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Unformatted text preview: Two special cases: a) Ifg(X1, X2) = (X1  u1)(X2  112), E(g(X1, X2)) is called the covariance of X1 and X2 , denoted C0V(X1,X2). This is the deﬁnitional formula for computing Cov(X1,X2).
A computing formula for the covariance is C0V(X1,X2) = E((X1— u1)(X2  uz )) : deﬁnition
' = [E(X1X2)l u1u2 rcomputing (See p.184 for the covariance). b) COV(X1,X2)/6162 is called the correlation coefﬁcient of X1 and X2 , denoted p.
Theorem: —1 s p S 1. p measures the "degree" of linearity in the jointh or pdf X1 and X 2 , and is sometimes called
the standardized covariance. (Make sure to distinguish between r and p, just as we distinguish between F (x) and a crf 0 give,
just as we distinguish between s and 6, just as we distinguish between X and 11, etc.). . Example: Return to problem 5.76 on p. 188). f(x1,x2)=:(X1+x%) 0<x1 <1 and0<x2 <1 = 0 elsewhere
1 6 1
uX1=Ix1—(x1+—)dxl=3/5 =O.6
0 5 3 —jx 9(x2+l)dx —3/5=06
HXZ 025 2 2 2  ll
E(X1X2) = jjx1x22(x1+ x%)dx2dx1= 7/20 = 0.35
00
The computing form therefore implies Cov(X1,X2) = (7/20)  (3/5)(3/5) = 1/ 100 = — 0.01 T 0 ﬁnd p, we need only ﬁnd 01 and 62.
Var(X1) = of = E(X12) — (3/5)2, and . 1 .
E(X12) = 1x12 gm +%)dx1 = (13/30) :> 612 = (13/30)—9/25 = 11/150
' 0 Similarly, 6% = 2/25 :> p = —_1/1—00— = —0.13 ./(11/150)(2/25) Some more results 1) If g(X1,X2)=X1+X2, then
E(X1+X2)= H1+H23Var(X1+X2)= Cy12 +5; +2COV(X13X2) E(X1 —X2) = u. u2,Var(X1+X2)= 6.2 +o§ —2Cov(X1.X2) 3) If X1 and X2 are independent random variables, then Cov(X1,X2) =0, and therefore p = 0. 4) If g(X1,X2) = ale + azXz, where a1 and a2 are known, given constants, then
E(31X1+ azxz) = a1111+ azliz Var(a1X1+ azXZ) = alzol2 + a§o§ + 2a1a2Cov(X1,X2) Example (For a gas station) X1 = amount of regular unleaded gasoline, in gals, purchased on a given day
X2 = amount of extra unleaded gasoline, in gals, purchased on a given day
X3 = amount of super unleaded gasoline, in gals, purchased on a given day
X4 = amount of diesel fuel, in gals, purchased on a given day. f (x1, x2,x3,x4) denotes the multivariate pdf (assume the k=4 RVs are continuous) of the 4 4
random variables. If they were assumed independent, then f (x1,x2, x3,x4) = Hf, (Xi) , Where
i=1 f, (xi) is the (marginal) pdf of X. Of course, independence is an unreasonable assumption in this
case. Suppose the price of regular is a1 = $2.70, the price of extra is a2 = $2.80, the price of super is a3
= $2.90, and the price of diesel is a4 = $2.70. Then the daily revenue is a random variable, X where X = 2.7X1+ 2.8X2 + 2.9X3 +2.7X4 is a g(X1,X2,X3,X4) , and a g(X1,X2,X3,X4) that is
a linear combination of X1, X2,X3,X4. Suppose u1 = 1200, u2 = 500, M = 200,u4 = 180. Then, 4
the expected daily revenue, E(X) is Zaiui = 2.7(1200) + 2.8(500) + 2.9(200) + 2.7(180) = 5706.
‘ i=1 Now suppose <31 =100,o2 = 80,63 = 50,64 = 60 , and suppose the six covariances are (Xi,XJ) Cov (Xi,Xj) pij
X1,X2 800 0.10
X1,X3  250  0.05
X1,X4 100 0.0167
X2,X3 120 0.03
X2,X4  60 — 0.0125
X3,X4 80 0.0267 4 4 4
Therefore, Zaizoi2 = 170345, and Z ZaiajCov(Xi ,Xj) = 5945.1 (i i jin the sum)
i=1 i=1j=1 :> a; =176290.1: 6x419.87 Note that if the four randOm variables were assumed independent then ox = V170345 21412.73. If Y is the mean amount of gallons purchased on a given day, then
Y =(X1+X2 +X3 +X4)/4 and
4 4 4 . 4 of 2; zeov(Xi,Xj)L/"‘ ‘i
“Y = 211i /4 = 520, Var(Y) = i=1 +L‘Fl—— = 1406.25 + 98.75 = 1505 , and therefore
i—l _ 16 16
the standard deviation of Y is 38.79. k k k k 
Note: 2 Zaiaj cov(Xi,XJ)=22 Z aiaj cov(Xi,XJ)
i=1j=1 \ i=1j=1
i¢j i<j ...
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 Spring '08
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