\left({\delta}_t-r\right)}{\theta r}, $$

(6.41)

We will show that the solution for s_t = μ_ta_t satisfying this approximation is given by

$$ {s}_t={s}_t^a+{\mu}_r^{\prime}\left({\delta}_t-r\right){a}^{\ast }={s}_t^a+{\mu}_r^{\prime}\left({\delta}_t-r\right)\left(1+{\left(b/r\right)}^{1/\theta}\right), $$

(6.42)

where a^∗ = x^∗ + s^∗ = 1 + (b/r)^1/θ is the stationary availability,

$$ {s}_t^a=\mu \left({x}_t+{s}_{t-1}^a\right)=\mu \sum \limits_{\tau =0}^{\infty }{\mu}^{\tau }{x}_{t-\tau } $$

(6.43)

is the amount of storage defined by the availability effect,

$$ {\mu}_r^{\prime }=-\frac{b^{1/\theta }{r}^{1/\theta -1}}{\theta {\left({b}^{1/\theta }+{r}^{1/\theta}\right)}^2} $$

is the derivative of μ with respect to r. Inserting this derivative into (6.42) implies

$$ {s}_t={s}_t^a-\frac{b^{1/\theta }{r}^{1/\theta -1}\left(1+{\left(b/r\right)}^{1/\theta}\right)}{\theta {\left({b}^{1/\theta }+{r}^{1/\theta}\right)}^2}\left({\delta}_t-r\right)={s}_t^a-\left(\mu /\theta r\right)\left({\delta}_t-r\right). $$

(6.44)

The linear approximation for supply process (6.16) is:

$$ {x}_t=\alpha {x}_{t-1}+\left(1-\alpha \right)+{\varepsilon}_t. $$

Iterating terms yields:

$$ {x}_t=\sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau }{\varepsilon}_{t-\tau }+\left(1-\alpha \right)\sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau }=1+\sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau }{\varepsilon}_{t-\tau }. $$

Inserting this into (6.43) implies:

$$ {s}_t^a=\sum \limits_{\tau =0}^{\infty }{\mu}^{\tau +1}\left(1+\sum \limits_{j=0}^{\infty }{\alpha}^j{\varepsilon}_{t-\tau -j}\right)=\frac{\mu }{1-\mu }+\sum \limits_{\tau =0}^{\infty }{\mu}^{\tau +1}\sum \limits_{j=0}^{\infty }{\alpha}^j{\varepsilon}_{t-\tau -j}. $$

(6.45)

Suppose that α > μ and rearrange the dual sum:

$$ \sum \limits_{\tau =0}^{\infty }{\mu}^{\tau +1}\sum \limits_{j=0}^{\infty }{\alpha}^j{\varepsilon}_{t-\tau -j}=\mu \sum \limits_{j=0}^{\infty }{\alpha}^j{\varepsilon}_{t-j}+{\mu}^2\sum \limits_{j=0}^{\infty }{\alpha}^j{\varepsilon}_{t-1-j}+\dots =\mu \left({\varepsilon}_t+\left(\mu +\alpha \right){\varepsilon}_{t-1}+\left({\mu}^2+\mu \alpha +{\alpha}^2\right){\varepsilon}_{t-2}\dots \right)=\mu \left({\varepsilon}_t+\left(\left(\mu /\alpha \right)+1\right)\alpha {\varepsilon}_{t-1}+\left({\left(\mu /\alpha \right)}^2+\left(\mu /\alpha \right)+1\right){\alpha}^2{\varepsilon}_{t-2}\dots \right)=\mu \sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau }{\varepsilon}_{t-\tau}\sum \limits_{j=0}^{\tau }{\left(\mu /\alpha \right)}^j $$

$$ =\mu \sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau}\frac{1-{\left(\mu /\alpha \right)}^{\tau +1}}{1-\mu /\alpha }{\varepsilon}_{t-\tau }=\frac{\mu }{\alpha -\mu }\ \sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau +1}\left(1-{\left(\mu /\alpha \right)}^{\tau +1}\right){\varepsilon}_{t-\tau }=\frac{\mu }{\alpha -\mu }\ \left(\alpha \sum \limits_{\tau =0}^{\infty }{\alpha}^{\tau }{\varepsilon}_{t-\tau }-\mu \sum \limits_{\tau =0}^{\infty }{\mu}^{\tau }{\varepsilon}_{t-\tau}\right)=\frac{\mu }{\alpha -\mu }\ \left(\alpha ln{x}_t-\mu ln{q}_t\right) $$

due to (6.23) and (6.25). Inserting this into (6.45) and then into (6.44) implies the solution for s_t:

$$ {s}_t={s}_t^a-\left(\frac{\mu }{\theta r}\right)\left({\delta}_t-r\right)={s}^{\ast }+\mu \frac{\mu ln{q}_t-\alpha ln{x}_t}{\mu -\alpha }-\left(\frac{\mu }{\theta r}\right)\left({\delta}_t-r\right), $$

since μ/(1 − μ) = (b/r)^1/θ = s^∗. One can show, similarly, that this formula holds for α < μ.

The solution for y_t is derived in the same way:

$$ {y}_t=1+\left(1-\mu \right)\frac{\mu ln{q}_t-\alpha ln{x}_t}{\mu -\alpha }+\left(\mu /\theta r\right)\left({\delta}_t-r\right). $$

The obtained solution for s_t and y_t satisfies condition (6.41), because the left-hand side of (6.41) equals (s_t/s^∗) −  ln y_t ≈ s_t/s^∗ − y_t =  − ((1 − μ)/θr)(δ_t − r) − (μ/θr)(δ_t − r) =  − (δ_t − r)/θr.

1.2 A.2 Proposition 6.2

Consider the equation for expected price growth (6.17), written as

$$ {E}_t\Delta \mathit{\ln}{p}_{t+1}=-r{\overset{\sim }{\delta}}_t. $$

(6.46)

Using the equation for price (6.28) we have:

$$ {E}_t\Delta \mathit{\ln}{p}_{t+1}=\theta \left(1-\mu \right)\frac{\mu {E}_t\Delta \mathit{\ln}{q}_{t+1}-\alpha {E}_t\Delta \mathit{\ln}{x}_{t+1}}{\alpha -\mu }-\mu {E}_t\Delta {\overset{\sim }{\delta}}_{t+1}. $$

(6.47)

From (6.16) and (6.24):

$$ {E}_t\Delta \mathit{\ln}{q}_{t+1}=\left(\mu -1\right)\mathit{\ln}{q}_t,\kern0.75em {E}_t\Delta \mathit{\ln}{x}_{t+1}=\left(\alpha -1\right)\mathit{\ln}{x}_t. $$

(6.48)

For the linear solution ${\overset{\sim }{\delta}}_t={\lambda}_q\mathit{\ln}{q}_t+{\lambda}_x\mathit{\ln}{x}_t$ we obtain:

$$ {E}_t\Delta {\overset{\sim }{\delta}}_{t+1}={\lambda}_q{E}_t\Delta \mathit{\ln}{q}_{t+1}+{\lambda}_x{E}_t\Delta \mathit{\ln}{x}_{t+1}={\lambda}_q\left(\mu -1\right)\mathit{\ln}{q}_t+{\lambda}_x\left(\alpha -1\right)\mathit{\ln}{x}_t. $$

(6.49)

Inserting (6.47), (6.48), (6.49) and (6.29) into (6.46) implies:

$$ {E}_t\Delta \mathit{\ln}{p}_{t+1}=\theta \left(1-\mu \right)\frac{\mu \left(\mu -1\right)\mathit{\ln}{q}_t-\alpha \left(\alpha -1\right)\mathit{\ln}{x}_t}{\alpha -\mu }-\mu \left({\lambda}_q\left(\mu -1\right)\mathit{\ln}{q}_t+{\lambda}_x\left(\alpha -1\right)\mathit{\ln}{x}_t\right)=-r\left({\lambda}_q\mathit{\ln}{q}_t+{\lambda}_x\mathit{\ln}{x}_t\right). $$

Gathering the terms with lnq_t and lnx_t in this equality implies equations for λ_q and λ_x, respectively:

$$ \theta \left(1-\mu \right)\frac{\mu \left(\mu -1\right)\mathit{\ln}{q}_t}{\alpha -\mu }-\mu {\lambda}_q\left(\mu -1\right)\mathit{\ln}{q}_t=-r{\lambda}_q\mathit{\ln}{q}_t $$

$$ \theta \left(1-\mu \right)\frac{\alpha \left(\alpha -1\right)\mathit{\ln}{x}_t}{\mu -\alpha }-\mu {\lambda}_x\left(\alpha -1\right)\mathit{\ln}{x}_t=-r{\lambda}_x\mathit{\ln}{x}_t, $$

which yield:

$$ {\lambda}_q=\frac{\theta {\left(1-\mu \right)}^2\mu }{\left(\alpha -\mu \right)\left(r+\mu \left(1-\mu \right)\right)},\kern0.75em {\lambda}_x=\frac{\theta \left(1-\mu \right)\left(1-\alpha \right)\alpha }{\left(\mu -\alpha \right)\left(r+\mu \left(1-\alpha \right)\right)}. $$

1.3 A.3 Equation (6.31)

Inserting ${\overset{\sim }{\delta}}_t={\lambda}_q\mathit{\ln}{q}_t+{\lambda}_x\mathit{\ln}{x}_t$ into the equation for price (6.28):

$$ \mathit{\ln}{p}_t=\theta \left(1-\mu \right)\frac{\mu ln{q}_t-\alpha ln{x}_t}{\alpha -\mu }-\mu \left({\lambda}_q\mathit{\ln}{q}_t+{\lambda}_x\mathit{\ln}{x}_t\right)=\left(\frac{\theta \left(1-\mu \right)\mu }{\alpha -\mu }-\mu {\lambda}_q\right)\mathit{\ln}{q}_t+\left(\frac{\theta \left(1-\mu \right)\alpha }{\mu -\alpha }-\mu {\lambda}_x\right)\mathit{\ln}{x}_t $$

we have it that

$$ {\eta}_q=\frac{\theta \left(1-\mu \right)\mu }{\alpha -\mu }-\mu {\lambda}_q=\frac{\theta \left(1-\mu \right)\mu }{\alpha -\mu}\left(1-\frac{\left(1-\mu \right)\mu }{r+\mu \left(1-\mu \right)}\right)=\frac{r\theta \left(1-\mu \right)\mu }{\left(\alpha -\mu \right)\left(r+\mu \left(1-\mu \right)\right)}={\lambda}_q\frac{r}{1-\mu }, $$

$$ {\eta}_x=\frac{\theta \left(1-\mu \right)\alpha }{\mu -\alpha }-\mu {\lambda}_x=\frac{\theta \left(1-\mu \right)\alpha }{\mu -\alpha}\left(1,-,\frac{\left(1-\alpha \right)\mu }{r+\mu \left(1-\alpha \right)}\right)=\frac{r\theta \left(1-\mu \right)\alpha }{\left(\mu -\alpha \right)\left(r+\mu \left(1-\alpha \right)\right)}={\lambda}_x\frac{r}{1-\alpha }. $$

1.4 A.4 Proposition 6.3

• 1. Equations (6.31), (6.47) and (6.17) imply:

  $$ \Delta \mathit{\ln}{p}_t={\eta}_q\Delta \mathit{\ln}{q}_t+{\eta}_x\Delta \mathit{\ln}{x}_t={\eta}_q\left(\mu -1\right)\mathit{\ln}{q}_t+{\eta}_x\left(\alpha -1\right)\mathit{\ln}{x}_t+\left({\eta}_q+{\eta}_x\right){\varepsilon}_t={E}_{t-1}\Delta \mathit{\ln}{p}_t+\left({\eta}_q+{\eta}_x\right){\varepsilon}_t=r-{\delta}_{t-1}+\left({\eta}_q+{\eta}_x\right){\varepsilon}_t. $$

From (6.29), (6.16), (6.24):

$$ {\overset{\sim }{\delta}}_t={\lambda}_q\mathit{\ln}{q}_t+{\lambda}_x\mathit{\ln}{x}_t={\lambda}_q\mu ln{q}_{t-1}+{\lambda}_x\alpha ln{x}_{t-1}+\left({\lambda}_q+{\lambda}_x\right){\varepsilon}_t=\mu \left({\lambda}_q\mathit{\ln}{q}_{t-1}+{\lambda}_x\mathit{\ln}{x}_{t-1}\right)+\left(\alpha -\mu \right){\lambda}_x\mathit{\ln}{x}_{t-1}+\left({\lambda}_q+{\lambda}_x\right){\varepsilon}_t=\mu {\overset{\sim }{\delta}}_{t-1}+\left(\alpha -\mu \right){\lambda}_x\mathit{\ln}{x}_{t-1}+\left({\lambda}_q+{\lambda}_x\right){\varepsilon}_t. $$

Consequently,

$$ \Delta {\delta}_t=r\left(1-\mu \right)+\left(\mu -1\right){\delta}_{t-1}+r\left(\alpha -\mu \right){\lambda}_x\mathit{\ln}{x}_{t-1}+r\left({\lambda}_q+{\lambda}_x\right){\varepsilon}_t=\left(1-\mu \right)\left(r-{\delta}_{t-1}\right)+r\left(\alpha -\mu \right){\lambda}_x\mathit{\ln}{x}_{t-1}+r\left({\lambda}_q+{\lambda}_x\right){\varepsilon}_t. $$

• 2. η_q + η_x=

  $$ ={\lambda}_q\frac{r}{1-\mu }+{\lambda}_x\frac{r}{1-\alpha }=\frac{r\theta \left(1-\mu \right)\mu }{\left(\alpha -\mu \right)\left(r+\mu \left(1-\mu \right)\right)}+\frac{r\theta \left(1-\mu \right)\alpha }{\left(\mu -\alpha \right)\left(r+\mu \left(1-\alpha \right)\right)}=\frac{r\theta \left(1-\mu \right)}{\left(\alpha -\mu \right)}\left(\frac{\mu }{r+\mu \left(1-\mu \right)},-,\frac{\alpha }{r+\mu \left(1-\alpha \right)}\right)=\frac{r\theta \left(1-\mu \right)}{\left(\alpha -\mu \right)}\cdot \frac{r\left(\mu -\alpha \right)+\mu \left(\mu \left(1-\alpha \right)-\alpha \left(1-\mu \right)\right)}{\left(r+\mu \left(1-\mu \right)\right)\left(r+\mu \left(1-\alpha \right)\right)}=\frac{r\theta \left(1-\mu \right)}{\left(\alpha -\mu \right)}\cdot \frac{r\left(\mu -\alpha \right)+\mu \left(\mu -\alpha \right)}{\left(r+\mu \left(1-\mu \right)\right)\left(r+\mu \left(1-\alpha \right)\right)}=-\frac{r\theta \left(1-\mu \right)\left(r+\mu \right)}{\left(r+\mu \left(1-\mu \right)\right)\left(r+\mu \left(1-\alpha \right)\right)}<0. $$

1.5 A.5 Equation (6.40)

From (6.26), (6.27), the current availability is

$$ {a}_t={s}_t+{y}_t={a}^{\ast }+\frac{\mu ln{q}_t-\alpha ln{x}_t}{\mu -\alpha }, $$

because a^∗ = s^∗ + 1. Hence, due to (6.16), (6.24):

$$ \Delta {a}_t=\frac{\mu ln{q}_t-\alpha ln{x}_t}{\mu -\alpha }-\frac{\mu ln{q}_{t-1}-\alpha ln{x}_{t-1}}{\mu -\alpha }={\varepsilon}_t-\frac{\left(1-\mu \right)\mu ln{q}_{t-1}-\left(1-\alpha \right)\alpha ln{x}_{t-1}}{\mu -\alpha } $$

$$ ={\varepsilon}_t-\frac{\left(1-\mu \right)\mu ln{q}_{t-1}-\left(1-\mu \right)\alpha ln{x}_{t-1}}{\mu -\alpha }-\frac{\left(1-\mu \right)\alpha -\left(1-\alpha \right)\alpha }{\mu -\alpha}\mathit{\ln}{x}_{t-1}={\varepsilon}_t+\left(1-\mu \right)\left({a}^{\ast }-{a}_{t-1}\right)+\alpha ln{x}_{t-1}=\left(1-\mu \right)\left({a}^{\ast }-{a}_{t-1}\right)+\mathit{\ln}{x}_t. $$