# Show bivariate normal distribution with non-diagonal covariance belongs to curved exponential family?

Question:

Suppose that $$(X_{i}, Y_{i})$$, $$i = 1, \dots ,n$$ are sampled i.i.d. from the two-dimensional normal distribution

$$\begin{bmatrix} X & Y \end{bmatrix} \sim \mathcal{N}\left( \begin{bmatrix} 0\\ 0\\ \end{bmatrix}, \begin{bmatrix} 1 & \theta\\ \theta & 1\\ \end{bmatrix} \right),$$

with $$\theta \in \Omega = (-1, 1)$$. Show that the joint density of $$(X_{i}, Y_{i})$$, $$i = 1, \dots ,n$$ is a 2-dimensional curved exponential family.

Attempt:

Since the density $$f_{X, Y}$$ can be expressed as

$$f_{X, Y}(x, y) = \frac{1}{2\pi\sqrt{1 - \theta^{2}}}\exp\bigg\{-\frac{1}{2(1 - \theta^{2})}(x^{2} + y^{2}) + \frac{\theta}{1 - \theta^{2}}xy\bigg\},$$

the joint density is

$$\prod_{i = 1}^{n}f_{X_{i}, Y_{i}}(x_{i}, y_{i}) = \left(\frac{1}{2\pi\sqrt{1 - \theta^{2}}}\right)^{n}\exp\bigg\{-\frac{1}{2(1 - \theta^{2})}\sum_{i = 1}^{n}(x_{i}^{2} + y_{i}^{2}) + \frac{\theta}{1 - \theta^{2}}\sum_{i = 1}^{n}x_{i}y_{i}\bigg\}$$

from which the natural parameter $$\eta(\theta)$$ is

$$\eta(\theta) = \begin{bmatrix} -\frac{1}{2(1 - \theta^{2})} & \frac{\theta}{1 - \theta^{2}} \end{bmatrix}$$

How to show the exponential family is curved? According to my understanding, we need to show $$\eta_{2}$$ is a non-linear function of $$\eta_{1}$$, although it is not obvious to me how.

• $\eta_2$ is a differentiable function of $\eta_1.$ If it is linear, then its derivative must be constant. Is the derivative constant?
– whuber
Apr 19 '20 at 17:05