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Does the family of distributions where $p(x,\theta)$ is the conditional frequency function of a binomial $\mathcal{B}(n,\theta)$, variable $X$, given that $X > 0$, belong to the exponencial family? Any help is appreciated. Thanks in advance!

MY ATTEMPT (EDITED)

As it is known, the binomial distribution is given by \begin{align*} p(x,\theta) = {{n}\choose{x}}\theta^{x}(1-\theta)^{n-x} \end{align*}

Thus we get the following expression for the conditional distribution \begin{align*} \textbf{P}(X = x|X > 0) & = \frac{\textbf{P}(X = x,X > 0)}{\textbf{P}(X > 0)}\\ & = \frac{\displaystyle{{n}\choose{x}}\theta^{x}(1-\theta)^{n-x}}{1 - \textbf{P}(X = 0)}\\ & = \frac{\displaystyle{{n}\choose{x}}\theta^{x}(1-\theta)^{n-x}}{1-(1 - \theta)^{n}} \end{align*}

This is as far as I can get. Can someone help me solve this exercise? Thanks in advance!

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    $\begingroup$ Can you show that ordinary binomial distributions belong to the exponential family? If so, what stops you using the same argument here? $\endgroup$ – Henry May 14 at 0:48
  • $\begingroup$ @user1337 Can you put in some steps to clairfy how $1-P(X=0)$ cancels a $\theta$ in the numerator? $\endgroup$ – Glen_b May 14 at 0:57
  • $\begingroup$ It was wrongly written. Thanks for the observation. $\endgroup$ – user1337 May 14 at 1:01
  • $\begingroup$ See en.m.wikipedia.org/wiki/…. $\endgroup$ – StubbornAtom May 14 at 6:57

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