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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 $A(\eta) =A(\eta)=\frac{1}{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}}$

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  • 1
    $\begingroup$ If each $X_i$ is Bernoulli with a parameter $\theta_i$ that depends on the index $i$, then they cannot be called iid, since they are not identically distributed. $\endgroup$
    – heropup
    Sep 26 '21 at 23:06
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    $\begingroup$ $\dfrac{\exp(\alpha-\beta t_i)}{1+\exp(\alpha + \beta t_i)}$ looks strange while $\dfrac{\exp(\alpha+\beta t_i)}{1+\exp(\alpha + \beta t_i)}$ or perhaps $\dfrac{\exp(\alpha-\beta t_i)}{1+\exp(\alpha - \beta t_i)}$ would look more natural to investigate as having simple log-odds $\endgroup$
    – Henry
    Sep 26 '21 at 23:13
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    $\begingroup$ Where have you written the joint pmf of $X_1,\ldots,X_n$? It is not the product of the $\theta_i$'s, but a product of the Bernoulli pmfs. Try again. $\endgroup$ Sep 27 '21 at 16:38
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    $\begingroup$ It's the independence of $X_1,\ldots,X_n$ that allows you to write $f(\boldsymbol x)=\prod_{i=1}^n \theta_i^{x_i} (1-\theta_i)^{1-x_i}$ where $x_i\in \{0,1\}$ for every $i$. Substitute the expression for $\theta_i$ and focus on the term $\prod_{i=1}^n (e^{\alpha+\beta t_i})^{x_i}=\prod_{i=1}^n e^{\alpha x_i+\beta t_ix_i}$. Why can't you turn this product into sums in the exponents? $\endgroup$ Sep 27 '21 at 19:16
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    $\begingroup$ In the final step the denominator should just be $\prod (1+e^{\alpha+\beta t_i})$. You can post a detailed answer below. $\endgroup$ Sep 28 '21 at 6:37
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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 $A(\eta) =A(\eta)=\frac{1}{\prod_{i=1}^n{(1+e^{\alpha+\beta t_i})}}$

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