MHB What is the Hessian method for determining concavity/convexity?

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The Hessian method for determining the concavity or convexity of a function involves calculating the Hessian matrix, which is the square matrix of second-order partial derivatives. To assess concavity, one examines the eigenvalues of the Hessian; if all eigenvalues are positive, the function is convex, while if all are negative, it is concave. The diagonal elements of the Hessian can provide insight, but the eigenvalues offer a definitive conclusion. This method is essential in optimization and economic modeling. Understanding the Hessian is crucial for analyzing the behavior of functions in various applications.
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Hello Everyone!

I'm trying to remember a quick method for determining whether a function is concave or convex. There was something that involved finding the Hessian of the function, and then looking at the diagonal elements, then, I completely forgot...

What's the rest of this method, I don't remember I even had to find the eigen values...

Thanks!
 
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OhMyMarkov said:
Hello Everyone!

I'm trying to remember a quick method for determining whether a function is concave or convex. There was something that involved finding the Hessian of the function, and then looking at the diagonal elements, then, I completely forgot...

What's the rest of this method, I don't remember I even had to find the eigen values...

Thanks!

Hi OhMyMarkov, :)

The method of using the Hessian of a function to determine the concavity/convexity is described >>here<<.

Kind Regards,
Sudharaka.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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