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I am trying to derive the differentiation of joint copula from this paper http://www.nicksun.fun/assets/ms_references/madsen2009.pdf, which is done in equation (4.3). To summarize

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I fail to understand why the derivative of $z'\Sigma^{-1}z$ is $z'(\Sigma^{-1} - I_n)z$? Am I missing something?

Here $z = [ \Phi^{-1} (F_1(y_1)), \ldots, \Phi^{-1} (F_n(y_n))]$, and $F_i$ are the marginal distributions. I understand that the product at the end is a result of chain rule, but I fail to understand the expression inside the exponent.

Hirak Sarkar
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1 Answers1

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I fail to understand why the derivative of $z'\Sigma^{-1}z$ is $z'(\Sigma^{-1} - I_n)z$? Am I missing something?

Yes, the second term is not the derivative of the first one. The term $z' I_n z$ stems from the chain rule. More precisely, if we differentiate $C(y; \Sigma)$ w.r.t. $y$ then the first application of the chain rule yields $$ \frac{\partial C(y; \Sigma)}{\partial y} = \frac{1}{\sqrt{(2\pi)^n}} \vert \Sigma \vert^{-1/2} \exp \left( - \frac 12 z' \Sigma^{-1} z \right) \frac{\partial z}{\partial y} $$ Now for $\partial z / \partial y$ we apply the chain rule once again. For the derivative of $\Phi^{-1}$ we use the formula for the derivative of the inverse, namely $$ (f^{-1})' (x) = \frac{1}{f' ( f^{-1} (x) )} . $$ (here the prime ' denotes derivative.) For $\Phi^{-1}$ the function $\Phi'$ is the density of the standard normal distribution. Combining this with the fact that the derivative of $F_i$ is $f_i$ we obtain $$ \frac{\partial z}{\partial y} = \sqrt{(2\pi)^n} \exp \left( \frac 12 z' I_n z \right) \prod_{i=1}^n f_i(y_i) , $$ where each of $n$ applications of the partial derivative yields one density of the standard normal distribution (and they can be combined via this matrix notation using $I_n$). Plugging this into the above equation gives the formula (4.3).

picky_porpoise
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