4.4Expectation 107
EXAMPLE 4.3h The joint density ofXandYis given by
f(x,y)={ 12
5 x(2−x−y)0<x<1, 0<y<^1
0 otherwiseCompute the conditional density ofX, given thatY=y, where 0<y<1.
SOLUTION For 0<x<1, 0<y<1, we have
fX|Y(x|y)=f(x,y)
fY(y)=f(x,y)
∫∞
−∞f(x,y)dx=x(2−x−y)
∫ 1
0 x(2−x−y)dx=x(2−x−y)
2
3 −y/2=6 x(2−x−y)
4 − 3 y■4.4Expectation
One of the most important concepts in probability theory is that of the expectation
of a random variable. IfX is a discrete random variable taking on the possible values
x 1 ,x 2 ,..., then theexpectationorexpected valueofX, denoted byE[X], is defined by
E[X]=∑ixiP{X=xi}In words, the expected value ofXis a weighted average of the possible values thatXcan
take on, each value being weighted by the probability thatXassumes it. For instance, if
the probability mass function ofXis given by
p(0)=^12 =p(1)then
E[X]= 0( 1
2)
+ 1( 1
2)
=^12is just the ordinary average of the two possible values 0 and 1 thatXcan assume. On the
other hand, if
p(0)=^13 , p(1)=^23