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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.

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

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