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Material Type: Notes; Class: Intro To Probability; Subject: Mathematics; University: University of Utah; Term: Summer 2008;
Typology: Study notes
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Theorem 15.1. Let g be a real-valued function of two variables, and (X, Y) have joint mass function f. If the sum converges then
E[g(X, Y)] =
x
y
g(x , y)f(x , y).
Corollary 15.2. For all a, b real,
E(aX + bY) = aEX + bEY.
Proof. Setting g(x , y) = ax + by yields
E(aX + bY) =
x
y
(ax + by)f(x , y)
=
x
ax
y
f(x , y) +
x
y
byf(x , y)
= a
x
xfX (x) + b
y
y
x
f(x , y)
= aEX + b
y
fY (y),
which is aEX + bEY.!
Theorem 15.3 (Cauchy–Schwarz inequality). If E(X 2 ) and E(Y 2 ) are finite, then |E(XY)| "
Proof. Note that (XE(Y 2 ) − YE(XY)) (^2)
= X^2 (E(Y 2 ))^2 + Y 2 (E(XY))^2 − 2 XYE(Y 2 )E(XY). Therefore, we can take expectations of both side to find that
E
The left-hand side is # 0. Therefore, so is the right-hand side. Solve to find that E(X 2 )E(Y 2 ) # (E(XY)) 2. [If E(Y 2 ) > 0, then this is OK. Else, E(Y 2 ) = 0, which means that P{Y = 0 } = 1. In that case the result is true, but tautologically.]!
Thus, if E(X 2 ) and E(Y 2 ) are finite, then E(XY) is finite as well. In that case we can define the covariance between X and Y to be Cov(X, Y) = E [(X − EX)(Y − EY)]. (12) Because (X − EX)(Y − EY) = XY − XEY − YEX + EXEY, we obtain the following, which is the computationally useful formula for covariance: Cov(X, Y) = E(XY) − E(X)E(Y). (13) Note, in particular, that Cov(X, X) = Var(X).
Theorem 15.4. Suppose E(X 2 ) and E(Y 2 ) are finite. Then, for all nonrandom a, b, c, d: (1) Cov(aX + b , cY + d) = acCov(X, Y); (2) Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
Proof. Let μ = EX and ν = EY for brevity. We then have Cov(aX + b , cY + d) = E [(aX + b − (aμ + b))(cY + d − (cν + d))] = E [(a(X − μ))(c(Y − ν))] = acCov(X, Y).
