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University of Illinois Fall 2006
ECE 413: Solutions to Problem Set 10
- (a) Let An denote the event that the mailman is not bitten on the n-th day. Then, {X = n} = Acn ∩ An− 1 ∩ An− 2 ∩ · · · ∩ A 1 and {X > n} = An ∩ An− 1 ∩ An− 2 ∩ · · · ∩ A 1. We are given that P {Acn|An− 1 ∩ An− 2 ∩ · · · ∩ A 1 } = 1 n + 1 and hence P {An|An− 1 ∩ An− 2 ∩ · · · ∩ A 1 } = n n + 1
Note also that P {A 1 } = P {Ac 1 } =
. Therefore,
P {X > n} = P {An ∩ An− 1 ∩ An− 2 ∩ · · · ∩ A 1 } = P {An|An− 1 ∩ · · · ∩ A 1 }P {An− 1 |An− 2 ∩ · · · ∩ A 1 } · · · P {A 2 |A 1 }P {A 1 } = n n + 1
×
n − 1 n
× · · · ×
×
n + 1
Hence pX (n) = P {X = n} = P {X > n − 1 } − P {X > n} =
n −^
n + 1 =^
n(n + 1) , n^ = 1,^2 ,.. ..
(b) E[X ] =
∑^ ∞
n=
n · P {X = n} =
∑^ ∞
n=
n ·
n(n + 1) =
∑^ ∞
n=
n + 1 =^ ∞^ from the fact that the harmonic series diverges.
(c) E[X ] =
∑^ ∞
n=
P {X > n} =
∑^ ∞
n=
n + 1
(d) Obviously, for n = 1, 2 ,.. ., P {Y = n} = P {Y = −n} =
2 n(n + 1)
. Hence,
E[Y] =
∑^ ∞
n=
n · P {X = n} +
∑^ ∞
n=
(−n) · P {X = −n} =
∑^ ∞
n=
n + 1 −
∑^ ∞
n=
n + 1 =^ ∞ − ∞^ and thus we see that E[Y] is undefined.
- A random variable uniformly distributed on [a, b] has mean a + b 2 and variance (b − a)^2 12 . Hence, we have that b − a = 6, and b + a = 2, giving a = − 2 , b = 4. The pdf is thus as shown below.
(a) By inspection, P {X < 0 } = 13.
(b) E[|X |] =
[ ∫ 0
− 2
−u du +
0
u du
]
[
−u^2 2
0
− 2
u^2 2
4
0
=^1
[2 + 8] =
- (a) It is easy to see that {Y > T } = ({X 1 > T } ∩ {X 2 > T } ∩ {X 3 > T }) ∪ ({X 1 > T } ∩ {X 2 > T } ∩ {X 3 ≤ T }) ∪ ({X 1 > T } ∩ {X 2 ≤ T } ∩ {X 3 > T }) ∪ ({X 1 ≤ T } ∩ {X 2 > T } ∩ {X 3 > T }) (b) Since P {Xi > T } = exp(−λT ), we get that P {Y > T } = (exp(−λT ))^3 + 3(exp(−λT ))^2 (1 − exp(−λT )) = 3 exp(− 2 λT ) − 2 exp(− 3 λT ).
Hence, E[Y] =
0
P {Y > T } dT =
0
3 exp(− 2 λT ) − 2 exp(− 3 λT ) dT =
2 λ
3 λ
λ−^1.
(c) Let p = exp(−λT ). Then, P {Y > T } = 3p^2 − 2 p^3 = 12 , which has only one solution p = 1/2 in the range 0 ≤ p ≤ 1. Thus, the median value of Y is the solution T to exp(−λT ) = 1/2, giving T = λ−^1 ln 2. (d) The MTBF λ−^1 of the single module is larger (!) than the MTBF of the TMR system. The median lifetimes are the same. The TMR system does not result in a more reliable system if MTBF and median lifetimes are the criteria. (e) If λ = − ln 0.999, then exp(−λ)) = 0.999. We have that P {X 1 > 1 } = exp(−λ)) = 0.999 while P {Y > 1 } = 3(0.999)^2 − 2(0.999)^3 = 0.999997002 so that the TMR system is likely to provide much more reliability for one unit of time. (f) The largest value of T for which P {Y > T } ≥ 0 .999 is −λ−^1 ln 0. 98163 ≈ 18 .53. Thus, the TMR system can be expected to work (with 99.9% reliability) for more than 18 units of time whereas the single module provides this level of reliability for just one unit of time. It is in matters such as these that the TMR systems shines... not in terms of comparisons of MTBF or median lifetimes.
- (a) The derivative of exp(−u^2 /2) is −u exp(−u^2 /2). Hence,
E[|X | ] = 2
0
u · √^1 2 π
exp(−u^2 /2) =
π
[
− exp(−u^2 /2)
∞
0
π
More generally, if X ∼ N (μ, σ^2 ), then E[|X − μ| ] = σ
π and is known as the absolute error. (b) Since t + x > t − x > 0, we have that (t + x)(t − x) = t^2 − x^2 > (t − x)^2 > 0. Hence,
exp(x^2 /2)Q(x) =
x
2 π
exp
t^2 − x^2 2
dt ≤
x
2 π
exp
(t − x)^2 2
dt =
- (a) P {X < 0 } = Φ
= Φ(5) = 1 − Q(5).
(b) P {− 10 < X < 5 } = Φ
2
2
= Φ(7.5) − Φ(0) = Φ(7.5) − 12 = 12 − Q(7.5).
(c) P { |X | ≥ 5 } = P { X ≤ − 5 }+P { X ≥ 5 } = Φ
2
2
1 − Q(2.5) + Q(7.5).
(d) P {X 2 − 3 X + 2 > 0 } = P {(X − 1)(X − 2) > 0 } = P {X < 1 } + P {X > 2 } = Φ
1 −(−10) 2
2 −(−10) 2
= Φ(5.5) + 1 − Φ(6) = 1 − Q(5.5) + Q(6).
- Let X ∼ N(0. 9 , 0. 0032 ) denote the width (in microns) of the trace.
(a) {X < 0. 9 − 0. 005 } or {X > 0 .9 + 0. 005 } for a trace to be deemed defective. P { |X − 0. 9 | > 0. 005 } = 2Φ(− 0. 005 / 0 .003) = 2Φ(− 1. 666.. .) = 2Q(1. 666.. .) ≈ 0 .095. (b) We need to find the maximum value of σ such that 2Q(0. 005 /σ) ≤ 10 −^2. Since Q(2.575) ≈ 0 .005, we get that 0. 0005 /σ > 2. 575 , that is, σ ≤ 0. 005 / 2. 575 ≈ 0 .00194.
