Ex 2.4: Rephrasing proposal
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As stated, Exercise 2.4 needs some repairs (just read through it once).
To mirror the sequencing of other examples and exercises, I suggest naming the parameter of interest first rather than second. Something along these lines:
Suppose that we have
$n$ data points,$x_1,\ldots,x_n$ , drawn independently from an exponential distribution with parameter$\lambda$ . We would like to know$\lambda$ . The likelihood function for our observations is
$p(x_1,\ldots,x_n|\lambda) = \lambda^n \exp\left(−\lambda\sum_{k=1}^n x_k\right).$
Starting with a Gamma prior distribution for$\lambda$ (see below), show that the posterior distribution for$\lambda$ is also a Gamma distribution. Provide formulas giving the posterior parameters$a^{*},b^{*}$ in terms of the prior parameters$a,b$ and the data. Use the following facts about Gamma distributions:$\ldots$
Yes, this is much better than what I wrote. I have rewritten this as:
Suppose that we perform
\begin{equation}
Prob(X=n)=\theta(1-\theta)^{ n-1}
\end{equation}
where
Let the prior on
The posterior distribution is a beta distribution with parameters
Conjugate forms 3
Suppose that we have
The \index{Exponential likelihood} exponential likelihood function is:
\begin{equation}
p(x_1,\dots,x_n | \lambda)=\lambda^n \exp (-\lambda \sum_{i=1}^n x_i )
\end{equation}
Starting with a Gamma prior distribution for
The \index{Gamma distribution} Gamma distribution is defined in terms of the parameters
\begin{equation}
\mathit{Gamma}(y | a,b)=\frac{b^a y^{a-1} \exp(-by)}{\Gamma(a)}
\end{equation}
The
\begin{equation}
\mathit{Gamma}(\lambda | a,b)=\frac{b^a \lambda^{a-1} \exp(-b\lambda)}{\Gamma(a)}
\end{equation}