Stat 5101 Lecture Notes - School of Statistics

102 Stat 5101 (Geyer) Course Notes(and, of course, the discrete case is analogous with the integrals replaced bysums). It is a useful mnemonic device to write (3.24) lining up the analogousbits in the numerator and denominator∫g(y)f(x, y) dyE{g(Y ) | x} = ∫f(x, y) dy .This looks a little funny, but it reminds us that the density in the numeratorand denominator is the same, and the variable of integration is the same. Theonly difference between the numerator and denominator is the function g(y)appearing in the numerator.If we plug in (3.22) instead of (3.21) for f(y | x) we get∫g(y)h(x, y) dyE{g(Y ) | x} = ∫(3.25)h(x, y) dywhere h(x, y) isanunnormalized joint density.These formulas make it clear that we are choosing the denominator so thatE(1 | x) = 1, which is the form the norm axiom takes when applied to conditionalprobability. That is, when we take the special case in which the functiong(y) is equal to one for all y, the numerator and denominator are the same.Example 3.4.5.Suppose X and Y have the unnormalized joint densityh(x, y) =(x+y)e −x−y ,x > 0,y>0,what is E(X | y)?Using (3.25) with the roles of X and Y interchanged and g the identityfunction we get∫xh(x, y) dxE(X | y) = ∫h(x, y) dx∫x(x + y)e −x−y dx= ∫(x + y)e−x−ydxUsing (3.23) the denominator is∫ ∞∫ ∞∫ ∞(x + y)e −x−y dx = e −y xe −x dx + ye −y e −x dx0=(1+y)e −yand the numerator is∫ ∞∫ ∞∫ ∞x(x + y)e −x−y dx = e −y x 2 e −x dx + ye −y xe −x dxHence0=(2+y)e −y00E(X | y) = 2+y1+y , y > 0.00