Optimal allocation of a fixed production under price uncertainty

Abstract

In this paper, we consider a model of production allocation in the context of the theory of the firm under uncertainty. This is the case of a firm that has just produced a known amount of an output and can allocate it to two possible ends: one with a certain price, the other with an uncertain price. We first establish conditions to determine whether the firm will make use of both ends or of only one of them. In particular, we find a limit value for the certain price (which we call the frontier price) below which the firm decides to allocate the total amount of production to the uncertain end. We then study comparative-static effects on the optimal output allocated to each end, and also on the frontier price. Finally, we analyze an application concerning the middleman who buys the firm’s output in the certain end. This is a pricing problem: namely obtaining the price in the certain end that the middleman must offer to the producer in order to attain a desired amount of output. In two specific cases, we also provide closed-form expressions for the optimal allocation to both ends and for the frontier price.

Appendices

Appendix 1

We collect some auxiliary lemmas in this appendix. The first is taken from Lippman and McCall (1982) (a partial result in their Theorem $2$ on p. 252). As the proof is simple, it is given here. Lemma 4 is a well-known result, at least in its non-strict version. We write here a simple proof for the strict case, which is the one needed.

Lemma 1

Let $\psi$ and $\phi$ be two real functions defined on $\mathbb {R}$ such that $\psi >0$ and $\phi$ is increasing. If $\xi =\psi \cdot \phi$, and $X$ is a non-degenerate real random variable such that the expectation $E\!\left[ X\, \psi (X) \right]$ is finite, then:

$$\begin{aligned} E\!\left[ X\, \xi (X) \right] \geqslant \phi (0)\, E\!\left[ X \,\psi (X) \right] , \end{aligned}$$

and the reverse inequality holds when $\phi$ is decreasing. In addition, if $\phi$ is strictly increasing or strictly decreasing, the corresponding inequalities also hold strictly.

Proof

When $t<0$, we can write: $\phi (t) \leqslant \phi (0)$ (since $\phi$ is increasing), and hence:

$$\begin{aligned} t \,\phi (t)\, \psi (t) \geqslant t\, \phi (0)\, \psi (t) \end{aligned}$$

(recalling that $\psi >0$). However, this inequality also holds when $t>0$. The result follows after a straightforward calculation. $\square$

Lemma 2

Define: $T(q,s)\equiv \mathop {\mathsf{{E} }}\nolimits \bigl [ u''\bigl ( sq + P(q_T-q) + B \bigr )\,(P-s) \bigr ]$. Depending on whether the firm exhibits DARA, CARA or IARA, the number $T(q,s)$ is greater than or equal to $0$, null, or smaller than or equal to $0$, respectively, in any of the following cases: $(q,s)=(q^*,p)$ and $(q,s)=(0,\alpha )$.

Proof

We have: $T(q^*,p) = \mathop {\mathsf{{E} }}\nolimits \bigl [ u''\bigl ( W(q^*) \bigr )\,(P-p) \bigr ] = \mathop {\mathsf{{E} }}\nolimits \bigl [ r_u\bigl ( W(q^*) \bigr ) \, u'\bigl ( W(q^*) \bigr )\,(p-P) \bigr ]$. Assume that the firm exhibits DARA. Set $\psi (t) = u'\bigl ( t(q^*-q_T) + p q_T + B \bigr )$ and $\phi (t) = r_u \bigl ( t(q^*-q_T) + p q_T + B \bigr )$; thus $\psi >0$, and $\phi$ is increasing. Taking $X=p-P$, with Lemma 1 we obtain:

$$\begin{aligned} T(q^*,p) \geqslant \phi (0)\, \mathop {\mathsf{{E} }}\nolimits \bigl [ u'\bigl ( W(q^*) \bigr )\,(p-P) \bigr ] = 0, \end{aligned}$$

according to the first-order condition. The proof for CARA or IARA is, mutatis mutandis, the same. And for the case $(q,s)=(0,\alpha )$, proceeding with the same functions $\psi$ and $\phi$, and the same random variable $X$, but writing $0$ and $\alpha$ instead of $q^*$ and $p$, respectively, we obtain:

$$\begin{aligned} T(0,\alpha ) \geqslant \phi (0)\, \mathop {\mathsf{{E} }}\nolimits \bigl [ u'\bigl ( W(0) \bigr )\,(\alpha -P) \bigr ] = 0, \end{aligned}$$

where the last factor is null according to the definition of $\alpha$. $\square$

Lemma 3

If $R_u<1$, then:

$$\begin{aligned} \mathop {\mathsf{{E} }}\nolimits [ u'(W) ] + (q_T-q) \mathop {\mathsf{{E} }}\nolimits \bigl [ u''(W) \, (P-p) \bigr ] > 0 \end{aligned}$$

and

$$\begin{aligned} \mathop {\mathsf{{E} }}\nolimits [ u'(W) ] - q \mathop {\mathsf{{E} }}\nolimits \bigl [ u''(W) \, (P-p) \bigr ] > 0 \end{aligned}$$

for all $q$.

Proof

As $R_u<1$, we have: $u'(W) + W \, u''(W) > 0$. If we write $W=W(q) = (q_T-q)(P-p)+W(q_T)$ and consider that the wealth $W$ is non-negative for every level of output, we obtain:

$$\begin{aligned} 0 < u'(W) + W \, u''(W)&= u'(W) + (q_T-q) u''(W)(P-p) + W(q_T) u''(W) \\&\quad \leqslant u'(W) + (q_T-q) u''(W)(P-p), \end{aligned}$$

and by taking expectations we conclude the result. In order to prove the second inequality, we simply write $W(q) = W(0) - q(P-p)$ and proceed in the same way.$\square$

Lemma 4

Let X and Y be two random variables, defined on an interval $\left[ 0,M\right]$, with the same finite expectation. Assume that $\int _0^t \bigl ( G(s) - F(s) \bigr ) \,ds \geqslant 0$ for all $t\in \left[ 0,M\right]$, with strict inequality for some $t_0$, where $F$ and $G$ are the distribution functions of $X$ and $Y$, respectively. If $v$ is a real function of class $\fancyscript{C}^2$ that is strictly concave on the interval $\left[ 0,M\right]$, then $\mathop {\mathsf{{E} }}\nolimits [v(X)] > \mathop {\mathsf{{E} }}\nolimits [v(Y)]$, and the reverse inequality holds if $v$ is strictly convex.

Proof

Set $H = F-G$, and $\tilde{H}(t) = \int _0^t H(s)\,ds$ for $0\leqslant t\leqslant M$. Thus $H(0)=\tilde{H}(0) =H(M)=0$, and also $\tilde{H}(M)=0$ since $\mathop {\mathsf{{E} }}\nolimits [X]=\mathop {\mathsf{{E} }}\nolimits [Y]$.¹² Therefore we can write:

$$\begin{aligned} \mathop {\mathsf{{E} }}\nolimits [v(X)] - \mathop {\mathsf{{E} }}\nolimits [v(Y)] = \int _0^M v(t) \, dH(t) = - \int _0^M v'(t)\, H(t)\, dt = \int _0^M v''(t) \, \tilde{H}(t)\, dt. \end{aligned}$$

For each $t,\,v''(t)\leqslant 0$ (due to the concavity of $v$) and $\tilde{H}(t)\leqslant 0$ (by hypothesis), and thus this last integral is positive or null. However, $\tilde{H}(t_0)<0$ and the function $\tilde{H}$ is continuous, and hence this function is strictly negative on some open interval $J$ around $t_0$. In addition, since the function $v$ is strictly concave, in any open subinterval of $\left[ 0,M\right]$ there exists a point where $v''$ is negative. The continuity of $v''$ lets us assert that $v''<0$ on some open interval included in $J$. With this observation, we can conclude that the integral is in fact strictly positive.$\square$

Appendix 2

In this appendix, for the two special cases analyzed in Sect. 3.5, we list the derivatives of $q^*$ and $\alpha$ with respect to the various parameters in order to verify the results in Propositions 3–8.

CARA utility and normal distribution of the random variable $P$ We have:

$$\begin{aligned}&\displaystyle \frac{dq^*}{dp} = \frac{1}{\sigma ^2 k} > 0; \qquad \frac{dq^*}{dk} = \frac{\mu -p}{\sigma ^2 k^2} > 0 \quad \text {and}\quad \frac{d\alpha }{dk} = -\sigma ^2 q_T < 0; \\&\displaystyle \quad \frac{dq^*}{d\mu } = -\frac{1}{\sigma ^2 k} < 0 \quad \text {and}\quad \frac{d\alpha }{d\mu } = 1; \\&\displaystyle \quad \frac{dq^*}{dq_T} = 1 > 0 \quad \text {and}\quad \frac{d\alpha }{dq_T} = -\sigma ^2 k<0; \qquad \frac{dq^*}{dB} = \frac{d\alpha }{dB} = 0; \\&\displaystyle \quad \frac{dq^*}{d\sigma } = \frac{2(\mu -p)}{\sigma ^3 k} > 0 \quad \text {and}\quad \frac{d\alpha }{d\sigma } = -2\sigma k q_T<0. \end{aligned}$$

This confirms the assertions in Propositions 3–8.

Quadratic utility In this paragraph, we write $D\equiv (\mu -p)^2+ \sigma ^2$. For the derivative $dq^*/dp$, we obtain:

$$\begin{aligned} \frac{dq^*}{dp} = \frac{ \bigl [ 2b\, (\mu -p) q_T + u'\bigl ( W(q_T) \bigr ) \bigr ] D - 2\, u'\bigl ( W(q_T) \bigr ) (\mu -p)^2 }{ 2b\,D^2 }, \end{aligned}$$

and considering that $u'\bigl ( W(q_T) \bigr ) = 2b\, D(q_T-q^*)/(\mu -p)$:

$$\begin{aligned} \frac{dq^*}{dp} = \frac{ (\mu -p)^2 q^* + \sigma ^2 (q_T-q^*) }{ (\mu -p) \,D } > 0, \end{aligned}$$

which confirms the result in Proposition 3. In order to verify Proposition 4, we highlight that an increase in the absolute risk aversion $r_u$, the utility remaining of the quadratic type, is equivalent to a decrease in the quotient $\gamma \equiv a/b$, as a result of the equality $r_u(t)=2/(\gamma -2t)$. Therefore:

$$\begin{aligned} \frac{dq^*}{d\gamma } = -\frac{\mu -p}{2D} <0 \quad \text {and}\quad \frac{d\alpha }{d\gamma } = \frac{ 2 \sigma ^2 q_T }{ [\gamma -2(\mu q_T+B)]^2 } >0, \end{aligned}$$

which is in accordance with the cited proposition. Finally, by taking into account that $r'_u=r^2_u$, we also have:

$$\begin{aligned}&\displaystyle \frac{d\alpha }{d\mu } = 1 - \sigma ^2 \, r^2_u(\mu q_T + B)\, q_T^2 < 1; \\&\displaystyle \frac{dq^*}{dq_T} \!=\! 1 + \frac{ p(\mu -p) }{ D } \!>\! 0 \quad \text {and}\quad \frac{d\alpha }{dq_T} \!=\! -\sigma ^2 \, r_u(\mu q_T + B)\, [ \mu \, r_u(\mu q_T + B) \, q_T + 1] \!<\! 0; \\&\displaystyle \frac{dq^*}{dB} = \frac{\mu -p}{ D } > 0 \quad \text {and}\quad \frac{d\alpha }{dB} = -\sigma ^2 \, r^2_u(\mu q_T + B)\, q_T < 0; \\&\displaystyle \frac{dq^*}{d\sigma } = \frac{ 2 \sigma (\mu -p) }{ D^2\, r_u(p q_T + B) } > 0 \quad \text {and}\quad \frac{d\alpha }{d\sigma } = -2\sigma \, r_u(\mu q_T + B)\, q_T < 0. \end{aligned}$$

This enables us to verify the results in Propositions 5–8.