# Statistics: bivariate normal distribution

See the attached file.

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.

PROPOSITION

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.

#### Solution Summary

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