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