*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **,
n
j−1
j−1 j
2n − 1
2 − 2 2n − 1
E( Xn ) =
P
,
+
P
,
2n
2n 2n
2n
2n
2n
j =1
2n − 1
j−1 j
j−1
=
−n
2n
2n
2
j =1
2n − 2
n 2 −1
1
= 2n
(j − 1)
2 j =1 1 (2n − 2)(2n − 1)
2 2n
2
1
1
1
1
= − n − n+1 + 2n
22
2
2
= implying
E(X ) = lim E(Xn ) = lim
n→∞ n→∞ 1
1
1
1
− n − n+1 + 2n
22
2
2 1
=.
2 Example 21.4. Consider ([0, 1], B1 , P) where B1 are the Borel sets of [0, 1] and P is the
uniform probability. Let Q1 = [0, 1] ∩ Q and consider the random variable Y : Ω → R given
by
ω , if ω ∈ [0, 1] \ Q1 ,
Y (ω ) =
0, if ω ∈ Q1 .
Note that Y is not a simple random variable, although it is non-negative. In order to compute
E(Y ) it is easiest to use Theorem 20.3. That is, let X (ω ) = ω for ω ∈ [0, 1] and note that
{ω : X (ω ) = Y (ω )} = Q1 \ {0}.
(The technical point here is that X (0) = Y (0) = 0. However, if ω ∈ Q1 with ω = 0, then
X (ω ) = Y (ω ).) Since P {Q1 \ {0}} = 0, we see that X and Y diﬀer on a set of probability
0 so that...

View
Full
Document