Page 23 - Probability, Random Variables and Random Processes

PROBABILITY 15

A u %=(s:s~Aors~(a)
But, by definition, there are no s E (a. Thus,
AU@=(S:SEA)=A
An0={s:s~Aands~@)
But, since there are no s E (a, there cannot be an s such that s E A and s E 0. Thus,
An@=@
Note that Eq. (1.55) shows that (a is mutually excIusive with every other event and including with
itself.

1.16. Show that the null (or empty) set is a subset of every set A.
From the definition of intersection, it follows that
(A n B) c A and (A n B) c B
for any pair of events, whether they are mutually exclusive or not. If A and B are mutually exclusive events,
that is, A n B = a, then by Eq. (1.56) we obtain
(acA and (a c B (1.57)
Therefore, for any event A,
@cA (1 .58)
that is, 0 is a subset of every set A.

1.17. Verify Eqs. (1 .1 8) and (1.1 9).

A, then s I$ U A, .
Suppose first that s E (1 ) )
That is, if s is not contained in any of the events A,, i = 1, 2, . . . , n, then s is contained in Ai for all
i = 1, 2, . . . , n. Thus

Next, we assume that

Then s is contained in A, for all i = 1,2, . . . , n, which means that s is not contained in Ai for any i = 1,
2, . . . , n, implying that

Thus,

This proves Eq. (1 .1 8).
Using Eqs. (1 .l8) and (1.3), we have

Taking complements of both sides of the above yields

which is Eq. (1 .l9).