Download Solutions to University Problem Set #6 in ECE 313, Summer 2003 and more Assignments Statistics in PDF only on Docsity!
of Illinois Page 1 of 4 Summer 2003
1. ω in Prob. Value of X (ω)
AcBcCc^ 1/8 0 ( a) Hence, X takes on values –1, 0, 1, 2, and 3. AcBcC 1/8 – AcBCc^ 1/8 2 (b) The pmf of X can be seen to be given by p X (–1) = 1/8, AcBC 1/8 1 p X (0) = 1/4 = p X (1) = p X (2), and p X (3) = 1/8. ABcCc^ 1/8 1 The CDF of X can be readily written down from this data. ABcC 1/8 0 ABCc^ 1/8 3 ABC 1/8 2
F(u) =
Note that the function is right continuous!
u
u
u
u
u
u
2.(a),(b) As studied in class, X is an exponential random variable with parameter μ. Hence, P{ X > τ} = exp(–μτ) and f X (u) = μ•exp(–μu) for u > 0, and 0 otherwise.
(c) The number of arrivals in the time interval (0, 6] is a Poisson random variable N (0,6] with parameter 6μ,
and we get P{ N (0,6] = 4} =
(6μ) 4 4! •exp(–6μ). (d) The number of arrivals in (4, 10] is N (4,10] is also a Poisson random variable with parameter 6μ, and
hence P{ N (4,10] = 2} =
(6μ) 2 2! •exp(–6μ). We are now asked for P[{ N (0,6] = 4}^ ∩^ { N (4,10] = 2}. Now, (0, 6] and (4, 10] are intersecting intervals of time, but we can partition them into the disjoint intervals (0, 4], (4, 6], and (6,10] and thus write P[{ N (0,6] = 4} ∩ { N (4,10] = 2} = P[{ N (0,4] = 4} ∩ { N (4,6] = 0} ∩ { N (6,10] = 2}]
- P[{ N (0,4] = 3} ∩ { N (4,6] = 1} ∩ { N (6,10] = 1}]
- P[{ N (0,4] = 2} ∩ { N (4,6] = 2} ∩ { N (6,10] = 0}]
= ^
(4μ) 4
4! •exp(–4μ)^ • [^ exp(–2μ)]^ •
2 2! •exp(–4μ)
+ ^
(4μ) 3
3! •exp(–4μ)^ •
1! exp(–2μ)^ •
1! •exp(–4μ)
+ ^
(4μ) 2
2! •exp(–4μ)^ •
2
2! •exp(–2μ)^ • [^ exp(–4μ)]
because occurrences in disjoint intervals are independent and hence we can multiply the probabilities.
Thus, P[{ N (0,6] = 4} ∩ { N (4,10] = 2} = ( )
256
3 •exp(–10μ)•
μ^6 + μ^5 +
3 16 μ
(c) P{ X > τ | A} is obviously 0 if τ ≥ 6, while for 0 ≤ τ < 6, P{ X > τ | A} = P{No arrivals in (0,τ] | 4 arrivals in (0,6]}
= { 4 ( 0 , 6 ]}
{ 0 ( 0 , ] 4 ( 0 , 6 ]}
P in
P in τ ∩ in
( )
{ 0 ( 0 , ]}{ 4 (, 6 ]}
P A
P in τ P in τ
[exp(–μτ)•exp(–μ(6–τ))•(μ(6–τ)) 4 /4!] exp(–6μ)•(6μ) 4 /4!
(6–τ) 4 6 4
4 which has value 1 at τ = 0 and approaches 0
as τ approaches 6.
3. X is N(500,100 2 ). Thus,
(a) P{ X < 700} = Φ((700–500)/100) = Φ(2) = 0.9772, so 97th percentile (b) Φ(1.645) ≈ 0.95 so a “score” of 564.5 puts you at the 95th percentile. (c) P{300 < X < 550} = Φ(550–500)/100) – Φ((300–500)/100) = Φ(0.5) – Φ(–2) = Φ(0.5) – (1 – Φ(2)) = Φ(0.5) + Φ(2) – 1 = 0.6915 + 0.9772 – 1 = 0.
of Illinois Page 2 of 4 Summer 2003
4.(a) d/du(exp–u 2 /2)) = –u exp(–u 2 /2). Hence,
E[| X |] = ⌡⌠
|u|•φ(u)du = 2⌡⌠
u•φ(u)du = 2
0
∞ u•
2 π
exp(-u 2 /2)du =
π • –exp(–u^
0
π < 1.
More generally, E[| X – μ|] = σ
π if the pdf of^ X^ is^ N(μ,σ
(^2) ). In statistical applications, E[| X – μ|] is
sometimes called the absolute error. (b) Since t + x > t – x > 0, (t + x)(t – x) = t 2 – x 2 > (t – x) 2. Hence, exp–(t^2 – x 2 ) < exp–(t – x) 2 and therefore
exp^
x 2
2 Q(x) =^ ⌡⌠
x
∞ 1 2 π
exp^ –
t^2 – x 2
2 dt <⌡⌠
x
∞ 1 2 π
exp^ –
(t – x) 2 2 dt = 1/2. What? Why 1/2?
5.(a) As in Problem 4(a), E[| X |] =
π (b) Since P{ X > 0} = 1/2, the conditional pdf is just the right half of the Gaussian pdf with a doubling in
height (to ensure that the area under the curve is 1), i.e. f X | X > 0(u| X > 0) =
2 π
exp(-u 2 /2)u > 0
0,u ≤ 0
(c) E[ Y ] =
0
∞ u•
2 π
exp(-u 2 /2)du =
π • –exp(–u^
0
π
6.(a) Obviously P{ Y = α} = P{ Y = –α} = 1/2.
(b) (1.29–1) = 0.29. (1.29 –1) 2 = 0.0841. (π/4–1) = –0.214…, (π/4–1) 2 = 0.046….
(–π/4–(–1)) = –0.214…, (–π/4–(–1)) 2 = 0.046…. Note that the error for + X is the same as that for – X.
(c) E[ Z ] = ⌡⌠
0
∞
(u–α) 2 ƒ(u)du + ⌡⌠
0
(u+α) 2 ƒ(u)du = ⌡⌠
∞
(u 2 + α^2 )ƒ(u)du – 4α⌡⌠
0
∞ uƒ(u)du = 1+α^2 –4α/ 2 π on
expanding out the quadratics, changing variables, and using the fact that E[ X^2 ] = σ^2 + μ^2 = 1. Note that uf(u) is a perfect integral. It is easy to show that E[ Z ] has minimum value 1–2/π at α = 2/π (d) From tables of Φ(•), we get P{ W = –3} = P{ W = +3} = Φ(–2.5) = 0.0062, P{ W = 0} = Φ(0.5) – Φ(–0.5) = 0.3830, P{ W = –1} = P{ W = +1} = Φ(1.5) – Φ(0.5) = 0.2417, and P{ W = –2} = P{ W = +2} = Φ(2.5) – Φ(1.5) = 0.0606. (e) P{ Z 2 = 1} = P{ W < 0} = 0.3085.
P{ Z 1 = 1} = P{ W = 2} + P{ W = 3} + P{ W = –1} + P{ W = –2} = 0. P{ Z 0 = 0} = P{ W = 2} + P{ W = 0} + P{ W = –2} = 0.5042 ≠ P{ Z 0 = 1}.
7.(a) The lifetime is an exponential random variable with parameter λ. Hence, the average lifetime is λ–1^ = (–ln 0.999)^ –1^ ≈^ 999.5 weeks which is more than 19 years! The^ median^ lifetime is T where P{ X > T} = exp(–λT) = 1/2, i.e. T = λ–1^ ln 2 ≈ 692.8 weeks which is only 13 years and a few months. (b) P{ X > 1} = exp(–λ) = 0.999 so even with this long-lived module, we can only be 99.9% sure that the thing is going to last for a week! (c) We can be sure that at least two of the three events { X 1 > t}, { X 2 > t}, { X 3 >t} must have occurred. In fact,
{ Y > t} is the (disjoint) union of the four events { 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}, and { X 1 ≤ t}∩{ X 2 > t}∩{ X 3 > t}.
(d) Since P{ X i > t} = exp(–λt), we readily find that the probability of the first event in (c) is exp(–3λt), while
the other three have probability exp(–2λt) [1–exp(–λt)]. Hence, P{ Y > t} = 3 exp(–2λt) – 2 exp(–3λt).
Furthermore, E[ Y ] = ⌡⌠
P{ Y > t} dt = ⌡⌠
3 exp(–2λt) – 2 exp(–3λt) dt = (3/2λ) – (2/3λ) = (5/6λ) < 1/λ
so that the average lifetime of the TMR system is actually smaller than that of the individual modules! To find the median lifetime, we set P{ Y > T} = 3 exp(–2λT) – 2 exp(–3λT) = 3p 2 – 2p 3 = 1/2. It is easy to
of Illinois Page 4 of 4 Summer 2003
10.(a) From the figure shown above, we see that the pdf is symmetric about u = 0. Hence, we get that
P{ X ≤ ln 2} =
2 +^ ⌡
ln 2 1 2 exp(–u) du =
2 +^
–^1
2 exp(–u)^
ln 2
- Notice that P{0^ ≤^ X^ ≤^ ln 2} =
(b) P{| X | ≤ ln 2 | X ≤ ln 2} = { ln 2 }
{| | ln 2 } { ln 2 }} ≤
X
X X
P
P
P{| X | ≤ ln 2} P{ X ≤ ln 2} =
2P{0 ≤ X ≤ ln 2} 3/4 =
(c) cos(π X /2) = 0 if X is an odd integer, and P{cos(π X /2) < 0} = … + P{–7 < X < –5} + P{-3 < X < –1} + P{1 < X < 3} + P{5 < X < 7} + … = 2[P{1 < X < 3} + P{5 < X < 7} + …} (by symmetry) = [exp(–1) – exp(–3)] + [exp(–5) – exp(–7)] + … = exp(–1)•
1+exp(–2) =^
2•cosh(1) = 0.324… on summing the geometric series.
(d) The minimum value of I is –1. F I (b) = P{ I ≤ b} = 0 if b < –1.
For b ≥ –1, F I (b) = P{ I ≤ b} = P{e V^ – 1 ≤ b} = P{ V ≤ ln(b+1)} = F V (ln(b+1)). But,
F V (a) =
ea/2,a ≤ 0, 1 – e–a/2,a > 0.
Thus, F I (b) =
b+ 2 –1^ ≤^ b^ ≤^ 0,
1 –
2(b+1),b > 0,
and f I (b) =
2 –1^ ≤^ b^ ≤^ 0, 1 2(b+1)^2
, b > 0.
Note that the pdf has constant value 1/2 in the range –1 ≤ b ≤ 0.
