Suppose there are five observations \(x = \left( {{x_1},.....,{x_5}} \right)\)and all are conditionally i.i.d.
\(Z = z\) with pdf \(g\left( {x|z} \right)\).
We will use the conditional version of the Bayes theorem to compute the conditional pdf of Z given\(\left( {{X_1},.....,{X_5}} \right) = \left( {{x_1},....{x_5}} \right)\)
Conditional pdf \({g_{345}}\left( {{x_3},{x_{4,}}{x_5}|{x_{1,}}{x_{2,z}}} \right)\)of\(\)\(Y = \left( {{x_3},{x_{4,}}{x_5}} \right)\)given Z=z and\(W = \left( {{x_1},{x_2}} \right)\)
Hence,
\(\begin{align}{g_1}\left( {y|z,w} \right) &= g\left( {{x_3}|z} \right)g\left( {{x_4}|z} \right)g\left( {{x_5}|z} \right)\\ &= {z^3}{e^{ - z\left( {{x_3} + {x_4} + {x_5}} \right)}}\end{align}\)
The conditional pdf is
\({f_2}\left( {z|w} \right) = \frac{1}{2}{\left( {2 + {x_1} + {x_2}} \right)^3}{z^2}{e^{ - z\left( {2 + {x_1} + {x_2}} \right)}}\)
The conditional pdf for the last two observations given the first two is
\({f_1}\left( {y|w} \right) = \frac{{60{{\left( {2 + {x_1} + {x_2}} \right)}^3}}}{{{{\left( {2 + {x_1} + {x_2} + ...{x_5}} \right)}^6}}}\)
Hence for all observations, the conditional probability is
\(\begin{align}{g_2}\left( {z|y,w} \right) &= \frac{{{z^3}{e^{ - z\left( {{x_3} + {x_4} + {x_5}} \right)}}\frac{1}{2}{{\left( {2 + {x_1} + {x_2}} \right)}^3}{z^2}{e^{ - z\left( {2 + {x_1} + {x_2}} \right)}}}}{{\frac{{60{{\left( {2 + {x_1} + {x_2}} \right)}^3}}}{{{{\left( {2 + {x_1} + ... + {x_5}} \right)}^6}}}}}\\ &= \frac{1}{{120}}{\left( {2 + {x_1} + .... + {x_5}} \right)^6}{z^5}{e^{ - z\left( {2 + {x_1} + ... + {x_5}} \right)}}\end{align}\)
