Chapter 12: Q1E (page 787)
Let \({\bf{f}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{ = cg}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) be a joint p.d.f for \(\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{,}}\)each \({x_{2\,}}\)let\({{\bf{h}}_{{\bf{2}}\,}}\left( {{{\bf{x}}_{{\bf{1}}\,}}} \right){\bf{ = g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) that is what we get by considering \({\bf{g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) as a function of \({{\bf{x}}_{{\bf{1}}\,}}\)for fixed \({{\bf{x}}_{2\,}}\)show that there is a multiplicative factor \({{\bf{c}}_{{\bf{2}}\,}}\)that does not depend on such that is the conditional p.d.f of \({{\bf{x}}_{{\bf{1}}\,}}\) given \(\left( {{{\bf{x}}_{{\bf{2}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\)
Short Answer
That is what we get by considering.
\({g_1}\left( {{x_1}\mid {x_2}} \right)\frac{{\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}} = \frac{{cg\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}}.\)
\({x_2}\,\,\,\)is fixed, and for each \({x_2}\,\,\,\)
\({h_2}\left( {{x_1}} \right)\, = g\left( {{x_1},{x_2}} \right),\)
\({c_2} = \frac{c}{{{f_2}\left( {{x_2}} \right)\,}}.\)