Show that E{$e^{\delta T(Y)}|\eta$} = $e^{A(\eta + \delta) - A(\eta)}$. Then, prove that E{$T(Y)|\eta$} = $\frac{\partial A(\eta)}{\partial \eta}$.

First, we can begin by recognizing the fact that a PDF/PMF must integrate (or sum) to one, and now we just have to transform $e^{\delta T(Y)}$ into some recognizable PDF/PMF in order to try to find its expectation. However, I'm at a loss getting this started. Any help please?

  • $\begingroup$ What is $T(Y)$? $\endgroup$ – Ami Tavory Feb 5 '18 at 0:11
  • 2
    $\begingroup$ @Ami Presumably the OP means to define their exponential family notation something like this: $f_Y(y|\theta) = \exp \left (\eta(\theta) \cdot T(y) - A(\theta) + B(y) \right )$. $\endgroup$ – Glen_b -Reinstate Monica Feb 5 '18 at 0:33

Let $$f_Y(y|\eta) =h(y)\exp(\eta T(y)-A(\eta))$$ then \begin{align} E\{ e^{\delta T(Y)}|\eta\} &=\int h(y)\exp((\eta+\delta)T(y)-A(\eta)) \, dy \\ &= \exp(A(\eta+\delta)-A(\eta)) \cdot \int h(y)\exp((\eta+\delta)T(y)-A(\eta+\delta)) \, dy \\ &=\exp(A(\eta+\delta)-A(\eta)) \end{align}

You might like to use the property of cumulant generating function to solve the last part.


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