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The Hessian matrix of the bivariate standard normal distribution (where both standard normal distributions are independent) is positive definite. Yet, it concaves downwards. If the Hessian is positive definite, shouldn't that mean that the function concaves upwards? (Positive definite Hessian implies increasing gradient implies concave upwards.)

Here are the calculations:

Let

\begin{equation} \phi_2(x, y) = \phi(x)\phi(y) = \frac{1}{2 \pi}e^{-\frac{x^2 + y^2}{2}} \end{equation}

Observe that $\frac{\partial \phi(x, y)}{\partial x} = -x\phi(x, y)$. The other derivatives follow similar patterns.

Then, the Hessian is

\begin{equation*} \begin{matrix} \begin{pmatrix} \phi_2(x, y)(x^2 - 1) & \phi_2(x, y)xy \\ \phi_2(x, y)xy & \phi_2(x, y)(y^2 - 1) \\ \end{pmatrix} \end{matrix} \end{equation*}

which means that the determinant is $\phi_2^2(x, y)(1 - x^2 - y^2)$. Therefore, in the unit circle where $x < 1$ and $y < 1$, the Hessian is positive definite (because the determinant of the Hessian is positive), which is strange because a positive definitive Hessian implies concaving upwards.

darkgbm
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1 Answers1

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Your criterion for positive definiteness is incorrect. A positive determinant is necessary but not sufficient for positive definiteness (Sylvester's criterion for positive definiteness of a symmetric matrix requires all leading principal minors to be positive.)

At the point $(0,0)$ the Hessian matrix for the bivariate standard normal is $ H:=\left(\begin{matrix}-1&0\\0&-1\end{matrix}\right) $. This matrix is in fact negative definite. This follows easily from the definition: Given a nonzero $x:=(a,b)^T$, we calculate $ x^THx=-(a^2+b^2)<0$.

grand_chat
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