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University of Illinois at Urbana-Champaign
ECE 534: RANDOM PROCESSES
Fall 2006
Probability Quiz Solutions
Monday, September 11, 2006
1 A group of Michigan Alumni at GoBlue Airlines just purchased trou- bled Air Illini. Air Illini currently only offers service to Springfield and Decatur. Because they are so unorganized, flights occur in a ran- dom manner and often get cancelled. The probability that a flight to Springfield or Decatur is cancelled is 1/5 and 1/10 respectively. Any particular flight goes to Springfield with probability 2/3 and Decatur with probability 1/3. a) What is the probability that a flight gets cancelled?
P (C) = P (C|S)P (S) + P (C|D)P (D)
= (1/5) × (2/3) + (1/10) × (1/3)
b) What is the expected number of flights before the first cancella- tion? The number of flights up to and including the first cancella- tion is a geometric random variable X with p = P (C) = 1/ 6. Geometric random variables with parameter p have expecta- tion 1 /p. The number of flights before the first cancellation is thus a random variable X˜ = X − 1. Thus
E[ X˜] = E[X − 1] = E[X] − 1 = 5.
c) What is the expected number of flights that occur before the first cancellation and after 3 non-cancellations? The geometric random variable is memoryless:
P (X > n + k|X > k) = P (X > n).
So the answer is the same as in b): 5.
d) Air Illini has 1000 flights per year. What is the probability that they have 1 or less cancellations in a year? The number of cancellations in a year Y is a binomial random variable with n = 1000 and p = P (C) = 1/ 6. For any k,
P (Y = k) =
n k
pk(1 − p)n−k.
The event E = { 1 or less cancellations in a year} is equivalent to the event {Y = 1 or Y = 0}. Thus
P (E) = P (Y = 1) + P (Y = 0)
=
(1/6)^1 (5/6)^999 +
(1/6)^0 (5/6)^1000
= 1000(1/6)(5/6)^999 + (5/6)^1000
= 200(5/6)(5/6)^999 + (5/6)^1000
= 201(5/6)^1000
3 Let X 1 and X 2 be continuous independent and identically distributed (i.i.d.) random variables with common cumulative distribution function (cdf) FX (x) and probability density function (pdf) fX (x) = (^) dxd FX (x). a) Let Z = X 2 − X 1. Find P (X 1 > x 1 , Z > z). Express your final answer in terms of the cdf FX (x).
P (X 1 > x 1 , Z > z) =
x 1
P (Z > z|X 1 = x′ 1 ) fX 1 (x′ 1 ) dx′ 1
x 1
P (X 2 − X 1 > z|X 1 = x′ 1 ) fX 1 (x′ 1 ) dx′ 1
x 1
P (X 2 − x′ 1 > z|X 1 = x′ 1 ) fX 1 (x′ 1 ) dx′ 1
x 1
P (X 2 > z + x′ 1 |X 1 = x′ 1 ) fX 1 (x′ 1 ) dx′ 1
= P (X 1 > x 1 , X 2 > z + x 1 ) = P (X 1 > x 1 ) P (X 2 > z + x 1 ) = (1 − FX (x 1 )) (1 − FX (z + x 1 ))
b) Find P (X 1 = X 2 ). Note that for any continuous random vari- able X, P (X = α) = 0 for all α. Define the random variable Z = X 2 − X 1 (already done for you in part a, as a hint). Since X 1 and X 2 are independent and continuous, so is Z. (Although not necessary to get credit on the problem, it can be shown that Z is continuous by the fol- lowing enclosed in the box):
FZ (z) = P (X 1 − X 2 ≤ z)
=
−∞
P (X 1 − X 2 ≤ z|X 2 = α) fX 2 (α) dα
−∞
P (X 1 − α ≤ z|X 2 = α) fX 2 (α) dα
−∞
P (X 1 ≤ α + z|X 2 = α) fX 2 (α) dα
−∞
FX 1 (α + z) fX 2 (α) dα
and thus (^) dzd FZ (z) exists:
fZ (z) =
d dz
FZ (z)
−∞
fX 1 (α + z) fX 2 (α) dα
Since Z is continuous, P (Z = α) = 0 for any α. Thus
P (X 1 = X 2 ) = P (X 2 − X 1 = 0) = P (Z = 0) = 0.
c) Find P (X 1 ≤ X 2 ). Note that minimal computation is required. Since X 1 and X 2 are iid, from symmetry P (X 1 ≤ X 2 ) = P (X 2 ≤ X 1 ). Thus
1 = P (X 1 ≤ X 2 ) + P (X 2 ≤ X 1 ) − P (X 1 = X 2 ) = 2 P (X 1 ≤ X 2 ) − 0
⇒ P (X 1 ≤ X 2 ) =
