5.5 Marginalized likelihood of an individual response vector

The marginalized likelihood observing a response vector \(\mathbf{Y}_i\) for student \(i\) is \[\begin{equation} L({\mathbf{Y}_i})=\sum_cL( {\mathbf{Y}_i}|{\alpha_i=\alpha_c})p(\mathbf{\alpha}_c) \end{equation}\]

Recall that \(L( {\mathbf{Y}_i}|{\mathbf{\alpha} _c})\) is

Code

Li.given.alpha.c

##                 00      10      01     11
## student 1  0.00081 0.00144 0.00144 0.0205
## student 2  0.00729 0.01296 0.01296 0.0051
## student 3  0.00081 0.00004 0.05184 0.0205
## student 4  0.00081 0.05184 0.00144 0.0205
## student 5  0.00729 0.01296 0.00036 0.0051
## student 6  0.00729 0.00036 0.01296 0.0051
## student 7  0.00729 0.01296 0.01296 0.0051
## student 8  0.00009 0.00016 0.00576 0.0819
## student 9  0.00081 0.05184 0.00144 0.0205
## student 10 0.00081 0.00144 0.00144 0.0205

Exercise 5.2 Let us assume \(p(00)=p(10)=\frac{1}{6}\) and \(p(01)=p(11)=\frac{1}{3}\). Verify that \(L({\mathbf{Y}_2})=.0094\).

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Note that

\[\begin{equation} L({\mathbf{Y}_i})=\sum_cL( {\mathbf{Y}_i}|{\alpha_i=\alpha_c})p(\mathbf{\alpha}_c) \end{equation}\]

Code

# population proportions
p.alphac <- c(1/6, 1/6, 1/3, 1/3)
L2 <- sum(Li.given.alpha.c[2, ] * p.alphac)
L2

## [1] 0.0094