Below you are given a probability density function (pdf) or a probability mass function (pmf) \pi (\theta ) of a random variable \theta. Each pdf/pmf is actually a member of a well-known parametric family whose parameter is denoted by either p or \lambda. However, in the specification below, the pdf/pmf is specified only up to a normalizing constant, and in terms of hyper-parameters a, b and c. Your task is to recognize the parametric family and express the parameter in terms of the hyper-parameters. In entering the expressions for the parameters, only the variables a, b, or c may be used.
( Note: In this problem, the distribution \textsf{Geom}(p) is assumed to be over the nonnegative integers. The more explicit specification for the geometric distribution is the number of failure until the first success in a sequence of i.i.d. Bernoulli(p) Trials. )
\pi (\theta ) \propto a^{1-\theta }(1-a)^{\theta } (for \theta \in \{ 0, 1\}, and it is known that a \in (0, 1))
\textsf{Ber}(p)
\textsf{Exp}(\lambda )
\textsf{Poiss}(\lambda )
\textsf{Geom}(p)
unanswered
\text {parameter}=
unanswered
\pi (\theta ) \propto c^{a\theta +b} (for \theta \in \mathbb {N} \cup \{ 0\}, and it is known that a \in (0, 1))
\textsf{Ber}(p)
\textsf{Exp}(\lambda )
\textsf{Poiss}(\lambda )
\textsf{Geom}(p)
unanswered
\text {parameter}=
unanswered
\pi (\theta ) \propto 100e^{a\theta +b} (for \theta \geq 0, and it is known that a < 0)
\textsf{Ber}(p)
\textsf{Exp}(\lambda )
\textsf{Poiss}(\lambda )
\textsf{Geom}(p)
unanswered
\text {parameter}=
For the first specification, the probability mass function \pi(\theta) is given by \pi(\theta) \propto a^{1-\theta}(1-a)^\theta for \theta \in {0, 1}. This is the probability mass function of the Bernoulli distribution, which represents a binary outcome (success or failure) with parameter p. Therefore, the correct answer is \textsf{Ber}(p) and the parameter is p = 1-a.
For the second specification, the probability mass function \pi(\theta) is given by \pi(\theta) \propto c^{a\theta+b} for \theta \in \mathbb{N} \cup {0}. This is the probability mass function of the geometric distribution, which represents the number of failures until the first success in a sequence of i.i.d. Bernoulli(p) trials. Therefore, the correct answer is \textsf{Geom}(p) and the parameter is p = 1-c.
For the third specification, the probability density function \pi(\theta) is given by \pi(\theta) \propto 100e^{a\theta+b} for \theta \geq 0. This is the probability density function of the exponential distribution, which represents the time between events in a Poisson process with rate parameter \lambda = 1/a. Therefore, the correct answer is \textsf{Exp}(\lambda) and the parameter is \lambda = 1/a.
For the first distribution, the given probability density function (pdf) can be recognized as the pdf of the Bernoulli distribution.
Parameter: p
Expression in terms of hyperparameters: p = 1 - a
For the second distribution, the given probability mass function (pmf) can be recognized as the pmf of the Geometric distribution.
Parameter: p
Expression in terms of hyperparameters: p = 1 - a
For the third distribution, the given probability density function (pdf) can be recognized as the pdf of the Exponential distribution.
Parameter: λ
Expression in terms of hyperparameters: λ = -a
For the fourth distribution, the given probability mass function (pmf) can be recognized as the pmf of the Geometric distribution.
Parameter: p
Expression in terms of hyperparameters: p = 1 - a