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This is a homework problem. I've derived the following distribution from an earlier part in the problem $$ f_{X_1,X_2}(x_1,x_2) = \dfrac{\Gamma(x_1+x_2+r)\alpha_1^{x_1}\alpha_2^{x_2}\theta^r}{\Gamma(r)x_1!x_2!(\alpha_1+\alpha_2+\theta)^{x_1+x_2+r}}$$

where $\alpha_1, \alpha_2, r, \theta>0$ and $x_1, x_2 = 0,1,...$.

I have to put this into an exponential family form and find what is the sufficient statistic for $(\alpha_1,\alpha_2,\theta)$, where $r$ is known and $r=1$. Here is my attempt:

$\begin{align*} f_{X_1,X_2}(x_1,x_2) &= \dfrac{\Gamma(x_1+x_2+1)\alpha_1^{x_1}\alpha_2^{x_2}\theta}{x_1!x_2!(\alpha_1+\alpha_2+\theta)^{x_1+x_2+1}}\\ &= \dfrac{\Gamma(x_1+x_2+1)}{x_1!x_2!}e^{(\log{\alpha_1})x_1+(\log{\alpha_2})x_2-(\log{\alpha_1+\alpha_2+\theta})(x_1+x_2+1)+\log{\theta}}\\ &= \dfrac{\Gamma(x_1+x_2+1)}{x_1!x_2!}e^{\left(\log{\tfrac{\alpha_1}{\alpha_1+\alpha_2+\theta}}\right)x_1+\left(\log{\tfrac{\alpha_2}{\alpha_1+\alpha_2+\theta}}\right)x_2-\log{\tfrac{\alpha_1+\alpha_2+\theta}{\theta}}}\\ &= h(x_1,x_2)e^{\sum_{j=1}^3\eta_1(\alpha_1,\alpha_2,\theta)T_j(x_1,x_2)-B(\alpha_1,\alpha_2,\theta)} \end{align*}$

So my sufficient statistic is $T(X_1,X_2)=(X_1,X_2,1)$???

I feel like I did something wrong.

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    $\begingroup$ Consider including the factor $\theta^{\ r}$ within $h(x_1,x_2)$ (and don't forget the $r$). Also, you can greatly simplify the logarithms using the rule $\log(a/b)=\log(a)-\log(b)$. $\endgroup$ – whuber Sep 20 '13 at 19:33
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    $\begingroup$ So is $r$ known? $\endgroup$ – Sam Livingstone Sep 20 '13 at 19:45
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    $\begingroup$ The exponent $x_1+x_2+r$ in the denominator of your first expression for $f_{X_1,X_2}(x_1,x_2)$ does not match the exponent $x_1+x_2+1$ in the second expression for same, and you also have $\Gamma(x_1+x_2+r)$ in the first expression by $\Gamma(x_1+x_2+1)$ in the second. I also note you've mis-arranged the parentheses in the second line of the second expression., although it doesn't seem to have affected your subsequent formulae. $\endgroup$ – jbowman Sep 20 '13 at 20:34
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    $\begingroup$ oops I forgot to say $r$ is known, suppose it is $r=1$. $\endgroup$ – bdeonovic Sep 21 '13 at 12:48

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