Monotone comparative statics

Let ${\displaystyle X\subseteq \mathbb {R} }$ and let ${\displaystyle f(\cdot ;s):X\rightarrow \mathbb {R} }$ be a family of functions parameterized by ${\displaystyle s\in S}$, where ${\displaystyle (S,\geq _{S})}$ is a partially ordered set (or poset, for short). How does the correspondence ${\displaystyle \arg \max \limits _{x\in X}f(x;s)}$ vary with ${\displaystyle s}$?

Standard comparative statics approach: Assume that set ${\displaystyle X}$ is a compact interval and ${\displaystyle f(\cdot ;s)}$ is a continuously differentiable, strictly quasiconcave function of ${\displaystyle x}$. If ${\displaystyle {\bar {x}}(s)}$ is the unique maximizer of ${\displaystyle f(\cdot ;s)}$, it suffices to show that ${\displaystyle f'({\bar {x}}(s);s')\geq 0}$ for any ${\displaystyle s'>s}$, which guarantees that ${\displaystyle {\bar {x}}(s)}$ is increasing in ${\displaystyle s}$. This guarantees that the optimum has shifted to the right, i.e., ${\displaystyle {\bar {x}}(s')\geq {\bar {x}}(s)}$. This approach makes various assumptions, most notably the quasiconcavity of ${\displaystyle f(\cdot ;s)}$.

While it is clear what it means for a unique optimal solution to be increasing, it is not immediately clear what it means for the correspondence ${\displaystyle \arg \max _{x\in X}f(x;s)}$ to be increasing in ${\displaystyle s}$. The standard definition adopted by the literature is the following.

Definition (strong set order):[1] Let ${\displaystyle Y}$ and ${\displaystyle Y'}$ be subsets of ${\displaystyle \mathbb {R} }$. Set ${\displaystyle Y'}$ dominates ${\displaystyle Y}$ in the strong set order (${\displaystyle Y'\geq _{SSO}Y}$) if for any ${\displaystyle x'}$ in ${\displaystyle Y'}$ and ${\displaystyle x}$ in ${\displaystyle Y}$, we have ${\displaystyle \max\{x',x\}}$ in ${\displaystyle Y'}$ and ${\displaystyle \min\{x',x\}}$ in ${\displaystyle Y}$.

In particular, if ${\displaystyle Y:=\{x\}}$ and ${\displaystyle Y':=\{x'\}}$, then ${\displaystyle Y'\geq _{SSO}Y}$ if and only if ${\displaystyle x'\geq x}$. The correspondence ${\displaystyle \arg \max _{x\in X}f(x;s)}$ is said to be increasing if ${\displaystyle \arg \max _{x\in X}f(x;s')\geq _{SSO}\arg \max _{x\in X}f(x;s)}$ whenever ${\displaystyle s'>_{S}s}$.

The notion of complementarity between exogenous and endogenous variables is formally captured by single crossing differences.

Definition (single crossing function): Let ${\displaystyle \phi :S\rightarrow \mathbb {R} }$. Then ${\displaystyle \phi }$ is a single crossing function if for any ${\displaystyle s'\geq _{S}s}$ we have ${\displaystyle \phi (s)\geq (>)\ 0\ \Rightarrow \ \phi (s')\geq (>)\ 0}$.

Definition (single crossing differences):[2] The family of functions ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$, ${\displaystyle f:X\times S\to \mathbb {R} }$, obey single crossing differences (or satisfy the single crossing property) if for all ${\displaystyle x'\geq x}$, function ${\displaystyle \Delta (s)=f(x';s)-f(x;s)}$ is a single crossing function.

Obviously, an increasing function is a single crossing function and, if ${\displaystyle \Delta (s)}$ is increasing in ${\displaystyle s}$ (in the above definition, for any ${\displaystyle x'>x}$), we say that ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey increasing differences. Unlike increasing differences, single crossing differences is an ordinal property, i.e., if ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey single crossing differences, then so do ${\displaystyle \{g(\cdot ;s)\}_{s\in S}}$, where ${\displaystyle g(x;s)=H(f(x;s);s)}$ for some function ${\displaystyle H(\cdot ;s)}$ that is strictly increasing in ${\displaystyle x}$.

Theorem 1:[3] Define ${\displaystyle F_{Y}(s):=\arg \max _{x\in Y}f(x;s)}$. The family ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey single crossing differences if and only if for all ${\displaystyle Y\subseteq X}$, we have ${\displaystyle F_{Y}(s')\geq _{SSO}F_{Y}(s)}$ for any ${\displaystyle s'\geq _{S}s}$.

Proof: Assume ${\displaystyle s'\geq _{S}s}$ and ${\displaystyle x\in F_{Y}(s)}$, and ${\displaystyle x'\in F_{Y}(s')}$. We have to show that ${\displaystyle \max\{x',x\}\in F_{Y}(s')}$ and ${\displaystyle \min\{x',x\}\in F_{Y}(s)}$. We only need to consider the case where ${\displaystyle x>x'}$. Since ${\displaystyle x\in F_{Y}(s)}$, we obtain ${\displaystyle f(x;s)\geq f(x';s)}$, which guarantees that ${\displaystyle x\in F_{Y'}(s')}$. Furthermore, ${\displaystyle f(x;s)=f(x';s)}$ so that ${\displaystyle x'\in F_{Y}(s)}$. If not, ${\displaystyle f(x;s)>f(x';s)}$ which implies (by single crossing differences) that ${\displaystyle f(x;s')>f(x';s')}$, contradicting the optimality of ${\displaystyle x'}$ at ${\displaystyle s'}$. To show the necessity of single crossing differences, set ${\displaystyle Y:=\{x,{\bar {x}}\}}$, where ${\displaystyle {\bar {x}}\geq x}$. Then ${\displaystyle F_{Y}(s')\geq _{SSO}F_{Y}(s)}$ for any ${\displaystyle s'\geq _{S}s}$ guarantees that, if ${\displaystyle f({\bar {x}};s)\geq (>)\ f(x;s)}$, then ${\displaystyle f({\bar {x}};s')\geq (>)\ f(x;s')}$. Q.E.D.

Application (monopoly output and changes in costs): A monopolist chooses ${\displaystyle x\in X\subseteq \mathbb {R} _{+}}$ to maximise its profit ${\displaystyle \Pi (x;-c)=xP(x)-cx}$, where ${\displaystyle P:\mathbb {R} _{+}\to \mathbb {R} _{+}}$ is the inverse demand function and ${\displaystyle c\geq 0}$ is the constant marginal cost. Note that ${\displaystyle \{\Pi (\cdot ,-c)\}_{(-c)\in \mathbb {R} _{-}}}$ obey single crossing differences. Indeed, take any ${\displaystyle x'\geq x}$ and assume that ${\displaystyle x'P(x')-cx'\geq (>)\ xP(x)-cx}$; for any ${\displaystyle c'}$ such that ${\displaystyle (-c')\geq (-c)}$, we obtain ${\displaystyle x'P(x')-c'x'\geq (>)\ xP(x)-c'x}$. By Theorem 1, the profit-maximizing output decreases as the marginal cost of output increases, i.e., as ${\displaystyle (-c)}$ decreases.

Single crossing differences is not a necessary condition for the optimal solution to be increasing with respect to a parameter. In fact, the condition is necessary only for ${\displaystyle \arg \max _{x\in Y}f(x;s)}$ to be increasing in ${\displaystyle s}$ for any ${\displaystyle Y\subset X}$. Once the sets are restricted to a narrower class of subsets of ${\displaystyle X}$, the single crossing differences condition is no longer necessary.

Definition (Interval):[4] Let ${\displaystyle X\subseteq \mathbb {R} }$. A set ${\displaystyle Y\subseteq X}$ is an interval of ${\displaystyle X}$ if, whenever ${\displaystyle x^{*}}$ and ${\displaystyle x^{**}}$ are in ${\displaystyle Y}$, then any ${\displaystyle x\in X}$ such that ${\displaystyle x^{*}\leq x\leq x^{**}}$ is also in ${\displaystyle Y}$.

For example, if ${\displaystyle X=\mathbb {N} }$, then ${\displaystyle \{1,2,3,4\}}$ is an interval of ${\displaystyle X}$ but not ${\displaystyle \{1,2,4\}}$. Denote ${\displaystyle [x^{*},x^{**}]=\{x\in X\ |\ x^{*}\leq x\leq x^{**}\}}$.

Definition (Interval Dominance Order):[5] The family ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey the interval dominance order (IDO) if for any ${\displaystyle x''>x'}$ and ${\displaystyle s'\geq _{S}s}$, such that ${\displaystyle f(x'';s)\geq f(x;s)}$, for all ${\displaystyle x\in [x',x'']}$, we have ${\displaystyle f(x'';s)\geq (>)\ f(x';s)\ \Rightarrow \ f(x'';s')\geq (>)\ f(x';s')}$.

Like single crossing differences, the interval dominance order (IDO) is an ordinal property. An example of an IDO family is a family of quasiconcave functions ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ where ${\displaystyle \arg \max _{x\in X}f(x,s)}$ increasing in ${\displaystyle s}$. Such a family need not obey single crossing differences.

A function ${\displaystyle f:X\times S\to \mathbb {R} }$ is regular if ${\displaystyle \arg \max _{x\in [x^{*},x^{**}]}f(x;s)}$ is non-empty for any ${\displaystyle x^{**}\geq x^{*}}$, where ${\displaystyle [x^{*},x^{**}]}$ denotes the interval ${\displaystyle \{x\in X\ |\ x^{*}\leq x\leq x^{**}\}}$.

Theorem 2:[6] Denote ${\displaystyle F_{Y}(s):=\arg \max _{x\in Y}f(x;s)}$. A family of regular functions ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obeys the interval dominance order if and only if ${\displaystyle F_{Y}(s)}$ is increasing in ${\displaystyle s}$ for all intervals ${\displaystyle Y\subseteq X}$.

Proof: To show the sufficiency of IDO, take any two ${\displaystyle s'\geq _{S}s}$, and assume that ${\displaystyle x'\in F_{Y}(s)}$ and ${\displaystyle x''\in F_{Y}(s')}$. We only need to consider the case where ${\displaystyle x'>x''}$. By definition ${\displaystyle f(x';s)\geq f(x;s)}$, for all ${\displaystyle x\in [x'',x']\subset Y}$. Moreover, by IDO we have ${\displaystyle f(x';s')\geq f(x'';s')}$. Therefore, ${\displaystyle x'\in F_{Y}(s')}$. Furthermore, it must be that ${\displaystyle f(x';s)=f(x'';s)}$. Otherwise, i.e., if ${\displaystyle f(x';s)>f(x'';s)}$, then by IDO we have ${\displaystyle f(x';s')>f(x'';s')}$, which contradicts that ${\displaystyle x''\in F_{Y}(s')}$. To show the necessity of IDO, assume that there is an interval ${\displaystyle [x'',x']}$ such that ${\displaystyle f(x';s)\geq f(x;s)}$ for all ${\displaystyle x\in [x'',x']}$. This means that ${\displaystyle x'\in \arg \max _{x\in [x'',x']}f(x;s)}$. There are two possible violations of IDO. One possibility is that ${\displaystyle f(x'';s')>f(x';s')}$. In this case, by the regularity of ${\displaystyle f(\cdot ;s')}$, the set ${\displaystyle \arg \max _{x\in [x'',x']}f(x;s')}$ is non-empty but does not contain ${\displaystyle x'}$ which is impossible since ${\displaystyle \arg \max _{x\in [x'',x']}f(x;s)}$ increases in ${\displaystyle s}$. Another possible violation of IDO occurs if ${\displaystyle f(x'';s')=f(x';s')}$ but ${\displaystyle f(x'';s)<f(x';s)}$. In this case, the set ${\displaystyle \arg \max _{x\in [x'',x']}f(x;s')}$ either contains ${\displaystyle x''}$, which is not possible since ${\displaystyle \arg \max _{x\in [x'',x']}f(x;s)}$ increases in ${\displaystyle s}$ (note that in this case ${\displaystyle x''\not \in \arg \max _{x\in [x'',x']}f(x;s')}$) or it does not contain ${\displaystyle x'}$, which also violates monotonicity of ${\displaystyle \arg \max _{x\in [x'',x']}f(x;s)}$. Q.E.D.

The next result gives useful sufficient conditions for single crossing differences and IDO.

Proposition 1:[7] Let ${\displaystyle X}$ be an interval of ${\displaystyle \mathbb {R} }$ and ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ be a family of continuously differentiable functions. (i) If, for any ${\displaystyle s'\geq _{S}s}$, there exists a number ${\displaystyle \alpha >0}$ such that ${\displaystyle f'(x;s')\geq \alpha f'(x;s)}$ for all ${\displaystyle x\in X}$, then ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey single crossing differences. (ii) If, for any ${\displaystyle s'\geq _{S}s}$, there exists a nondecreasing, strictly positive function ${\displaystyle \alpha :X\rightarrow \mathbb {R} }$ such that ${\displaystyle f'(x;s')\geq \alpha (x)f'(x;s)}$ for all ${\displaystyle x\in X}$, then ${\displaystyle \{f(\cdot ;s)\}_{s\in S}}$ obey IDO.

Application (Optimal stopping problem):[8] At each moment in time, agent gains profit of ${\displaystyle \pi (t)}$, which can be positive or negative. If agent decides to stop at time ${\displaystyle x}$, the present value of his accumulated profit is

${\displaystyle V(x;-r)=\int _{0}^{x}e^{-rt}\pi (t)dt,}$

where ${\displaystyle r>0}$ is the discount rate. Since ${\displaystyle V'(x;-r)=e^{-rx}\pi (x)}$, the function ${\displaystyle V}$ has many turning points and they do not vary with the discount rate. We claim that the optimal stopping time is decreasing in ${\displaystyle r}$, i.e., if ${\displaystyle r'>r>0}$ then ${\displaystyle \arg \max _{x\geq 0}V(x;-r)\geq _{SSO}\arg \max _{x\geq 0}V(x;-r')}$. Take any ${\displaystyle r'<r}$. Then, ${\displaystyle V'(x;-r)=e^{-rx}\pi (x)=e^{(r'-r)x}V'(x;-r').}$ Since ${\displaystyle \alpha (x)=e^{(r'-r)x}}$ is positive and increasing, Proposition 1 says that ${\displaystyle \{V(\cdot$ ;-r)\}_{(-r)<0}} obey IDO and, by Theorem 2, the set of optimal stopping times is decreasing.

The above results can be extended to a multi-dimensional setting. Let ${\displaystyle (X,\geq _{X})}$ be a lattice. For any two ${\displaystyle x}$, ${\displaystyle x'}$ in ${\displaystyle X}$, we denote their supremum (or least upper bound, or join) by ${\displaystyle x'\vee x}$ and their infimum (or greatest lower bound, or meet) by ${\displaystyle x'\wedge x}$.

Definition (Strong Set Order):[9] Let ${\displaystyle (X,\geq _{X})}$ be a lattice and ${\displaystyle Y}$, ${\displaystyle Y'}$ be subsets of ${\displaystyle X}$. We say that ${\displaystyle Y'}$ dominates ${\displaystyle Y}$ in the strong set order (${\displaystyle Y'\geq _{SSO}Y}$ ) if for any ${\displaystyle x'}$ in ${\displaystyle Y'}$ and ${\displaystyle x}$ in ${\displaystyle Y}$, we have ${\displaystyle x\vee x'}$ in ${\displaystyle Y'}$ and ${\displaystyle x\wedge x'}$ in ${\displaystyle Y}$.

Examples of the strong set order in higher dimensions.

• Let ${\displaystyle X=\mathbb {R} }$ and ${\displaystyle Y:=[a,b]}$, ${\displaystyle Y':=[a',b']}$ be some closed intervals in ${\displaystyle X}$. Clearly ${\displaystyle (X,\geq )}$, where ${\displaystyle \geq }$ is the standard ordering on ${\displaystyle \mathbb {R} }$, is a lattice. Therefore, as it was shown in the previous section ${\displaystyle Y'\geq _{SSO}Y}$ if and only if ${\displaystyle a'\geq a}$ and ${\displaystyle b'\geq b}$;
• Let ${\displaystyle X=\mathbb {R} ^{n}}$ and ${\displaystyle Y}$, ${\displaystyle Y'\subset X}$ be some hyperrectangles. That is, there exist some vectors ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle a'}$, ${\displaystyle b'}$ in ${\displaystyle X}$ such that ${\displaystyle Y:=\{x\in X\ |\ a\leq x\leq b\}}$ and ${\displaystyle Y':=\{x\in X\ |\ a'\leq x\leq b'\}}$, where ${\displaystyle \geq }$ is the natural, coordinate-wise ordering on ${\displaystyle \mathbb {R} ^{n}}$. Note that ${\displaystyle (X,\geq )}$ is a lattice. Moreover, ${\displaystyle Y'\geq _{SSO}Y}$ if and only if ${\displaystyle a'\geq a}$ and ${\displaystyle b'\geq b}$;
• Let ${\displaystyle (X,\geq _{X})}$ be a space of all probability distributions with support being a subset of ${\displaystyle \mathbb {R} }$, endowed with the first order stochastic dominance order ${\displaystyle \geq _{X}}$. Note that ${\displaystyle (X,\geq _{X})}$ is a lattice. Let ${\displaystyle Y:=\Delta ([a,b])}$, ${\displaystyle Y':=\Delta ([a',b'])}$ denote sets of probability distributions with support ${\displaystyle [a,b]}$ and ${\displaystyle [a',b']}$ respectively. Then, ${\displaystyle Y'\geq _{SSO}Y}$ with respect to ${\displaystyle \geq _{X}}$ if and only if ${\displaystyle a'\geq a}$ and ${\displaystyle b'\geq b}$.

Definition (Quasisupermodular function):[10] Let ${\displaystyle (X,\geq _{X})}$ be a lattice. The function ${\displaystyle f:X\to \mathbb {R} }$ is quasisupermodular (QSM) if

${\displaystyle f(x)\geq (>)\ f(x\wedge x')\ \Rightarrow \ f(x\vee x')\geq (>)\ f(x').}$

The function ${\displaystyle f}$ is said to be a supermodular function if ${\displaystyle f(x\vee x')-f(x')\geq f(x)-f(x\wedge x').}$ Every supermodular function is quasisupermodular. As in the case of single crossing differences, and unlike supermodularity, quasisupermodularity is an ordinal property. That is, if function ${\displaystyle f}$ is quasisupermodular, then so is function ${\displaystyle g:=H\circ f}$, where ${\displaystyle H}$ is some strictly increasing function.

Theorem 3:[11] Let ${\displaystyle (X,\geq _{X})}$ is a lattice, ${\displaystyle (S,\geq _{S})}$ a partially ordered set, and ${\displaystyle Y}$, ${\displaystyle Y'}$ subsets of ${\displaystyle X}$. Given ${\displaystyle f:X\times S\to \mathbb {R} }$, we denote ${\displaystyle \arg \max _{x\in Y}f(x;s)}$ by ${\displaystyle F_{Y}(s)}$. Then ${\displaystyle F_{Y'}(s')\geq _{SSO}F_{Y}(s)}$ for any ${\displaystyle s'\geq _{S}s}$ and ${\displaystyle Y'\geq _{SSO}Y}$

Proof: ${\displaystyle (\Leftarrow )}$. Let ${\displaystyle Y'\geq _{SSO}Y}$, ${\displaystyle s'\geq _{S}s}$, and ${\displaystyle x'\in F_{Y'}(s')}$, ${\displaystyle x\in F_{Y}(s)}$. Since ${\displaystyle x\in F_{Y}(s)}$ and ${\displaystyle Y'\geq _{SSO}Y}$, then ${\displaystyle f(x;s)\geq f(x'\wedge x;s)}$. By quasisupermodularity, ${\displaystyle f(x'\vee x;s)\geq f(x';s)}$, and by the single crossing differences, ${\displaystyle f(x'\vee x;s')\geq f(x';s')}$. Hence ${\displaystyle x'\vee x\in F_{Y'}(s')}$. Now assume that ${\displaystyle x'\wedge x\not \in F_{Y}(s)}$. Then ${\displaystyle f(x;s)>f(x'\wedge x;s)}$. By quasisupermodularity, ${\displaystyle f(x'\vee x;s)>f(x';s)}$, and by single crossing differences ${\displaystyle f(x'\vee x;s')>f(x';s')}$. But this contradicts that ${\displaystyle x'\in F_{Y'}(s')}$. Hence, ${\displaystyle x'\wedge x\in F_{Y}(s)}$.

${\displaystyle (\Rightarrow )}$. Set ${\displaystyle Y':=\{x',x'\vee x\}}$ and ${\displaystyle Y:=\{x,x'\wedge x\}}$. Then, ${\displaystyle Y'\geq _{SSO}Y}$ and thus ${\displaystyle F_{Y'}(s)\geq _{SSO}F_{Y}(s)}$, which guarantees that, if ${\displaystyle f(x;s)\geq (>)\ f(x'\wedge x;s)}$, then ${\displaystyle f(x'\vee x;s)\geq (>)\ f(x';s)}$. To show that single crossing differences also hold, set ${\displaystyle Y:=\{x,{\bar {x}}\}}$, where ${\displaystyle {\bar {x}}\geq x}$. Then ${\displaystyle F_{Y}(s')\geq _{SSO}F_{Y}(s)}$ for any ${\displaystyle s'\geq _{S}s}$ guarantees that, if ${\displaystyle f({\bar {x}};s)\geq (>)\ f(x;s)}$, then ${\displaystyle f({\bar {x}};s')\geq (>)\ f(x;s')}$. Q.E.D.

Application (Production with multiple goods):[12] Let ${\displaystyle x}$ denote the vector of inputs (drawn from a sublattice ${\displaystyle X}$ of ${\displaystyle \mathbb {R} _{+}^{l}}$) of a profit-maximizing firm, ${\displaystyle p\in \mathbb {R} _{++}^{l}}$ be the vector of input prices, and ${\displaystyle V}$ the revenue function mapping input vector ${\displaystyle x}$ to revenue (in ${\displaystyle \mathbb {R} }$). The firm's profit is ${\displaystyle \Pi (x;p)=V(x)-p\cdot x}$. For any ${\displaystyle x'}$, ${\displaystyle x\in X}$, ${\displaystyle x'\geq x}$, ${\displaystyle V(x')-V(x)+(-p)(x'-x)}$ is increasing in ${\displaystyle (-p)}$. Hence, ${\displaystyle \{\Pi (\cdot ;p)\}_{p\in \mathbb {R} _{++}^{l}}}$ has increasing differences (and so it obeys single crossing differences). Moreover, if ${\displaystyle V}$ is supermodular, then so is ${\displaystyle \Pi (\cdot ;p)}$. Therefore, it is quasisupermodular and by Theorem 3, ${\displaystyle \arg \max _{x\in X}\Pi (x;p)\geq _{SSO}\arg \max _{x\in X}\Pi (x;p')}$ for ${\displaystyle p'\geq p}$.