Recall that the exponential distribution with parameter \lambda is given by the pdf by
\displaystyle \displaystyle f_\lambda (y) = \lambda e^{-\lambda y}.
Let \theta =\lambda. Rewrite f_\lambda (y) in the form
\displaystyle \displaystyle f_\theta (y) = h(y) \exp \left( \eta (\theta ) T(y) -B(\theta ) \right),
and enter \eta (\theta ),\, T(y),\, B(\theta ) below.
These functions are not unique. To get unique answers, let h(y)=1,\, and let the coefficient of y in T(y) be +1.
T(y)=\quad
unanswered
\eta (\theta )=\quad
unanswered
B(\theta )=\quad
unanswered
If instead of h(y)=1,\, we had used \widetilde{h}(y)=C for some constant C, then what is \widetilde{B}(\theta ) in terms of B(\theta ) and C? That is, find \widetilde{B}(\theta ) such that the pdf f_\theta (y) of Y\sim \textsf{Exp}(\theta ) is
\displaystyle \displaystyle f_\theta (y) \, =\, \widetilde{h}(y) \exp \left( \eta (\theta ) T(y) -\widetilde{B}(\theta ) \right).
(Enter B for B(\theta ) and C for C. Your answer should be in terms of only C and B(\theta ). Enter “ln" for the natural logarithm.)
\widetilde{B}(\theta )=\quad
T(y) = y
η(θ) = -1/θ
B(θ) = 0
If h(y) is replaced by h̃(y) = C, then the new B̃(θ) can be found by:
B̃(θ) = B(θ) - ln(C)
To rewrite the pdf of the exponential distribution in the form given, we need to determine the functions T(y), η(θ), and B(θ).
We start by considering that T(y) should have a coefficient of +1 in front of y. Therefore, we can set T(y) = y.
Next, we need to determine η(θ) and B(θ). We can do this by expressing the given pdf in the exponential family form:
f_θ(y) = θe^(-θy)
Comparing this to the form f_θ(y) = h(y)exp(η(θ)T(y) - B(θ)), we can see that:
η(θ) = -θ
B(θ) = 0
Therefore, we have:
T(y) = y
η(θ) = -θ
B(θ) = 0
If instead of h(y) = 1, we had used \widetilde{h}(y) = C for some constant C, the new B(\theta) would be obtained by multiplying the coefficient in front of y in T(y) by ln(C):
\widetilde{B}(\theta) = B(\theta) + ln(C)
So, in terms of B(\theta) and C, we would have:
\widetilde{B}(\theta) = B(\theta) + ln(C)