<p>I don't think this proposition holds without assuming that $g$ is also increasing.
Take $f(x_1,x_2)=x_1x_2$ (actually any supermodular function) and $g(z)=-z$ (decreasing and convex). Then, $g(f(x_1,x_2))=-x_1x_2$ which is neither increasing in its arguments or supermodular.</p>

<p>If you assume also that $g$ is increasing, you get $g'(f(x))\geq 0$ straightforwardly.</p>

<p>You can easily show supermodularity using the conddition on the second derivatives. Denote $x=(x_1,...,x_n)$. Supermodularity is equivalent for function f to $$\frac{\partial^2 f}{\partial x_i \partial x_j}(x)\geq 0$$ for any $i\ne j$.
If you twice-differentiate $g \otimes f$ ($\otimes$ is the compound operator),
$$
\frac{\partial^2 g \otimes f}{\partial x_i \partial x_j}(x)=\frac{\partial^2 f}{\partial x_i \partial x_j}(x) . g'(f(x)) + \frac{\partial f}{\partial x_i}(x) .\frac{\partial f}{\partial x_j}(x) .g''(f(x))
$$
You can check that $\frac{\partial^2 g \otimes f}{\partial x_i \partial x_j}(x)\geq 0$ so that $g \otimes f$ is supermodular since $f$ is supermodular, $g$ is increasing and $g$ is convex.</p>

CC BY-SA 3.0
2016-06-03T12:31:24.880Z
2016-06-03T12:31:24.880Z
GuiWil
<blockquote>
  <p>If $f:\mathbb{R}^n \to \mathbb{R} $ is twice continuously differentiable, has a non-negative gradient, and is supermodular, and $g: \mathbb{R} \to \mathbb{R}$ is twice continuously differentiable and convex, then $g(f(x))$ is increasing and supermodular</p>
</blockquote>

<p>To show increasing I think we can just show $\displaystyle \frac{d}{dx} g(f(x)) &gt; 0$</p>

<p>$$ \displaystyle \frac{d}{dx} g(f(x) = g'(f(x)) f'(x) $$</p>

<p>Since $f(x)$ has a non-negative gradient $f'(x) \geq 0$, but I'm not sure how to show $g'(f(x)) &gt; 0$ as given convex tells us $g''(x) &gt; 0$</p>

<p>For supermodularity, let $l, l' \in \mathbb{R}^n$; $g(f(x))$ will be supermodular if the following inequality holds:</p>

<p>$$ \displaystyle (g \circ f)(l \vee l') + (g \circ f)(l \wedge l') \geq (g \circ f)(l) + (g \circ f)(l') $$</p>

<p>I think if we had $l_i \leq l_i', \; \forall i$, then we will have equality as $(g \circ f)(l \vee l') = (g \circ f)(l') $ and $ (g \circ f)(l \wedge l') = (g \circ f)(l)$, but don't know what to do for the other cases.</p>

<mathematical-economics>
Show a compound function is increasing and supermodular given its components
Sunhwa
# Show a compound function is increasing and supermodular given its components

(Asked by: username_0 on 2016-06-02)

> If $f:\mathbb{R}^n \to \mathbb{R} $ is twice continuously differentiable, has a non-negative gradient, and is supermodular, and $g: \mathbb{R} \to \mathbb{R}$ is twice continuously differentiable and convex, then $g(f(x))$ is increasing and supermodular

To show increasing I think we can just show $\displaystyle \frac{d}{dx} g(f(x)) > 0$

$$ \displaystyle \frac{d}{dx} g(f(x) = g'(f(x)) f'(x) $$

Since $f(x)$ has a non-negative gradient $f'(x) \geq 0$, but I'm not sure how to show $g'(f(x)) > 0$ as given convex tells us $g''(x) > 0$

For supermodularity, let $l, l' \in \mathbb{R}^n$; $g(f(x))$ will be supermodular if the following inequality holds:

$$ \displaystyle (g \circ f)(l \vee l') + (g \circ f)(l \wedge l') \geq (g \circ f)(l) + (g \circ f)(l') $$

I think if we had $l\_i \leq l\_i', \; \forall i$, then we will have equality as $(g \circ f)(l \vee l') = (g \circ f)(l') $ and $ (g \circ f)(l \wedge l') = (g \circ f)(l)$, but don't know what to do for the other cases.

**Tags:** mathematical-economics

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## 1 answers

### Answer by username_1 on 2016-06-03. Score: 1 | Accepted answer

I don't think this proposition holds without assuming that $g$ is also increasing.
Take $f(x\_1,x\_2)=x\_1x\_2$ (actually any supermodular function) and $g(z)=-z$ (decreasing and convex). Then, $g(f(x\_1,x\_2))=-x\_1x\_2$ which is neither increasing in its arguments or supermodular.

If you assume also that $g$ is increasing, you get $g'(f(x))\geq 0$ straightforwardly.

You can easily show supermodularity using the conddition on the second derivatives. Denote $x=(x\_1,...,x\_n)$. Supermodularity is equivalent for function f to $$\frac{\partial^2 f}{\partial x\_i \partial x\_j}(x)\geq 0$$ for any $i\ne j$.
If you twice-differentiate $g \otimes f$ ($\otimes$ is the compound operator),
$$
\frac{\partial^2 g \otimes f}{\partial x\_i \partial x\_j}(x)=\frac{\partial^2 f}{\partial x\_i \partial x\_j}(x) . g'(f(x)) + \frac{\partial f}{\partial x\_i}(x) .\frac{\partial f}{\partial x\_j}(x) .g''(f(x))
$$
You can check that $\frac{\partial^2 g \otimes f}{\partial x\_i \partial x\_j}(x)\geq 0$ so that $g \otimes f$ is supermodular since $f$ is supermodular, $g$ is increasing and $g$ is convex.


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