Midterm 3 Solutions
Math 5010–1, Spring 2005
1. Let X denote a random variable whose probability density function (pdf) is given by
f (x) = c|x|−3 ,
|x| > 1.
(a) Plot f , and find c.
Solution: To find c compute
it equal to one, viz.,
Z ∞
1 = 2c
R∞
−∞
f (x) dx and set
2. We toss a fair coin repeatedly, and independently every time, until the second ‘H’ occurs. Let X denote the
number of tosses needed to accomplish this.
(a) Find the probability mass function of X.
Solution: The possible values are k = 2, 3, . . ., and
P{X = k} is the probability that you get one head
in the first k − 1 trials, and a head on the kth. This
is p times the probability that Binomial(k − 1, p)
is equal to one; i.e.,
k−1
P{X = k} = p
p(1 − p)k−2
1
2
k−2
= (k − 1)p (1 − p)
x−3 dx = c.
k = 2, 3, . . . .
1
(b) Find E[X].
So c = 1 .
Solution:
(b) Compute P{X > 5}.
R
Solution: P{X > 5} = 5∞ x−3 dx = 1/50 .
∞
E[X] =
P{|X| ≤ 2} =
(d) Compute E[X].
Solution: E[X] = 0 by symmetry.
(e) Does E[X 2 ] exist? Is it finite? (These are two
different questions.)
Solution: Yes it exists, but is infinite. Note that
Z n 2
x
1
x3
dx =
Z n
1
1
x
dx = ln n.
E[X 2 ] = 2 lim ln n = ∞ .
k−2
.
To simplify this, first note that for 0 < r < 1,
!
∞
∞
2
d
∑ k(k − 1)rk−2 = dr2 ∑ rk
k=2
k=0
2
1
d
= 2
dr
1−r
d
1
=
dr (1 − r)2
2
=
.
(1 − r)3
Therefore,
Therefore, by symmetry,
n→∞
∑ k(k − 1)p (1 − p)
k=2
(c) Compute P{|X| ≤ 2}.
Solution: By
symmetry,
R 2 −3
2 1 x dx = 3/4 .
2
2
2p2
.
E[X] = 3 =
p
p