Unformatted text preview: We give another deﬁnition of the expectation in the continuous case. First
suppose X is nonnegative and bounded above by a constant M . Deﬁne Xn (ω )
to be k/2n if k/2n ≤ X (ω ) < (k +1)/2n . We are approximating X from below
by the largest multiple of 2−n . Each Xn is discrete and the Xn increase to
X . We deﬁne E X = limn→∞ E Xn .
Let us argue that this agrees with the ﬁrst deﬁnition in this case. We have
E Xn =
k /2 n = k
P(Xn = k/2n ) =
2n
k
2n k /2 n k
P(k/2n ≤ X < (k + 1)/2n )
2n (k+1)/2n (k+1)/2n f (x)dx =
k/2n k/2n k
f (x)dx.
2n If x ∈ [k/2n , (k + 1)/2n ), then x diﬀers from k/2n by at most 1/2n . So the
last integral diﬀers from
(k+1)/2n xf (x)dx
k/2n by at most (1/2n )P(k/2n ≤ X < (k + 1)/2n ) ≤ 1/2n , which goes to 0 as
n → ∞. On the other hand,
(k+1)/2n M xf (x)dx =
k/2n xf (x)dx,
0 which is how we deﬁned the expectation of X .
We will not prove the following, but it is an interesting exercise: if Xm
is any sequence of discrete random variables that increase up to X , then
limm→∞ E Xm will have the same value E X .
To show linearity, if X and Y are bounded positive random variables,
then take Xm discrete increasing up to X and Ym discrete increasing up to
Y . Then Xm + Ym is discrete and increases up to X + Y , so we have
E (X + Y ) = lim E (Xm + Ym ) = lim E Xm + lim E Ym = E X + E Y.
m→∞ m→∞ m→∞ If X is not bounded or not necessarily positive, we have a similar deﬁnition;
we will not do the details. This second deﬁnition of expectation is mostly
useful for theoretical purposes and much less so for calculations.
Similarly to the discrete case, we have
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 Spring '10
 ansan
 Probability theory, xm, )/2n, k/2n

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