<p>As @Herr K. stated, the beginning point is being able to take a derivative. The Hessian is a matrix equivalent to a second order derivative sometimes denoted as <span class="math-container">$\nabla^{2}$</span>. Start by finding the gradient, <span class="math-container">$\nabla$</span> which is a vector of first order derivatives of every <em>variable</em> in the function. In your case, this would be<br>
<span class="math-container">$$ \nabla=
  \left[ {\begin{array}{cc}
   \frac{\partial{f}}{\partial{K}}  \\
   \frac{\partial{f}}{\partial{L}} \\
  \end{array} } \right] =
  \left[ {\begin{array}{cc}
   aAK^{a-1}L^{b}  \\
   bAK^{a}L^{b-1} \\
  \end{array} } \right] $$</span>
The Hessian, would be the next (second) derivative of each of the above with respect to both variables and it takes the form:
<span class="math-container">$$ \nabla^{2}=
  \left[ {\begin{array}{cc}
   \frac{\partial^{2}f}{\partial{K}^2} &amp; \frac{\partial^{2}f}{\partial{K}\partial{L}} \\
   \frac{\partial^{2}f}{\partial{L}\partial{K}} &amp; \frac{\partial^{2}f}{\partial{L}^2} \\
  \end{array} } \right] $$</span>
Using whatever this would be in your case (left to you to compute) you can take the <em>determinant</em> of the resulting matrix and, with values for your parameters <span class="math-container">$a$</span> and <span class="math-container">$b$</span> (or at least their signs like with the other variables and parameters) you can determine whether the <em>determinant</em> is positive or negative and determine the concavity. If it is positive definite or positive semi-definite, this would imply either strict or non-strict convexity, respectively. If you don't know how to take a determinant it is simply dont by using a matrix say</p>

<p><span class="math-container">$$ A=
  \left[ {\begin{array}{cc}
   a_{1} &amp; a_{2} \\
   a_{3} &amp; a_{4} \\
  \end{array} } \right] $$</span>
and then taking its determinant using the operation (without explanation for sake of brevity):
<span class="math-container">$$ A=
   {\begin{vmatrix}
   a_{1} &amp; a_{2} \\
   a_{3} &amp; a_{4} \\
  \end{vmatrix} } = a_{1}a_{4}-a_{2}a_{3} $$</span>
Using this with the above, you should be able to figure the rest out and figure out whether or not its concave! (hint: for <span class="math-container">$a,b\geq 0$</span> and <span class="math-container">$a+b\leq1$</span> it is concave)</p>

CC BY-SA 4.0
2019-08-16T16:21:54.117Z
" for a,b≤0 it is concave" This is wrong. a ≥ 0, b ≥ 0, and a + b ≤ 1 are required for the function to be concave on the stated domain( "the set of points $(K,L)$ with $K≥0$ and $L≥0$")
user18214
CC BY-SA 4.0
2019-08-16T21:43:18.990Z
Oh oops my mistake, thank you for pointing that out.
Brennan
CC BY-SA 4.0
2019-08-15T18:18:14.560Z
2019-08-16T21:44:03.043Z
2019-08-16T21:44:03.043Z
Brennan
<p>The function is <span class="math-container">$f(K,L)= AK^{a}L^{b}$</span> on the set of points <span class="math-container">$(K,L)$</span> with <span class="math-container">$K\geq 0$</span> and <span class="math-container">$L\geq 0$</span>, assuming <span class="math-container">$A&gt;0$</span>
How do I find the Hessian and check for concavity?</p>

CC BY-SA 4.0
2019-08-15T16:26:15.243Z
Do you know how to take a derivative?
Herr K.
<mathematical-economics>
How do I derive Hessian of this function and check for concavity?
Grangerluck
# How do I derive Hessian of this function and check for concavity?

(Asked by: username_0 on 2019-08-15)

The function is $f(K,L)= AK^{a}L^{b}$ on the set of points $(K,L)$ with $K\geq 0$ and $L\geq 0$, assuming $A>0$
How do I find the Hessian and check for concavity?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: Do you know how to take a derivative?  

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## 1 answers

### Answer by username_2 on 2019-08-15. Score: 7

As @Herr K. stated, the beginning point is being able to take a derivative. The Hessian is a matrix equivalent to a second order derivative sometimes denoted as $\nabla^{2}$. Start by finding the gradient, $\nabla$ which is a vector of first order derivatives of every *variable* in the function. In your case, this would be  
$$ \nabla=
\left[ {\begin{array}{cc}
\frac{\partial{f}}{\partial{K}} \\
\frac{\partial{f}}{\partial{L}} \\
\end{array} } \right] =
\left[ {\begin{array}{cc}
aAK^{a-1}L^{b} \\
bAK^{a}L^{b-1} \\
\end{array} } \right] $$
The Hessian, would be the next (second) derivative of each of the above with respect to both variables and it takes the form:
$$ \nabla^{2}=
\left[ {\begin{array}{cc}
\frac{\partial^{2}f}{\partial{K}^2} & \frac{\partial^{2}f}{\partial{K}\partial{L}} \\
\frac{\partial^{2}f}{\partial{L}\partial{K}} & \frac{\partial^{2}f}{\partial{L}^2} \\
\end{array} } \right] $$
Using whatever this would be in your case (left to you to compute) you can take the *determinant* of the resulting matrix and, with values for your parameters $a$ and $b$ (or at least their signs like with the other variables and parameters) you can determine whether the *determinant* is positive or negative and determine the concavity. If it is positive definite or positive semi-definite, this would imply either strict or non-strict convexity, respectively. If you don't know how to take a determinant it is simply dont by using a matrix say

$$ A=
\left[ {\begin{array}{cc}
a\_{1} & a\_{2} \\
a\_{3} & a\_{4} \\
\end{array} } \right] $$
and then taking its determinant using the operation (without explanation for sake of brevity):
$$ A=
{\begin{vmatrix}
a\_{1} & a\_{2} \\
a\_{3} & a\_{4} \\
\end{vmatrix} } = a\_{1}a\_{4}-a\_{2}a\_{3} $$
Using this with the above, you should be able to figure the rest out and figure out whether or not its concave! (hint: for $a,b\geq 0$ and $a+b\leq1$ it is concave)

> **Comments:**
> **username_3**: " for a,b≤0 it is concave" This is wrong. a ≥ 0, b ≥ 0, and a + b ≤ 1 are required for the function to be concave on the stated domain( "the set of points $(K,L)$ with $K≥0$ and $L≥0$")  
> **username_2**: Oh oops my mistake, thank you for pointing that out.  

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