18.5. Linearity of Expectation 773
(c)LetV Df1;2;3gand.R;S;T/take the following triples of values with equal
probability,
.1;1;1/;.2;1;1/;.1;2;3/;.2;2;3/;.1;3;2/;.2;3;2/:Verify that
1.Ris independent ofST,- The eventŒRDSçis not independent ofŒSDTç.
3.SandThave a uniform distribution.
Problem 18.3.
LetR,S, andTbe mutually independent indicator variables.
In general, the event thatSDT is not independent ofRDT. We can explain
this intuitively as follows: suppose that bothRandTare more likely to equal 1
than to equal 0, butSis equally likely to be 0 or 1, which implies that it is equally
likely as not thatSDT. On the other hand, knowing thatRDSmakes it more
likely than not thatSD 1 , and knowing thatSD 1 , makes it more likely than not
thatSDT. So knowing thatRDSmakes it more likely than not thatSDT.
Now prove rigorously (without any appeal to intuition) that the eventsŒRDSç
andŒSDTçare independent iff eitherRis uniform^4 , orT is uniform, orSis
constant^5.
Problems for Section 18.3
Practice Problems
Problem 18.4.
SupposeR,S, andTbe mutually independent random variables on the same prob-
ability space with uniform distribution on the rangeŒ1;3ç.
LetMDmaxfR;S;Tg. Compute the values of the probability density function,
PDFM, ofM.
(^4) That is, PrŒRD1çD1=2.
(^5) That is, PrŒSD1çis one or zero.