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Question

To work our way up to GLMs, we will begin by defining exponential family distributions. We say that a class of distributions is in the exponential family if it can be written in the form

\(p(y ; \eta)=b(y) \exp \left(\eta^{T} T(y)-a(\eta)\right)\)

Here, \(\eta\) is called the natural parameter (also called the canonical parameter) of the distribution; T(y) is the [...] (for the distributions we consider, it will often be the case that T(y) = y); and \(a(\eta)\) is the log partition function. The quantity \(e^{-a(\eta)}\) essentially plays the role of a normalization constant, that makes sure the distribution \(p(y;\eta)\) sums/integrates over y to 1.

Answer

sufficient statistic


Question

To work our way up to GLMs, we will begin by defining exponential family distributions. We say that a class of distributions is in the exponential family if it can be written in the form

\(p(y ; \eta)=b(y) \exp \left(\eta^{T} T(y)-a(\eta)\right)\)

Here, \(\eta\) is called the natural parameter (also called the canonical parameter) of the distribution; T(y) is the [...] (for the distributions we consider, it will often be the case that T(y) = y); and \(a(\eta)\) is the log partition function. The quantity \(e^{-a(\eta)}\) essentially plays the role of a normalization constant, that makes sure the distribution \(p(y;\eta)\) sums/integrates over y to 1.

Answer
?

Question

To work our way up to GLMs, we will begin by defining exponential family distributions. We say that a class of distributions is in the exponential family if it can be written in the form

\(p(y ; \eta)=b(y) \exp \left(\eta^{T} T(y)-a(\eta)\right)\)

Here, \(\eta\) is called the natural parameter (also called the canonical parameter) of the distribution; T(y) is the [...] (for the distributions we consider, it will often be the case that T(y) = y); and \(a(\eta)\) is the log partition function. The quantity \(e^{-a(\eta)}\) essentially plays the role of a normalization constant, that makes sure the distribution \(p(y;\eta)\) sums/integrates over y to 1.

Answer

sufficient statistic

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