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    Statistics: bivariate normal distribution

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    Provide a proof of the following theorem (see attached for full description):

    The Case of Normal Random Variables
    When the X 1- 's form a random sample from a normal distribution, f and T 0 are both
    normally distributed. Here is a more general result concerning linear combinations.
    The proof will be given toward the end of the section.
    If X 1, X 2, ..., X ,2 are independent, normally distributed rv's (with possibly different
    means and/or variances), then any linear combination of the X i 's also has a normal
    distribution. In particular, the difference X 1 — X 2 between two independent, normally
    distributed variables is itself normally distributed.
    Exarnple 6.15 (Exaniple 6.12 continued)
    The total revenue from the sale of the three grades of gasoline on a particular day
    was Y: 3.5X 1+ 3.65X 2 + 3.8X 3, and we calculated ,u Y = 6465 and (assuming
    independence) 0 Y = 493.83. If the X i 's are normally distributed, the probability that
    revenue exceeds 5000 is
    P(Y > 5000) = P (Z > ) = P(Z > -2.96?)
    = 1 - <I>(—2.96T) = .9985
    The CLT can also be generalized so it applies to certain linear combinations. Roughly
    speaking, if n is large and no individual term is likely to contribute too much to the
    overall value, then Yhas approximately a normal distribution.
    Proofs for the Case n =2
    For the result concerning expected values, suppose that X 1 and X 2 are continuous
    with joint pdff[x 1, x 2). Then
    Effli-X1 + I12-X2] = _f:,,(='i1I1 + a2I2].f(I1!I2) I511 512
    = iii I1f(I1= I2) 512 I511 + fl: I2f(I1= I2) 511 512
    = a1z1f_x;1[z1) J11 + ag zgf_x;2(zg) dz;
    = a1E(X1) + rz2E(X2)
    Summation replaces integration in the discrete case. The argument for the variance
    result does not require specifying whether either variable is discrete or continuous.
    Recalling that V(Y) =E[(Y— ,u Y)2],
    Vffli-X1 + I12-X2) = Eflfli-X1 + I12-X2 — (@1111 + flzilzllzl
    = Efflif-X1 — Fill + "if-X2 — Fzlz + 3='11='12(X1 — Filf-X2 — #2)}
    The expression inside the braces is a linear combination of the variables Y1 = (X 1 —
    H Q2, Y2 =(X 2 - H 2)2, and Y3 =(X 1 - H 1)(X 2 - H 2), $0 carrying the E °PBFa1i°I1
    through to the three terms gives ='1iV(X1) + ='1§V(X:) + 3='11='1:@=='(X1,X:) as required.

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    Solution Summary

    Prove that a bivariate normal distribution could be written as a sum of linearly independent random variables.