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A statistical model for the data set $\bf{y}$ is an exponential family with canonical parameter vector $\theta = (\theta_1,.. \theta_k)$ and canonical statistic $\bf{t(y)} $=$(t_1(\boldsymbol{y}),..t_k(\boldsymbol{y}) ) $ if it has structure

$f(\boldsymbol{y};\theta) = a(\theta)h(\boldsymbol{y})e^{\theta^T\bf{t(y)}} $

under Certain regularity conditions the distribution of $\bf{t}$ has probability function or density function

$f(\boldsymbol{t};\theta) = a(\theta) g(\bf{t})e^{\theta^T \boldsymbol{t}}$

where $g(\boldsymbol{t}) = \sum_{\boldsymbol{t(y)=t}} h(\boldsymbol{y}) $

In the discrete case, the probability function follows immediately by summation over the possible outcomes of $\boldsymbol{y}$ for given $\boldsymbol{t(y) = t}$

Can someone derive this, because I tried but I am not getting there....

$P(t(\boldsymbol{y})=\boldsymbol{t}) = \sum_{t(\boldsymbol{y})=\boldsymbol{t}} P(t(\boldsymbol{y}),\boldsymbol{Y} = \boldsymbol{y}) = \sum_{t(\boldsymbol{y}) = \boldsymbol{t}}P(t(\boldsymbol{y})|\boldsymbol{ Y} = \boldsymbol{y})\cdot a(\theta)h(\boldsymbol{y})e^{\theta^T \boldsymbol{t}} = a(\theta)e^{\theta^T \boldsymbol{t}} \cdot\sum_{t(y) = t} P(t(\boldsymbol{y})|\boldsymbol{Y}=\boldsymbol{y})\cdot h(\boldsymbol{y})$

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  • $\begingroup$ Show us what you tried. This could be a self study question requiring the self study tag. I think canonical statistic is the same as natural parameters. $\endgroup$ – Michael Chernick May 24 '17 at 16:15
  • $\begingroup$ This proof in the book is not given, its "up to the reader to check it", I tried what I thought would work.. @MichaelChernick $\endgroup$ – J.doe May 24 '17 at 17:01
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You're pretty much there. Here are all your steps, with the intended answer added at the end. \begin{align*} P(t(\boldsymbol{y}) = \boldsymbol{t}) &= \sum_{t(\boldsymbol{y}) =\boldsymbol{t}} P(t(\boldsymbol{y})= \mathbf{t}),\boldsymbol{Y} = \boldsymbol{y}) \tag{you} \\ &=\sum_{t(\boldsymbol{y}) = \boldsymbol{t}}P(t(\boldsymbol{y})= \mathbf{t})|\boldsymbol{ Y} = \boldsymbol{y})\cdot a(\theta)h(\boldsymbol{y})e^{\theta^T \boldsymbol{t}} \tag{you}\\ &= a(\theta)e^{\theta^T \boldsymbol{t}} \cdot\sum_{t(y) = t} P(t(\boldsymbol{y}) = \mathbf{t})|\boldsymbol{Y} =\boldsymbol{y})\cdot h(\boldsymbol{y})\tag{you} \\ &= a(\theta) g(\boldsymbol{t}) e^{\theta^T \boldsymbol{t}} \tag{goal} \end{align*} The last line follows because that conditional probability is $1$. If you're conditioning on the fact that $Y=y$, and you're summing over all $y$s such that $t(y) = t$, then that probability must be $1$.

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