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ECE 534 RANDOM PROCESSES SPRING 2008
SOLUTIONS TO PROBLEM SET 1
Ex.1. Exercise 1
- The sample space Ω = {HH, HT, T H, T T }, where T represents a tail and H a head.
- The event space F is the power set of Ω, i.e. all the subsets of Ω,
F ={∅, Ω, {HH}, {HT }, {T H}, {T T }, {HH, HT }, {HH, T H}, {HH, T T }, {HT, T H}, {HT, T T }, {T H, T T }, {HH, HT, T H}, {HH, HT, T T }, {HH, T H, T T }, {HT, T H, T T }}
- Note in each toss, the probability of seeing a head is PH = 12 × 12 + 12 = 34 and the probability of seeing a tail is PT = 1 − PH = 14. Furthermore, the two tosses are independent. So the probability measure P on the event space F is the product measure across the two tosses, and is given by (in the same order as above):
{ 0 , 1 ,
- By independence of the tosses,
P ({toss 1 = H}|{toss 2 = T }) = PH = 3/ 4
Ex.2. Exercise 2 (a)
A, Ac^ independent ⇐⇒ P (A ∩ Ac) = P (A)P (Ac) ⇐⇒ 0 = P (∅) = P (A)P (Ac) ⇐⇒ P (A) = 0 or P (A) = 1
Note that we cannot say A = ∅ or A = Ω since there can be nonempty sets whose probability measure equals 0.
(b)
A, Ac^ independent given C ⇐⇒ P (A ∩ Ac|C) = P (A|C)P (Ac|C) ⇐⇒ 0 = P (∅) = P (A|C)P (Ac|C) ⇐⇒ P (A|C) = 0 or P (A|C) = 1 ⇐⇒ P (A ∩ C) = 0 or P (A ∩ C) = P (C)
Because P (C) 6 = 0 and by use of Bayes’s rule. For the same reason as in (a), we cannot say A ∩ C = ∅ or C ⊂ A.
1.3. Congestion at output ports (a) One possibility is Ω = { 1 , 2 ,... , 8 }^4 = {(d 1 , d 2 , d 3 , d 4 ) : 1 ≤ di ≤ 8 for 1 ≤ i ≤ 4 }, where the packets are assumed to be numbered one through four, and di is the output port of packet i. Let F be all the subsets of Ω, and for any event A, let P [A] = | 8 A 4 |. (b)
P {X 1 = k 1 ,... , X 8 = k 8 } =
k 1 k 2 · · · k 8
where
k 1 k 2 ···k 8
= (^) k 1 !k 2 4!!···k 8! is the multinomial coefficient. (c) One way to do this problem is to note that Xj =
i=1 Xij^ , where^ Xij^ = 1 if packet^ i^ is routed to output port j, and Xij = 0 otherwise. Suppose j 6 = j′. Then Xij Xij′^ ≡ 0, and so also, E[Xij Xij′ ] = 0. Thus, Cov(Xij , Xij′ ) = 0 − 182 = − 641. Also, Cov(Xij , Xi′j′ ) = 0 if i 6 = i′. Thus,
Cov(Xj , Xj′^ ) = Cov(
∑^4
i=
Xij ,
∑^4
i′=
Xi′j′^ )
∑^4
i=
∑^4
i′=
Cov(Xij , Xi′j′ )
∑^4
i=
Cov(Xij , Xij′^ ) = 4(−
(d) Consider the packets one at a time in order. The first packet is routed to a random output port. The second is routed to a different output port with probability 78. Given the first two packets are routed to different output ports, the third packet is routed to yet another output port with probabilty 68. Similarly, given the first three packets are routed to distinct output ports, the fourth packet is routed to yet another output port with probability 58. The answer is thus 8 · 7 · 6 · 5 84 =^
105 256 ≈^0.^410. (e) The event is not true if and only if there are either exactly 3 packets assigned to one output port or all four packets assigned to one output port. There are 4 · 8 · 7 possibilities for exactly three packets to be assigned to one output port, since there are four choices for which packet is not with the other three, eight choices of output port for the group of three, and given that, seven choices of output port for the fourth packet. There are 8 possibilities for all four packets to be routed to the same output port. Thus, some output port has three or more packets assigned to it with probability 4 ·^88 ·7+8 4 = 4 ·7+1 83 = 51229 ≈ 0 .0566. Thus, P {Xi ≤ 2 for all i} = 1 − 51229 ≈ 0 .9434. 1.5. Conditional probability of failed device given attempt (a) P [first attempt fails]=0.2+(0.8)(0.1)=0. (b) P [server is working | first attempt fails ] = P [server working, first attempt fails]/P [first attempt fails] =(0.8)(0.1)/0.28≈ 0. (c) P [second attempt fails | first attempt fails ] =P [first two attempts fail]/P [first attempt fails] = [0.2 + (0.8)(0.1)^2 ]/ 0. 28 ≈0. (d) P [server is working | first and second attempts fail ] =P [server is working and first two attempts fail]/P [first two attempts fail] = (0.8)(0.1)^2 /[0.2 + (0.8)(0.1)^2 ] ≈0.
1.7. Conditional lifetimes and the memoryless property of the geometric distribution (a) P [X > 3] = 1 − p(3) = 0.8, P [X > 8 |X > 5] = P^ [{X> P 8 [X>}∩{5]X> 5 }]= P P^ [[X>X>8]5] = (^0).^040 = 0. (So a five year old working battery is not equivalent to a new one!) (b) P [Y > 3] = P [miss first three shots] = (1 − p)^3. On the other hand,
P [Y > 8 |Y > 5] =
P [{Y > 8 } ∩ {Y > 5 }]
P [Y > 5]
P [Y > 8]
P [Y > 5]
(1 − p)^8 (1 − p)^5 = (1 − p)^3.
(A player that has missed five shots is equivalent to a player just starting to take shots.) (c) Y has a geometric distribution. (Part (b) illustrates the fact that the geometric distribution is the memoryless lifetime distribution on the positive integers. The exponential distribution is the
(b) By inspection, fX (x) =
- 5 x 0 ≤ x ≤ 1
- 5 1 ≤ x ≤ 2 0 .5(3 − x) 2 ≤ x ≤ 3 0 else (c) Since the density of X is symmetric about 1.5 and the mean exists, E[X] = 1. 5. E[X^2 ] = 0 .5[
0 x
(^3) dx + ∫^2 1 x
(^2) dx + ∫^3 2 x
(^2) (3 − x)dx] = 0.5[ 1 4 +^
7 3 +^
11 4 ] =^
8 3 , so Var(X) =^
8 3 −^ (^
3 2 )
12.^ A slicker way to find the variance is to observe the X has the same distribution as U 1 + 2U 2 , where U 1 and U 2 are independent and uniformly distributed over [0, 1], so Var(X) = Var(U 1 ) + 4Var(U 2 ) = 125. (d) If 0 ≤ x ≤ 1, the conditional density of Y given X = x is the uniform density over the interval
[0, x 2 ]. That is, for 0 < x ≤ 1: fY |X (y|x) =
x 0 ≤^ y^ ≤^
x 2 0 else (e) By inspection, if 1 ≤ x ≤ 2, the conditional density of Y given X = x is the uniform density
over the interval [ x− 2 1 , x 2 ]. That is, for 1 < x ≤ 2: fY |X (y|x) =
2 x− 2 1 ≤ y ≤ x 2 0 else (f) E[Y |X = x] is well defined over the support of fX , namely, over the interval [0, 3]. For each X in this interval, the conditional density of Y give X = x is a uniform density, so the conditional
mean is the midpoint of the interval. Therefore. E[Y |X = x] =
x/ 4 0 ≤ x ≤ 1 (x − 0 .5)/ 2 1 ≤ x ≤ 2 (x + 1)/ 4 2 ≤ x ≤ 3 undefined x 6 ∈ [0, 3]
1.29. Transformation of densities (a)
0
0 (u^ −^ v) (^2) dudv = ∫^1 0
0 (u (^2) − 2 uv + v (^2) )dudv = 1 6 , so^ c^ = 6. (b) The map from the u, v plane to the x, y plane given by x = u^2 and y = u^2 v^2 maps the unit square [0, 1] × [0, 1] into the triangular region 0 ≤ y ≤ x ≤ 1 in one-to-one fashion. The inverse
mapping is given by u = v^1 /^2 and v = (y/x)^1 /^2. Also,
∣∣ ∂(x,y) ∂(u,v)
2 u 0 2 uv^2 2 u^2 v
∣ = 4u
(^3) v = 4xy 1 / (^2).
Therefore,
fXY (x, y) = fU V (u, v)
∣∣^ ∂(x, y) ∂(u, v)
− 1
6(x^1 /^2 − (y/x)^1 /^2 )^2 4 xy^11 / 2 if 0 ≤ y ≤ x ≤ 1 0 else
Ex.11. Exercise 11 (a) By definition, ΦX (u) = E[ejuX^ ]. In order to do the calculation, we use the inverse Fourier Transform (most of the useful ones are available in undergraduate linear systems books). Here we are using the definition f (t) = (^21) π
−∞
a π(a^2 +ω^2 ) e
j ωt (^) dω and the formula F− (^1) { 2 α α^2 +ω^2 }^ =^ e
−α|t|. Thus, we have ΦX (u) = e−a|u|.
(b) Since X and Y are independent,
ΦZ (u) = ΦX+Y (u) = E[eju(X+Y^ )] = E[eju(X)eju(Y^ )] = ΦX (u)ΦY (u) = e−(a+b)|u|
Now, using the formula for the Cauchy density, we conclude that
fZ (z) =
a + b π((a + b)^2 + x^2 )
