# The second derivative of an integration of multivariate normal with matrix form

I have a function with the form like this, $F(\beta) = ln (\int_{X\beta}^{\infty} \phi(u|0,\Sigma) du)$. Here $X, \Sigma$ are matrices. $\beta$ and $u$ are vectors. $\phi$ is the multivariate normal density function.

I need to get the second derivative of this function.

I'm poor in doing matrix calculus. But, I derive the result for the first derivative as follows,

$\frac{\partial F(\beta)}{\partial \beta} = \frac{1}{\int_{X\beta}^{\infty} \phi(u|0,\Sigma) du} \frac{\partial \phi(u|0,\Sigma)}{\partial \beta} =\frac{1}{\int_{X\beta}^{\infty} \phi(u|0,\Sigma) du} \frac{\partial \sqrt{(2\pi)^k |\Sigma|}exp(-\frac{1}{2}(X\beta)^T\Sigma^{-1}X\beta)}{\partial \beta}= \frac{1}{\int_{X\beta}^{\infty} \phi(u|0,\Sigma) du} \sqrt{(2\pi)^k |\Sigma|}exp(-\frac{1}{2}(X\beta)^T\Sigma^{-1}X\beta) \frac{\partial -\frac{1}{2} (X\beta)^T\Sigma^{-1}X\beta}{\partial \beta}= \frac{1}{\int_{X\beta}^{\infty} \phi(u|0,\Sigma) du} \sqrt{(2\pi)^k |\Sigma|}exp(-\frac{1}{2}(X\beta)^T\Sigma^{-1}X\beta) (-\frac{1}{2}X^T\Sigma^{-1}X\beta-\frac{1}{2}(X\beta)^T\Sigma^{-1}X)$

Is my first derivative correct? Can anyone show me the result for the second derivative?

## 1 Answer

Restating the problem \eqalign{ y &= X\beta \cr \phi(y) &= \phi_o\,\exp(-\tfrac{y^T\Sigma^{-1}y}{2}) \cr p = \exp(F) &= \int_y^\infty\phi(u)\,du \cr } Since the variable $y$ only appears in the lower limit of integration, and not explicitly in the kernel inside the integral, Leibniz's rule is simply \eqalign{ dp &= -\phi(y)\,dy = -\phi(y)\,X\,d\beta \cr } We can also calculate the differential of $p$ directly in terms of $F$, with the help of the Hadamard $(\odot)$ product
\eqalign{ dp &= p\odot dF = {\rm diag}(p)\,dF = P\,dF \cr } Equating the two expression allows us to find the gradient wrt $\beta$ \eqalign{ dF &= -\phi(y)\,P^{-1}X\,d\beta\cr \frac{\partial F}{\partial\beta} &= -\phi(X\beta)\,P^{-1}X \cr &= -\phi_o\,\exp\Big(\tfrac{-\beta^TX^T\Sigma^{-1}X\beta}{2}\Big)\,{\rm diag}\big(\exp(F)\big)^{-1}X \cr\cr } So that's the gradient; the Hessian will be much more complicated. Are you sure you need it?

• thanks. Hessian is exactly what I need. This function is actually part of a multinational probit likelihood function. I need this to calculate the variance of $\beta$. I didn't understand why $p=diag(p)$ in Equation 5. Can you enlighten my on that? Is the Hessian the following: $\frac{\partial^2 F}{\partial\beta^2} = -\phi_o\,\exp\Big(\tfrac{-\beta^TX^T\Sigma^{-1}X\beta}{2}\Big)\,(\beta^T(X^T\Sigma^{-1}X+X^T(\Sigma^{-1})^TX)){\rm diag}\big(\exp(F)\big)^{-1}X + -\phi_o\,\exp\Big(\tfrac{-\beta^TX^T\Sigma^{-1}X\beta}{2}\Big)\,{\rm diag}\big(\exp(F)\big)^{-2}\phi(X\beta)XX$ – Ding Li Jun 8 '18 at 4:35
• @DingLi Since $(F,\beta)$ are vectors, the gradient is already a matrix, and the Hessian will be a $3^{rd}$ order tensor. You cannot write it using standard matrix/vector notation, you must use index notation and work it out element by element. – greg Jun 8 '18 at 14:55