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ECE 534: Random Processes Spring 2010
Solutions to Problem Set 3
2.11 Convergence in distribution to a nonrandom limit Suppose P [X = c] = 1 and limn→∞ Xn = X d. Let > 0. It suffices to prove that P [|Xn − X| ≤ ] → 1 as n → ∞. Note that P [|Xn − X| ≤ ] ≥ P [c − < Xn ≤ c + ] = Fn(c + ) − Fn(c − ). Since c − is a continuity point of FX and FX (c − ) = 0, it follows that Fn(c − ) → 0. Similarly, Fn(c + ) → 1. Thus Fn(c + ) − Fn(c − ) → 1, so that P [|Xn − X| ≤ ] → 1. Therefore convergence in probability holds. Note: A slightly different approach would be to prove that for any > 0, there is an n so large that P [|Xn − c| ≤ ] ≥ 1 − .
2.12 Convergence of a minimum In this problem, the sequence (Xn) converges to zero in all four senses. The most direct way to prove this is to show a.s. convergence and augment convergence in the m.s. sense with the boundedness argument. The answer given at the end of the notes uses an alternative proof. Firstly it shows convergence in probability. The fact that for each ω the sequence (Xn(ω)) is nonincreasing and lower bounded gives us the existence of an a.s. limit. Finally we can claim that (Xn) converges a.s. to zero by applying Prop. 2.1.13 (e) (A sequence of random variables can have only one limit). Two common mistakes appear in the homework:
- Claiming the a.s. convergence by simply proving P [ω : limn→∞ Xn(ω) > 0]. However, this is not sufficient since the limit of (Xn(ω)) may not even exists. We need to say that the sequence is nonincreasing and lower bounded to eliminate this possibility.
- Claiming that limn→∞ Xn = 0 a.s. by saying Xn is nonincreasing and lower bounded by zero. Now this is not true since there exists ω ∈ Ω such that (Xn(ω)) does not converge to zero. We need to show that this event has probability zero.
Below is the proof of a.s. convergence. For each ω, the sequence (Xn(ω)) is nonincreasing and bounded below by zero and thus has a limit. Then
P
[
ω : lim n→∞ Xn(ω) > 0
]
= P
[ ∞
m=
ω : lim n→∞ Xn(ω) >
m
}]
∑^ ∞
m=
P
[
ω : lim n→∞ Xn(ω) >
m
]
∑^ ∞
m=
P
[
Un >
m for all n
]
Hence (Xn) converges a.s. to zero.
ECE 534 HW3 Spring 2010
2.13 Convergence of a product
(a) Examine Sn = ln Xn. The sequence Sn, n ≥ 1 is the sequence of partial sums of the independent and identically distributed random variables ln Uk. Observe that E[ln Uk] =
0 ln(u)^
1 2 du^ =^
1 2 (x^ ln^ x^ −^ x)|
2 0 = ln 2^ −^1 ≈ −^0 .306.^ Therefore, by the strong law of large numbers, lim{ n→∞ S nn = ln 2 − 1 a.s., i.e., P (S) = 1 where S = ω : limn→∞ Sn n(ω )= ln 2 − 1
. This means that, given ω ∈ S and > 0, there is a finite N such that | S nn − (ln 2 − 1)| ≤ for all n ≥ N. Equivalently, ( 2 e1+
)n ≤ Xn ≤
e^1 −
)n for n ≥ N.
For sufficiently small , both ends approach 0 as n goes to infinity. Hence
P
[
n^ lim→∞ Xn^ = 0
]
≥ P
[
n^ lim→∞
Sn n = ln 2 − 1
]
Conclude that limn→∞ Xn = 0 a.s., which implies that also limn→∞ Xn = 0 p. and limn→∞ Xn = 0 d. It remains to check for convergence in the m.s. sense. If Xn were to converge in the m.s. sense, it would have to converge to the same random variable in probability. But Xn does not converge to zero in the mean square. One way to see that Xn does not converge to zero in the m.s. sense is to note that E[Xn] = 1 for all n, so limn→∞ E[Xn] 6 = 0. Another way to see that Xn does not converge to zero in the m.s. sense is to observe
E[|Xn − 0 |^2 ] = E[U 12... U (^) n^2 ] = E[U 12 ] · · · E[U (^) n^2 ] = (
)n^6 → 0.
In summary, Xn converges to zero in the a.s., p., and d. senses, but does not converge in the m.s. sense.
2.17 Sums of i.i.d. random variables, III
(a) ΦXi,n (u) = E
[
ejuXi,n
]
= 1 + λn (eju^ − 1) so ΦYn (u) =
1 + λn (eju^ − 1)
)n .
(b) Since limn→∞
1 + αn
)n = eα, it follows that limn→∞ ΦYn (u) = eλ(e ju−1)
. This limit as a function of u is the characteristic function of a random variable Y with the Poisson distribution with mean λ.
(c) Thus, (Yk) converges in distribution, and the limiting distribution is the Poisson distri- bution with mean λ. There is not enough information given in the problem to determine whether Yn converges in any of the stronger senses (p., m.s., or a.s.), because the given information only describes the distrbitution of Yn for each n but gives nothing about the joint distribution of the Yn’s. Note that Yn has a binomial distribution for each n.
