For the two following pmfs with one parameter \theta that are written in the form
\displaystyle \displaystyle f_\theta (y) \displaystyle = \displaystyle h(y) e^{w(\theta ,y)},
first decompose \, w(\theta ,y)\, as
\displaystyle \displaystyle w(\theta ,y) \displaystyle = \displaystyle \eta (\theta )T(y)-B(\theta ),
then enter the product \eta (\theta )T(y) below. Select the distribution that f_\theta defines.
For \, f_\theta (y)\, =\, e^{w(\theta ,y)}\, where
\displaystyle \displaystyle \, w(\theta ,y)=y\ln (\theta )+(1-y)\ln (1-\theta )\,
and \, y=0,1,\, \theta \in (0,1):
\eta (\theta ) T(y)=\quad
unanswered
What distribution does the pmf f_\theta (y) define?
\mathcal{N}(\theta ,1)
\mathcal{N}(1,\theta )
\textsf{Ber}(\theta )
\textsf{Poiss}(\theta )
none of the above
unanswered
For \, \displaystyle f_\theta (y)\, =\, \frac{1}{y!} e^{w(\theta ,y)}\, where \, w(\theta ,y)=-\theta +y\ln (\theta ),\, and \, y=0,1,2,\ldots ,\, \theta \in (0,1):
\eta (\theta ) T(y)=\quad
unanswered
What distribution does the pmf f_\theta (y) define?
\mathcal{N}(\theta ,1)
\mathcal{N}(1,\theta )
\textsf{Ber}(\theta )
\textsf{Poiss}(\theta )
none of the above
For the first pmf, we have:
\[\displaystyle \displaystyle w(\theta ,y)=y\ln (\theta )+(1-y)\ln (1-\theta )\]
We can decompose \(\displaystyle w(\theta ,y)\) as:
\[\displaystyle \displaystyle w(\theta ,y)=y\ln (\theta )+(1-y)\ln (1-\theta )=-\ln(1-\theta)+y\ln \left(\frac{\theta}{1-\theta}\right)\]
So, we have:
\[\displaystyle \displaystyle \eta (\theta )T(y)=-\ln(1-\theta)\]
This corresponds to the logarithm of the odds ratio in the Bernoulli distribution. Therefore, the distribution that \(\displaystyle f_{\theta}(y)\) defines is \textsf{Ber}(\(\displaystyle \theta\)).
For the second pmf, we have:
\[\displaystyle \displaystyle w(\theta ,y)=-\theta +y\ln(\theta)\]
We can decompose \(\displaystyle w(\theta ,y)\) as:
\[\displaystyle \displaystyle w(\theta ,y)=-\theta +y\ln(\theta)=-(-\theta+y\ln(\theta))\]
So, we have:
\[\displaystyle \displaystyle \eta (\theta )T(y)=-\theta\]
This corresponds to the parameter value in the Poisson distribution. Therefore, the distribution that \(\displaystyle f_{\theta}(y)\) defines is \textsf{Poiss}(\(\displaystyle \theta\)).
For the first pmf, we have:
\[
w(\theta, y) = y \ln(\theta) + (1-y) \ln(1-\theta)
\]
To decompose it, we let:
\[
\eta(\theta) = \ln(\theta) \quad \text{and} \quad T(y) = y+(1-y) = 1
\]
So, the product of \(\eta(\theta)\) and \(T(y)\) is:
\[
\eta(\theta) T(y) = \ln(\theta) \cdot 1 = \ln(\theta)
\]
The distribution that this pmf \(f_\theta(y)\) defines is:
\textsf{Ber}(\(\theta\))
For the second pmf, we have:
\[
w(\theta, y) = -\theta + y \ln(\theta)
\]
To decompose it, we let:
\[
\eta(\theta) = -1 \quad \text{and} \quad T(y) = y
\]
So, the product of \(\eta(\theta)\) and \(T(y)\) is:
\[
\eta(\theta) T(y) = -1 \cdot y = -y
\]
The distribution that this pmf \(f_\theta(y)\) defines is:
\textsf{Poiss}(\(\theta\))