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University of Illinois Fall 2003
ECE 313: Solutions to Problem Set
- (a)
μ = E[X] =
∫ (^3) 2
xf (x)dx
=
∫ (^3) 2
x 3 − 2 dx
=
E[X^2 ] =
∫ (^3) 2
x^2 3 − 2 dx
=
Hence,
σ^2 = V ar[X] = E[X^2 ] − μ^2 =
(b) From Chebyshev’s inequality,
P (|X − μ| ≥ k) ≤ σ^2 k^2 For k = 2σ, we have P (|X − μ| ≥ 2 σ) ≤
σ^2 4(σ^2 )
Thus, the upper bound given by Chebyshev’s inequality is 0.25. (c) P (|X − μ| ≥ 2 σ) = P {(X − μ) ≥ 2 σ, X ≥ μ} + P {(μ − X) > 2 σ, X < μ}
= P {X ≥ (2.5 + .577)} + P {X < (2. 5 − .577)}
= P (X ≥ 3 .077) + P (X < 1 .923) = 0
σ Y^2 = V ar[Y ] = V ar(αX + β) = α^2 σ^2 X = 9α^2 Thus, if σ Y^2 = 5^2 ,
α = ±
√ σ^2 Y 9
Also, μY = E[Y ] = E[αX + β] = αμX + β Therefore, for μY = 2
β = 2 − α · 1 =
1 3 ,^ α^ =^
5 3 11 3 ,^ α^ =^ −
5 3
- (a) FY , the cumulative distribution function of Y , is given by
FY (y) = P {αX + β ≤ y}
P
{ X ≤ y−αβ
} = FX
( y−β α
) , (α > 0)
P
{ X ≥ y−αβ
} = 1 − FX
( y−β α
) , (α < 0)
Differentiation yields that the density function of Y is
fY (y) =
|α| fX
( (^) y − β α
)
1 |α| ,^2 ≤^
y−β α ≤^3
0 , otherwise where
y − β α ≤ 3 is equivalent to
2 α + β ≤ y ≤ 3 α + β, (α > 0)
2 α + β ≥ y ≥ 3 α + β, (α < 0)
This shows that
Y follows uniform distribution over the interval
(2α + β, 3 α + β), (α > 0)
(3α + β, 2 α + β), (α < 0)
(b) Solving the equations
2 α + β = 0
3 α + β = 1
( for^ α >^ 0),
3 α + β = 0
2 α + β = 1
( for^ α <^ 0)
we have α = 1, β = − 2 , given α > 0 , and α = − 1 , β = 3, given α < 0.
- Proof by Induction Base case: For k = 1,
E[X^1 ] = −xe−λx
∣∣ ∣∣ ∞ 0
∫ (^) ∞ 0
e−λxdx
= 0 − e−λx λ
∣∣ ∣∣
∞ 0 =
λ
λ^1
- (a) Standard Gaussian RV: Z ∼ N (0, 1)
P (X < 0) = P
( X − (−1)
) = P
( Z <
)
(with Calculator1) = Φ
)
(with Calculator 2) = 1 − Q
)
(b)
P (− 10 < X < 5) = P
X − (−1)
)
= P
< Z < 3
) = Φ(3) − Φ
)
(with Calculator 1) = Φ(3) + Φ
) − 1
= (1 − Q(3)) +
( 1 − Q
)) − 1
(with Calculator 2) = 1 − Q(3) − Q
)
(c)
P (|X| ≥ 5) = 1 − P (|X| ≤ 5) = 1 − P (− 5 ≤ X ≤ 5) = 1 − P (− 2 ≤ Z ≤ 3) = 1 − (Φ(3) − Φ(−2)) = 1 − Φ(3) + 1 − Φ(2) (with Calculator 1) = 2 − Φ(3) − Φ(2) (with Calculator 2) = Q(3) + Q(2)
- Let g(x) = ax^ for some constant a. Then g(s)g(t) = asat^ = as+t^ = g(s + t). This proves the forward part. Proof of the converse: If g(s + t) = g(s)g(t), then g( (^) n^2 ) = g( (^1) n + (^1) n ), and repeating this yields g(mn ) = gm( (^1) n ). Also, g(1) = g(
n
n
︸ ︷︷ n︸ n times
) = gn( (^) n^1 ), or g( (^1) n ) = (g(1))^1 /n.
Hence g(mn ) = (g(1))m/n. Since all rational numbers can be expressed in the form mn , this proves the converse for all rational numbers. To extend this result to any arbitrary real number x, we consider the sequence of rational numbers rn such that limn→∞ rn = x. Since g(x) is continuous, we can write limn→∞ g(rn) = g(limn→∞ rn) = g(x). But
nlim→∞ g(rn) = limn→∞(g(1))rn^ = (g(1))limn→∞^ rn^ = (g(1))x Therefore, g(x) = (g(1))x, for all real numbers x.
