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University of Illinois, Urbana–Champaign Fall 2007 ECE 313: Solutions to Homework 8
- Let r.v. U ∼ Exp(3) and V ∼ Exp(6), then fX = 23 fU + 13 fV.
(a) Checking for non-negative property,∫ fX (x) ≥ 0 for all x. The total probability is ∞ 0 fX^ (b)db^ = 2/3 + 1/3 = 1. Therefore,^ fX^ is a valid pdf. (b)
E[X] =
E[U ] +
E[V ] =
E[X^2 ] =
E[U 2 ] +
E[V 2 ] =
var(X) = E[X^2 ] − (E[X])^2 =
(c) P (|X| < 2) = P (0 ≤ X < 2) = 23 P (0 ≤ U < 2) + 13 P (0 ≤ V < 2) = 1 − 23 e−^6 − 13 e−^12 =
9983
(a) The pdf for the continuous part of X is
fX (b) =
3 1 < b <^2 0 otherwise
The pmf for the discrete part of X is
pX (b) =
3 b^ = 2, b^ = 1 0 otherwise
Note that the
1 fX^ (b)db^ +^ pX^ (1) +^ pX^ (2) = 1. Then, we have
E[X] =
1
b 3
db +
E[X^2 ] =
1
b^2 3
db +
var(X) = E[X^2 ] − (E[X])^2 =
E[X] can also be obtained by summing up the shaded area.
Fx(b)
0 2 b
(b) E[3X^2 − 4] = 3E[X^2 ] − 4 = 10/3 = 3. 3333
- (a) According to the law of total probability,
−∞ ce
− 2 |x|dx = ∫^0 −∞ ce
2 xdx + ∫^ ∞ 0 ce
− 2 xdx = c = 1. Thus, c = 1. (b) By symmetry, E[X] = 0.
var(X) = E[X^2 ] = 2
0
x^2 e−^2 xdx = 1/ 2
E[Y ] = E[3X − 2] = − 2 var(Y ) = 9var(X) = 9/ 2
(c)
P (Y > 4) = P (3X − 2 > 4) = P (X > 2) =
2
e−^2 xdx =
e−^4 2
P (Y > 0) = P (3X − 2 > 0) = P (X >
e−^4 /^3 2
P (Y > 4 |Y > 0) =
P (Y > 4)
P (Y > 0)
= e−^8 /^3 = 0. 0695
(d) Since X is a continuous r.v except X = 0,
P (Y ≥ 4 |Y ≥ 0) = P (Y > 4 |Y > 0) = e−^8 /^3
P (Y < 4 |Y ≥ 0) = 1 − P (Y ≥ 4 |Y ≥ 0) = 1 − e−^8 /^3 = 0. 9305
- (a)
E[|X − a|] =
∫ (^) a
0
(a − x)
A
dx +
∫ A
a
(x − a)
A
dx =
A^2 2 −^ aA^ +^ a
2
A
A
((a −
A
)^2 +
A^2
Letting the square part be zero to obtain minimum, we have a = A/2. (b)
E[|X − a|] =
∫ (^) a
0
(a − x)λe−λxdx +
∫ A
a
(x − a)λe−λxdx = a +
2 e−aλ λ
λ
To find a that minimizes E[|X − a|], first we need to find the critical points by taking derivative of the RHS and let it equals zero. We got equation 1 − 2 e−aλ^ = 1. Solving for a, we got a = ln 2 λ. The second derivative 2λe−aλ^ > 0 for a = ln 2 λ. Thus it is the minimum.
- (a) Let X be the life time of one light bulb. X follows exponential distribution with Poisson rate λ = 10 per hour. Since exponential distribution is memoryless, the probability that the bulb is working at 11pm only depends on the two hours after last check at 9pm, which equals to the probability p that a bulb continuously work during the past two hours, s.t. p = P (X > 2) =
2 10 e
− 10 tdt = e− (^20). The probability that all 5 bulbs are working at 11pm is p^5 = e−^100 since they are independent.
(b) The probability that only two bulbs are still operating is
2
p^2 (1 − p)^3 = 10e−^40 (1 − e−^20 )^3 = 4. 25 × 10 −^17. (c) According to memoryless property, the probability that a bulb that was working at 11pm can last until midnight is P (X > 1) = e−^10. Thus the probability of two bulbs working is e−^20.
