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ECE/CS 313: Probability with Engineering Applications Fall 2002
Problem Set 10 Solutions
- Generation of random variables with specified probability density function By integration, FX (x) = x^2 for 0 ≤ x ≤ 1. It is desired that P [g(U ) ≤ x] = x^2. For fixed x, let u be the value so that g(u) = x. The event {g(U ) ≤ x} is thus equivalent to {U ≤ u}, which has probability u. Thus, we want u = x^2 , or equivalently, x =
u. That is, g(u) =
u for 0 ≤ u ≤ 1 does the trick. (Note: In general, for each x, the cumulative probability distribution function desired for g(U ) determines value u such that g(u) = x. In other words, the inverse of the function g is specified, from which we can find g.)
- Normal approximation to the binomial distribution The number of people S favoring the proposition is a binomial random variable with parameters (100, 0.65). It has mean μ = 65 and variance σ^2 = 100(0.65)(1 − 0 .65) = 22.75. The DeMoivre-Laplace limit theorem tells us that S is well approximated by a normal random variable with parameters μ and σ^2. (a) P [S ≥ 50] = P [ √S 22 −^65. 75 ≥ √^5022 −.^6575 ] ≈ 1 − Φ(− 3 .14) = Φ(3.14) = 0. 9992
(b) P [60 ≤ S ≤ 70] = P [ √^6022 −.^6575 ≤ √S 22 −^65. 75 ≥ √^7022 −.^6575 ] =≈ Φ(1.048) − Φ(− 1 .048) = 2Φ(1.048) − 1 = 0. 706
(c)P [S ≤ 75] = P [ √S 22 −^65. 75 ≤ √^7522 −.^6575 ] =≈ Φ(2.096) = 0. 9778
- The Poisson process
(a) P [N = n] = e
−μT (^) (μT )n n! (b) P [N 1 = i] = e
−μτ (^) (μτ )i i!
(c) P [N 2 = j] = e
−μ(T −τ )(μ(T −τ ))j j! (d)
P [N 1 = i|N = n] =
P [N 1 = i, N = n] P [N = n]
P [N 1 = i, N 2 = j] P [N = n]
P [N 1 = i]P [N 2 = j] P [N = n]
n! i!j!
τ T
)i(
T − τ T
)j
which can also be written as P [N 1 = i|N = n] =
(n i
pi(1 − p)n−i, where p = (^) Tτ. As a function of i, the answer is the pmf of a binomial probability. (e)
P [N = n|N 1 = i] =
P [N = n, N 1 = i] P [N 1 = i]
P [N 1 = i, N 2 = j] P [N 1 = i]
P [N 1 = i]P [N 2 = j] P [N 1 = i]
= P [N 2 = j] =
eμ(T^ −τ^ )(μ(T − τ ))j j!
which can also be written as P [N = n|N 1 = i] = e
−μ(T −τ )(μ(T −τ ))n−i (n−i)! That is, given^ N^1 =^ i, the total number of arrivals is i plus a random number of arrivals. The random number of arrivals has the Poisson distribution with mean μ(T − τ ).
- Waiting times for a Poisson process
(a) P [N (t) = i] = e
− 2 t(2t)i i! for nonnegative integers^ i^ (Poisson distribution, mean 2t) (b) P [E 1 ] = P [N (3.5) = 0] = e
− (^7) (7) 0 0! =^ e
P [E 2 ] = 0.00091 since E 2 is the same event as E 1
P [E 3 ] = P [N (3.5) ≤ 2] = P [N (3.5) = 0] + P [N (3.5) = 1] + P [N (3.5) = 2] = e−^7 (1 + 7 + 7
2 2 ) = 0.^0296 P [E 4 ] = 0.0296 since E 4 is the same event as E 3
P [E 5 ] = P [N (t) ≤ 2] = P [N (t) = 0] + P [N (t) = 1] + P [N (t) = 2] = e−^2 t(1 + 2t + (2t)
2 2 ) P [E 6 ] = 1 − P [E 5 ] = 1 − e−^2 t(1 + 2t + (2t)
2 2 ) (c) Differentiating yields f (t) = e−^2 t(2(1 + 2t + (2t)
2 2 )^ −^2 −^4 t) =^ e
− 2 t (2t)^2 2 ) which indeed is the gamma density with parameters (i = 3, μ = 2). (d) The expected time between arrivals is 1/μ, so the expected time until the 10th^ arrival is 10/μ = 5 minutes.
- Hypothesis testing for the variance of a Gaussian variate (a)
Λ(x) =
1 b
√ 2 π e
− 2 xb^22
1 a√ 2 π e
− 2 xa^22
a b
e−^
x^2 2 b^2 +^
x^2 2 a^2 =
a b
e
x 22 ( 1 a^2 −^
1 b^2 )
(b) The ML rule is to choose H 1 when Λ(X) > 1. Thus, by taking the natural logarithm of both sides of
this inequality we obtain the rule: If (ln a b ) + X
2 2 (^
1 a^2 −^
1 b^2 )^ >^ 0 choose^ H^1. After a bit of algebra, we derive the rule that H 1 should be chosen when
ln
b a
X^2
a^2
b^2
X^2
b^2 − a^2 a^2 b^2
2 a^2 b^2 b^2 − a^2
ln
b a
< X^2 ⇒ K = ab
2(ln b − ln a) b^2 − a^2
< |X|
(c) The MAP rule is to choose H 1 when Λ(X) > π π^01. After a bit of algebra, we derive the rule that H 1
should be chosen when
ln
bπ 0 aπ 1
X^2
a^2
b^2
X^2
b^2 − a^2 a^2 b^2
⇒ K = ab
2(ln(bπ 0 ) − ln(aπ 1 )) b^2 − a^2
< |X|
(d) pf alse alarm = P {|X| > K | H 0 } =
∫ −K
−∞ f^0 (u)du^ +^
K f^0 (u)du^ = Φ(−K/a) + 1^ −^ Φ(K/a) = 2Q(K/a). pmiss = P {|X| < K | H 1 } =
∫ K
−K f^1 (u)du^ = Φ(K/b)^ −^ Φ(−K/b) = 1^ −^2 Q(K/b).
- Hypothesis testing for the mean of a Laplace variate (a) It is helpful to explicitly note that
|u − 1 | =
−u + 1 : u < 1 u − 1 : u ≥ 1 ,^ |u^ + 1|^ =
−u − 1 : u < − 1 u + 1 : u ≥ − 1.
Thus,
f 1 (u) =
2 e
u− (^1) : u < 1 1 2 e
−u+1 (^) : u ≥ 1 f 0 (u) =
2 e
u+1 (^) : u < − 1 1 2 e
−u− (^1) : u ≥ − 1
(b)
Λ(u) =
e−|u−^1 | e−|u+1|^
eu−^1 eu+1^ =^ e
− (^2) : u < − 1
eu−^1 e−u−^1 =^ e
2 u (^) : − 1 ≤ u < 1
e−u+ e−u−^1 =^ e
(^2) : 1 < u
(c) The likelihood ratio Λ(u) is nondecreasing and it crosses 1 at u = 0. Thus, the ML decision rule is to decide H 1 is true if X > 0 and decide H 0 otherwise. (d)
pf alse alarm =
0
f 0 (u)du =
0
e−u−^1 du =
e−^1
0
e−udu =
e−^1
By symmetry, we see that pmiss is also given by pmiss = 12 e−^1 (of course, one could always compute the
integral
−∞
1 2 e
−|u− 1 |du = ∫^0 −∞
1 2 e
(u−1)du = 1 2 e
− 1 ∫^0
−∞
1 2 e
udu = 1 2 e
(e) The MAP decision rule is to choose H 1 if Λ(u) > π π^01 = 2, and choose H 0 otherwise. Note that the
solution of e^2 u^ = 2 is u = ln 2 2 , which is between -1 and 1. Thus, the MAP decision rule is to choose H 1 if
X ≥ K and choose H 0 otherwise, where K = ln 2 2 = ln
(f)
pf alse alarm =
ln
√ 2
f 0 (u)du =
e−^1
ln
√ 2
e−udu =
e−^1 (−e−∞^ + e−^ ln^
√ 2 ) =
e−^1.
Similarly,
pmiss =
∫ (^) ln √ 2
−∞
f 1 (u)du =
e−^1
∫ (^) ln √ 2
−∞
eudu =
e−^1 eln^
√ 2
e−^1 =
e−^1.
