Recall from the slides that the Gamma distribution can be reparameterized using the two parameters a, the shape parameter, and \mu, the mean. The pdf looks like
\displaystyle \displaystyle f_{(a,\mu )}(y) \displaystyle = \displaystyle \frac{1}{\Gamma (a)}\left(\frac{a}{\mu }\right)^ a \, y^{a-1}\, e^{-\frac{ay}{\mu }}
Let {\boldsymbol \theta }=\begin{pmatrix} a\\ \mu \end{pmatrix}\, and rewrite this as the pdf of a 2-parameter exponential family. Enter {\boldsymbol \eta }({\boldsymbol \theta })\cdot \mathbf{T}(\mathbf{y}) below.
{\boldsymbol \eta }({\boldsymbol \theta })\cdot \mathbf{T}(\mathbf{y})=\quad
{\boldsymbol \eta }({\boldsymbol \theta })\cdot \mathbf{T}(\mathbf{y})= \begin{pmatrix} \ln\left(\frac{a}{\mu}\right) \\ a-1 \end{pmatrix} \cdot \begin{pmatrix} y \\ 1 \end{pmatrix} = \ln\left(\frac{a}{\mu}\right) \cdot y + (a-1) \cdot 1 = y \ln\left(\frac{a}{\mu}\right) + a - 1
To rewrite the Gamma distribution as a 2-parameter exponential family, we need to express the pdf in the form:
\[
f(y;\boldsymbol{\theta}) = h(y) \exp \left\{ \boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(y) - A(\boldsymbol{\theta}) \right\}
\]
where:
- \(f(y;\boldsymbol{\theta})\) is the pdf of the Gamma distribution,
- \(h(y)\) is a function only involving \(y\),
- \(\boldsymbol{\eta}(\boldsymbol{\theta})\) is a vector of parameters,
- \(\mathbf{T}(y)\) is a vector of sufficient statistics of \(y\),
- \(A(\boldsymbol{\theta})\) is the log-normalizing constant.
From the given pdf of the Gamma distribution:
\[
f_{(a,\mu)}(y) = \frac{1}{\Gamma(a)} \left(\frac{a}{\mu}\right)^a y^{a-1} e^{-\frac{ay}{\mu}}
\]
we can determine \(\boldsymbol{\eta}(\boldsymbol{\theta})\) and \(\mathbf{T}(y)\) as follows:
- \(\boldsymbol{\eta}(\boldsymbol{\theta})\) is a vector containing the natural parameters, which are the logarithms of the ratio of parameters in the pdf. In this case, \(\boldsymbol{\theta} = \begin{pmatrix} a \\ \mu \end{pmatrix}\), so \(\boldsymbol{\eta}(\boldsymbol{\theta}) = \begin{pmatrix} \ln \left(\frac{a}{\mu}\right) \\ 1 \end{pmatrix}\).
- \(\mathbf{T}(y)\) is a vector containing the sufficient statistics, which correspond to the powers of \(y\) in the pdf. In this case, \(\mathbf{T}(y) = \begin{pmatrix} y^{a-1} \\ y \end{pmatrix}\).
To determine \(\boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(y)\), we simply take the dot product of the two vectors:
\(\boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(y) = \ln \left(\frac{a}{\mu}\right) \cdot y^{a-1} + 1 \cdot y\)
Thus, the answer is:
\(\boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(y) = \ln \left(\frac{a}{\mu}\right) \cdot y^{a-1} + y\)