ECE/CS 313: Probability with Engineering Applications Fall 2002
Problem Set 7 Solutions
1. Maximum variance Bernoulli and binomial random variables
(a) The variance of the Bernoulli random variable Xis p(1 −p) which is zero at each end of the interval
of possible values for p, [0,1]. The function is positive in between. The derivative of the function is 1 −2p
which is zero precisely when p=1
2. So p=1
2maximizes the variance, and the maximum value is 1
4.
(b) The variance of the binomial random variable Yis np(1 −p), a constant ntimes larger than the variance
in part (a). So by part (a), the variance of Yis also maximized by p=1
2, and the maximum variance is n
4.
2. Some calculations with the Poisson distribution
(a) Parameter nis the number of leaves on the tree at the beginning of the one minute observation interval,
and pis the probability that any given leaf will fall off during the observation interval.
(b) The estimator ˆ
λML is the value of λthat maximizes P[X= 12] = e−λλ12
12! . Equivalently, it maximizes
e−λλ12, or equivalently it maximizes the natural log of this quantity, −λ+ 12 ln(λ). This expression tends
to −∞ as λ→0 or λ→ ∞. The derivative −1 + 12
λis zero only when λ= 12. Thus, ˆ
λML = 12.
(c)
P[X= 2|X≤3] = P[{X= 2} ∩ {X≤3}]
P[X≤3] =P[{X= 2}]
P[X≤3] =λ2/2
1 + λ+λ2
2+λ3
6
(The factors e−λcancel out.)
3. On the shape of the Poisson distribution
“Proposition 6.10If Xis a Poisson random variable with parameter λ, where λ > 0, then as kgoes from 0
to ∞,P{X=k}first increases monotonically and then decreases monotonically, reaching its largest value
when kis the largest integer less than or equal to λ.
Proof: We prove the proposition by considering P{X=k}/P{x=k−1}and determining for what values
of kit is greater than or less than 1. Now,
P[{X=k}]
P[{X=k−1}]=
e−λλk
k!
e−λλk−1
(k−1)!
=λ
k
Hence P{X=k} ≥ P{X=k−1}if and only if k≤λ, and the proposition is proved.”
(This is essentially Proposition 6.1 and its proof in the limit n→ ∞,p→0 with np =λfixed.)
4. Poisson approximation and the birthday paradox
(a) There are 365 −kdates left to choose from, so the probability is 365−k
365 .
(b) 1 ·364
365 ·363
365 ···365−k+1
365
k P [all kdistinct] Poisson approximation
10 0.883052 0.884009
20 0.588562 0.594195
30 0.293684 0.303680
(c) P[E12] = 1
365 because given the birthday of the first student, the chances of the second student having
the same birthday is 1
365 .
P[E12E13 ] = 1
3652because given the birthday of the first student, the chances of both the second and third
student having the same birthday as the first student is 1
3652.
P[E12E13 E23] = 1
3652since the event E12E13E23 is identical to the event E12E13.
Yes, the events Eij are pairwise independent since P[EijEi0j0] = 1
3652=P[Eij]P[Ei0j0] whenever ij and i0j0
are distinct pairs with i < j and i0< j0.
No, the events are not independent since P[E12E13E23] = 1
36526=P[E12]P[E13 ]P[E23].
(d) The suggested approximation is P[X=k]≈exp(−k
2/365) = exp(−k(k−1)
2×365 ). Numerical values are
indicated in the table.
