Download Solutions to Problem Set #7 in ECE 413, Fall 2004, University of Illinois and more Assignments Statistics in PDF only on Docsity!
of Illinois Page 1 of 3 Fall 2004
- E[Points] = 10(1/10) + 5(2/10) + 3(2/10) = 2. 6
- (a)
E[|X − a|] =
∫ A
a
x − a A dx^ +
∫ (^) a
0
a − x A dx^ =^
A
a −
a^2 A
d da
(·) =^2 a A
− 1 = 0 ⇒ a = A/ 2
(b)
E[|X − a|] =
0
(a − x)λe−λxdx +
∫ A
a
(x − a)λe−λxdx
= a(1 − e−λa) + ae−λa^ + e
−λa λ
λ
−λa λ
− ae−λa
Dierentiation yields that the minimum is attained at a where
e−λa^ = 1/ 2 or a = log(2/λ)
Since the exponential distribution is memoryless, the probability that the radio will be working after an additional years is 1 − F (8) = e−^1.
E[Xk] =
0
xkλe−λxdx
= λ−k
0
(λx)kλe−λxdx
= λ−kΓ(k + 1) = k!/λk
- Using Equation (6.3) on p. 222 of Ross,
E[X] = B(a^ + 1, b) B(a, b)
= Γ(a^ + 1) Γ(a + b + 1)
Γ(a + b) Γ(a)
= a a + b E[X^2 ] =
B(a + 2, b) B(a, b) =^
Γ(a + 2) Γ(a + b + 2)
Γ(a + b) Γ(a) =^
(a + 1)a (a + b + 1)(a + b)
Thus,
V ar(X) = E[X^2 ] − (E[X])^2 = (a^ + 1)a (a + b + 1)(a + b)
− a
2 (a + b)^2
= ab (a + b + 1)(a + b)^2
of Illinois Page 2 of 3 Fall 2004
E[X] =
0
x2(e−^3 x^ + e−^6 x)dx = 2
0
xe−^3 xdx +
0
xe−^6 xdx
Likewise,
E[X^2 ] =
0
x^2 2(e−^3 x^ + e−^6 x)dx = 2
0
x^2 e−^3 xdx +
0
x^2 e−^6 xdx
Hence, V ar(X) = E[X^2 ] − (E[X])^2 =
54 −^
182 =^
Note that P (|X| < 2) = P (X < 2) since X is positive valued. Therefore,
P (|X| < 2) =
0
2(e−^3 x^ + e−^6 x)dx = 2
0
e−^3 xdx +
0
e−^6 xdx
3 e
− 2 / 3 +^1
6 e
− 2 / 6
E[X] =
−∞
xf (x)dx + 1 · P (X = 1) +
1
xf (x)dx + 2 · P (X = 2) +
2
xf (x)dx
1
x 3 dx^ + 2^ ·^
E[X^2 ] =
−∞
x^2 f (x)dx + 1 · P (X = 1) +
1
x^2 f (x)dx + 4 · P (X = 2) +
2
x^2 f (x)dx
1
x^2 3
dx + 4 · 1 3
=^22
Hence, V ar(X) = E[X^2 ] − (E[X])^2 = 367.
Note that P (|X − 1 | < 1) = P (0 < X < 2). Therefore,
P (|X − 1 | < 1) =
0
f (x)dx + P (X = 1) +
1
f (x)dx =
0
3 dx^ +
