Take a function $\Phi(x,y): \mathbb{R}^K\times \mathbb{R}^K\rightarrow \mathbb{R}$. Here the arguments of $\Phi$ are two $K\times 1$ vectors, $x$ and $y$.
Consider the cross-partial derivative of $\Phi(x,y)$ with respect $x$ and $y$ and evaluate it at $(x,y)=(a,b)$ (assume that such derivative exists). We denote it by
$$
\frac{\partial \Phi(a,b)}{\partial x \partial y}
$$
I'm looking for the analogue of this derivative when $x$ and $y$ can only take integer values.
My hypothesis is that the analogue of the derivative above could be
$$
[1] \hspace{1cm}\frac{\partial \Phi(a,b)}{\partial x \partial y} \approx \Phi(a+1,b+1)+\Phi(a,b)-\Phi(a+1,b)-\Phi(a,b+1)
$$
Is this correct?
Also, what would be the interpretation of the result? I.e., suppose that algebraic sum gives me a positive value: what does that mean?
Following a comment below, the concept of finite differences here is helpful. In particular
$$
[2] \hspace{1cm} \frac{\partial \Phi(a,b)}{\partial x \partial y} \approx \frac{\Phi(a+1,b+1)+\Phi(a-1,b-1)-\Phi(a-1,b+1)-\Phi(a+1,b-1)}{4}
$$
I this correct? I am particularly interested in the sign of the numerator. Could you help me to interpret this sign?
Best Answer
To make my comment more precise, the expression you have wrote above would be the multivariable forward finite difference.
As for interpretation, I guess you could just say that it's an approximation of the ordinary derivative of $\Phi$ at $(a,b)$ if $\Phi$ were defined on $\mathbb{R}^2$; you can't say much more about the accuracy about the approximation unless you know more about $\Phi$. It might help if you give more context for the problem.