Show that the joint distribution of $X_1, X_2, \dots, X_n$ belongs to a two-parameter exponential family.

Question

Let $X_1, X_2, \dots, X_n$ be ind $ \sim\text{Ber}(\theta_i)$ where

\begin{equation} \theta_i = P(X_i=1)=\frac{\exp(\alpha+\beta t_i)}{1+\exp(\alpha + \beta t_i)} \end{equation}

where $t_1, t_2, \dots, t_n$ are known constants. Show that the joint distribution of $X_1, X_2, \dots, X_n$ belongs to a two-parameter exponential family.

$\bf{My Attempt (WIP) :}$
We can start by finding the joint pmf of $X_1, X_2, \dots, X_n$

\begin{align*} f(\mathbf{X}) &= \prod_{i=1}^n \theta_i^{x_i} (1-\theta_i)^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (1-\frac{e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (\frac{1+e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}}-\frac{e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (\frac{1}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i + 1 - x_i}} \\ &=\prod_{i=1}^n \frac{e^{\alpha x_i+\beta t_ix_i}}{1+e^{\alpha+\beta t_i}}\\ &=\frac{e^{\alpha\sum_{i=1}^n x_i+\beta\sum_{i=1}^n t_i x_i} }{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}} \end{align*}

This is exponential family with $T(x) = (\sum x_i, \sum t_i x_i)$, $\eta= (\alpha, \beta)$, and $A(\eta) =A(\eta)=\frac{1}{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}}$

Answer 1

We can start by finding the joint pmf of $X_1, X_2, \dots, X_n$

\begin{align*} f(\mathbf{X}) &= \prod_{i=1}^n \theta_i^{x_i} (1-\theta_i)^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (1-\frac{e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (\frac{1+e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}}-\frac{e^{\alpha+\beta t_i}}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i}} (\frac{1}{1+e^{\alpha+\beta t_i}})^{1-x_i}\\ &=\prod_{i=1}^n \frac{(e^{\alpha+\beta t_i})^{x_i}}{ (1+e^{\alpha+\beta t_i})^{x_i + 1 - x_i}} \\ &=\prod_{i=1}^n \frac{e^{\alpha x_i+\beta t_ix_i}}{1+e^{\alpha+\beta t_i}}\\ &=\frac{e^{\alpha\sum_{i=1}^n x_i+\beta\sum_{i=1}^n t_i x_i} }{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}} \end{align*}

This is exponential family with $T(x) = (\sum x_i, \sum t_i x_i)$, $\eta= (\alpha, \beta)$, and $A(\eta) =A(\eta)=\frac{1}{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}}$