Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

(Jacob Rumans) #1
2.4. Independent Random Variables 117

2.3.9.Five cards are drawn at random and without replacement from an ordinary
deck of cards. LetX 1 andX 2 denote, respectively, the number of spades and the
number of hearts that appear in the five cards.

(a)Determine the joint pmf ofX 1 andX 2.

(b)Find the two marginal pmfs.

(c)What is the conditional pmf ofX 2 ,givenX 1 =x 1?

2.3.10.LetX 1 andX 2 have the joint pmfp(x 1 ,x 2 ) described as follows:

(x 1 ,x 2 ) (0,0) (0,1) (1,0) (1,1) (2,0) (2,1)
p(x 1 ,x 2 ) 181 183 184 183 186 181

andp(x 1 ,x 2 ) is equal to zero elsewhere. Find the two marginal probability mass
functions and the two conditional means.
Hint: Write the probabilities in a rectangular array.

2.3.11.Let us choose at random a point from the interval (0,1) and let the random
variableX 1 be equal to the number that corresponds to that point. Then choose
a point at random from the interval (0,x 1 ), wherex 1 is the experimental value of
X 1 ; and let the random variableX 2 be equal to the number that corresponds to
this point.


(a)Make assumptions about the marginal pdff 1 (x 1 ) and the conditional pdf
f 2 | 1 (x 2 |x 1 ).

(b)ComputeP(X 1 +X 2 ≥1).

(c)Find the conditional meanE(X 1 |x 2 ).

2.3.12.Letf(x)andF(x) denote, respectively, the pdf and the cdf of the random
variableX. The conditional pdf ofX,givenX>x 0 ,x 0 a fixed number, is defined
byf(x|X>x 0 )=f(x)/[1−F(x 0 )],x 0 <x, zero elsewhere. This kind of conditional
pdf finds application in a problem of time until death, given survival until timex 0.


(a)Show thatf(x|X>x 0 )isapdf.

(b)Letf(x)=e−x, 0 <x<∞, and zero elsewhere. ComputeP(X> 2 |X>1).

2.4 IndependentRandomVariables.....................


LetX 1 andX 2 denote the random variables of the continuous type that have the
joint pdff(x 1 ,x 2 ) and marginal probability density functionsf 1 (x 1 )andf 2 (x 2 ),
respectively. In accordance with the definition of the conditional pdff 2 | 1 (x 2 |x 1 ),
we may write the joint pdff(x 1 ,x 2 )as


f(x 1 ,x 2 )=f 2 | 1 (x 2 |x 1 )f 1 (x 1 ).
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