Section 5.3
Suppose X1 , X2 , … , Xn are independent random variables (which may
be either of the discrete type or of the continuous type) each from the
same distribution. Then we say that the collection of independent and
identically distributed random variables X1 , X2 , … , Xn is a random
sample of size n from the common distribution.
Important Theorems in the Text:
Theorem 5.3-1
(See Class Exercises #1(a)(b) in Section 5.2 and #1(e) in Section 5.3.)
Theorem 5.3-2 (See Class Exercise #10 in Sections 4.1, 4.2, 4.3.)
Theorem 5.3-3 (See Class Exercises #13 & 15 in Sections 4.1, 4.2, 4.3.)
1. An urn contains six chips, one $3 chip, two $2 chips, and three $1
chips. Two chips are selected at random and with replacement. The
following random variables are defined:
X1 = dollar value of the first chip selected,
X2 = dollar value of the second chip selected.
(a) Explain why X1 , X2 can be treated as a random sample of size 2
from a common distribution, and find the p.m.f. for that
common distribution.
Since the selection is done with replacement, X1 and X2 are
independent and identically distributed. The p.m.f. for that
4–x
common distribution is
f(x) = —— if x = 1, 2, 3.
6
1. - continued
(b) Find the joint p.m.f. of (X1 , X2).
The joint p.m.f. is
(4 – x1)(4 – x2)
f(x1) f(x2) = ——————
36
if x1 = 1, 2, 3 and x2 = 1, 2, 3 .
(c) Find the mean and variance for each of X1 and X2 .
3
2
1
E(X1) = E(X2) = E(X) = (1) — + (2) — + (3) —
6
6
6
3
2
1
E(X12) = E(X22) = E(X2) = (1) — + (4) — + (9) —
6
6
6
Var(X1) = Var(X2) = Var(X) = (10/3) – (5/3)2 = 5/9
5
= —
3
10
= —
3
(d) Find P(X1 + X2 < 4) two ways: from the p.m.f. of the common
distribution and also from the joint p.m.f.
P(X1 = 1 , X2 = 1) + P(X1 = 1 , X2 = 2) + P(X1 = 2 , X2 = 1) =
(3/6)(3/6) + (3/6)(2/6) + (2/6)(3/6) = 21/36 = 7/12
or
(3)(3)/36 + (3)(2)/36 + (2)(3)/36 = 21/36 = 7/12
(e) Find E(X1 + X2) two ways: from the p.m.f. of the common
distribution and also from the joint p.m.f.
5
5
10
E(X1 + X2) = E(X1) + E(X2) = — + — = —
3
3
3
or
9
6
3
6
E(X1 + X2) = (1+1) — + (1+2) — + (1+3) — + (2+1) — +
36
36
36
36
4
2
3
2
1
10
(2+2) — + (2+3) — + (3+1) — + (3+2) — + (3+3) — = —
36
36
36
36
36
3
1. - continued
(f) Find E[1 / (X1 + X2)].
1
1
E ——— = ——
X1 + X2
1+1
1
——
2+1
9
1
— + ——
36
1+2
6
1
— + ——
36
2+2
6
1
— + ——
36
1+3
4
1
— + ——
36
2+3
3
— +
36
2
— +
36
1
3
1
2
1
1
73
—— — + —— — + —— — = ——
3 + 1 36
3 + 2 36
3 + 3 36
270
(g) Let Y = X1 + X2 . Find the m.g.f. of Y, and find the p.m.f. of Y.
MY(t) = E(etY) = E[e t(X1 + X2 ) ] = E[e tX1 e tX2 ] =
E[e tX1 ] E[e tX2 ] = [(1/2)et + (1/3)e2t + (1/6)e3t ] 2 =
(1/4)e2t + (1/3)e3t + (5/18)e4t + (1/9)e5t + (1/36)e6t
The p.m.f. of Y is g(y) =
1/4
1/3
5/18
1/9
1/36
if y = 2
if y = 3
if y = 4
if y = 5
if y = 6
2. Suppose W, X, and Y are mutually independent random variables
each having an exponential distribution with E(W) = 1, E(X) = 5, and
E(Y) = 10.
(a) Explain why W, X, and Y cannot be treated as a random sample
of size 3.
Since W, X, and Y, are not identically distributed (i.e., they
do not share a common distribution), these three random
variables cannot be treated as a random sample.
(b) Find the joint p.d.f. of (W, X, Y).
The joint p.d.f. is f(w,x,y) = e– w
e– x / 5 e– y / 10
—— —— =
5
10
– w – x / 5 – y / 10
e
——————
50
if w > 0, x > 0, y > 0
(c) Find P[max(W, X, Y) < 2].
P[max(W,X,Y) < 2] = P(W < 2  X < 2  Y < 2 ) =
2
2
2
–x/5
– y / 10
e
e
P(W < 2) P(X < 2) P(Y < 2) = e– w dw —— dx —— dy =
5
10
0
0
(1 – e– 2)(1 – e– 2 / 5) (1 – e– 2 / 10)
0
2. - continued
(d) Find the mean and variance of 4X + 3Y − 5W .
E(4X + 3Y − 5W) = 4E(X) + 3E(Y) − 5E(W) =
4(5) + 3(10) − 5(1) = 45
Var(4X + 3Y − 5W) = 42Var(X) + 32Var(Y) + 52Var(W) =
42(25) + 32(100) + 52(1) = 1325
(e) Find E(XY – W2).
E(XY − W2) = E(X)E(Y) − E(W2) =
E(X)E(Y) − {Var(W) + [E(W)]2} =
(5)(10) − {1 + 12} = 48