University Problem Set #4: Solutions ECE 313
of Illinois Page 1 of 3 Fall 2001
1.(a) The number of weeks that your investment doubles in value is a binomial random variable Y with
parameters (5,1/2). Since the investment halves in value during the remaining 5–Y weeks, and each
halving cancels one doubling, we have that X = 32•22Y–5. The possible values of X are 1, 4, 16, 64, 256,
and 1024, corresponding to Y = 0, 1, 2, 3, 4, 5 respectively.
(b) P{X = 1} = P{Y = 0} = 1/32. P{X = 4} = P{Y = 1} = 5/32. P{X = 16} = P{Y = 2} = 10/32.
P{X = 64} = P{Y = 3} = 10/32. P{X = 256} = P{Y = 4} = 5/32. P{X = 1024} = P{Y = 5} = 1/32.
(c) E[X] = 1•(1/32) + 4•(5/32) + 16•(10/32) + 64•(10/32) + 256•(5/32) + 1024•(1/32) = 97.65625.
The TV commercial understates the performance — undoubtedly a first!
(d) P{X < 32} = P{X = 1} + P{X = 4} + P{X = 16} = 1/2.
(e) There is a 50% chance of losing money on this investment. Most people do not mind making investments
if the amount at risk is small but the payoff from a win is enormous, e.g. lottery tickets are a losing bet for
most buyers, but people don’t mind an almost sure loss of a dollar because all they see is the huge payoff.
Matters are considerably different when the amounts at risk are large, and most people tend to choose more
conservatively. This explains why the local 7-11 sells lottery tickets but Neiman-Marcus does not.
2. P{X is even} = P{X = 0} + P{X = 2} + … + P{X = M} where M = N if N is even and M = N – 1 if N is
odd. Hence, P{X is even} = qN + ( )
N
2qN–2p2 + … + ( )
N
MqN–MpM . Note that the last term is pN if N
is even, and it is NqpN–1 if N is odd (whatever happened to ( )
N
M??)
Now, (q + p)N = qN + ( )
N
1qN–1p + ( )
N
2qN–2p2 + … + ( )
N
N–1 qpN–1 + pN while
(q – p)N = qN – ( )
N
1qN–1p + ( )
N
2qN–2p2 – …+ (–1)N–1( )
N
N–1 qpN–1 + (–1) NpN.
We add these equations together and note alternate terms cancel.. Also if N is odd, the last terms cancel,
while if N is even, the next–to–last terms cancel. In either case, we see that the sum is just 2P{X is even},
i.e. P{X is even} = [(q+p)N+(q – p)N]/2 = [1+(1 – 2p)N]/2. Exercise: when is P{X is even} < 1/2?
3.(a) P{W = n} = ( )
N–K
n(0.8)n(0.2)N–K–n, for n = 0, 1, 2, … , N–K.
(b) For fixed N and n, the answer to part (a) is a function of K, say f(K). Then,
f(K)
f(K–1) = ( )
N–K
n(0.8)n(0.2)N–K–n
( )
N–(K–1)
n(0.8)n(0.2)N–(K–1)–n = (N–K)(N–K–1)…(N–K–n+1)
(N–(K–1))(N–(K–1)–1)…(N–(K–1)–n+1)0.2 ≥ 1 if and only
if N–K–n+1
N–K+1 ≥ 0.2, that is, K ≤ N – 1.25n + 1. Since K is an integer, the maximum-likelihood estimate
can be expressed as ^
K = N – 1.25n + 1. For N = 100 and n = 8, N – 1.25n + 1 = 91 is an integer.
Hence, f(91)/f(90) = 1 and either 90 or 91 can be taken to be the maximum-likelihood estimate of K. The
formula we derived gives 91 as the estimate but it is perfectly legitimate to use 90 instead.
(c) The examiner’s estimate ~
K is C – 0.25n = N – n – 0.25n which has value 90 if N = 100 and n = 8
while ^
K is 91. Thus, the examiner’s estimate is smaller than the answer obtained in part (b), but still
perfectly defensible as a maximum-likelihood estimate as described in part (b) (she is just a tough grader!).
In fact, ~
K = ^
K except when 0.25n happens to be an integer (because N – n – 0.25n = N – 1.25n + 1
except when 0.25n is an integer) and in this special case, the examiner is opting for the lower value.
For the case N = 100 and n = 10, we get ^
K = 100 – 12.5 + 1 = 88.5 = 88 while ~
K is C – 0.25n
= 90 – 2.5 = 90 – 2 = 88 is the same. The unlucky student who blew all 10 guesses suffers the further
indignity of having a score of 88 on the exam even though the answers to 90 questions were known (and
recorded correctly on the answer sheet!)
(d) If N = 100 and K = 90, W is a binomial random variable with parameters (10,0.8) and hence P{W = n} is
maximized at n = 0.8×(10+1) = 8. If this maximum-probability event actually occurs, then as discussed
in parts (b) and (c), ~
K = 90 which correctly estimates K, while ^
K = 91 slightly overestimates K.
(e) If W = 4, the guessing penalty reduces the score by just 1 — the student gets a 95 while knowing only 90
answers whereas the unlucky one who blew all 10 guesses got only an 88 while knowing 90 answers! This
is the origin of the saying “Some people have all the luck!”
(f) It is easily verified that if C equals 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, and 90,
then the examiner’s estimate ~
K is 100, 99, 98, 97, 95, 94, 93, 92, 90, 89, and 88. The probability of
underestimating K is P{W = 9} + P{W = 10} = 0.37580964… while the probability of correctly
estimating K is P{W = 8} = 0.30198989… The probability of overestimating K can be easily found as
