Math 5010-1, Spring 2005
Assignment 10
Problems
#1, p. 379. Let H = {the coin lands heads}. Let X denote her winning, and D the number of
dots that are rolled. Then,
1
1
E[X | H] + E[X | Hc ]
2
2
1
1
= E[2D | H] + E[D/2 | Hc ]
2
2
1
= E[D | H] + E[D | Hc ].
4
E[X] =
But the outcome of the die is independent of the outcome of the coin. So, E[D | H] =
E[D] = 3.5, and E[D | Hc ] = 3.5, for the same reasons. Therefore, E[X] = 3.5 +
(3.5/4) ≈ 4.38.
#3, p. 379. By independence, the joint pdf of (X, Y ) is
f (x, y) = 1
0 ≤ x, y ≤ 1.
Therefore,
Z
α
1
Z
1
|x − y|α dy dx
E [|X − Y | ] =
0
Z
0
1
Z
Z
α
1
Z
1
(x − y) dy dx +
=
0
=
x
0
2
1+α
0
1
Z
x1+α dx =
0
(y − x)α dy dx
x
2
.
(1 + α)(2 + α)
#17, p. 381. Let Ii = 1 if your ith guess is correct; else let Ii = 0. You ought to check that
N = I1 + · · · + In .
(a) E[N ] = E[I1 ] + · · · + E[In ]. Now E[Ii ] is equal to the probability that the ith
guess is correct; this probability is 1/n for all i = 1, . . . , n. Therefore, E[N ] =
(1/n) + · · · + (1/n) = 1, as asserted. This is the gist of the hint.
(b) Suppose we are told the first i − 1 cards, and we wish to make the ith guess. Our
choice is then, “any one of the remaining n − (i − 1) cards with equal probability.”
So the probability that we get the ith one right is then
E[Ii ] =
1
1
.
n−i+1
Therefore,
E[N ] =
n
X
n
X
E[Ii ] =
i=1
i=1
n
X1
1
=
≈
n − i + 1 j=1 j
Z
n
1
dx
= ln n.
x
#50, p. 385. First we compute the marginal pdf of Y :
∞
Z
fY (y) =
0
e−x/y e−y
dx = e−y
y
y > 0.
That is, Y is exponential with parameter λ = 1.
Next we find the conditional pdf of X given that Y = y:
fX|Y (x|y) =
=
f (x, y)
fY (y)
e−x/y
y
x > 0.
So, conditionally on {Y = y}, X is exponentially distributed with parameter λ = 1/y.
Therefore,
Z ∞
2
E[X | Y = y] =
x2 fX|Y (x|y) dx
−∞
Z ∞
dx
=
x2 e−x/y
y
0
Z ∞
= y2
z 2 e−z dz
(z = x/y)
0
2
= y Γ(3) = 2y 2 .
#75, p. 389. Recall that M (t) = exp(λ(et −1)) is the moment generating function of a Poisson with
parameter λ. Therefore, X is Poisson with parameter 2. However, MY (t) = (1/4)10
is not a moment generating function. For instance, MY (0) must be one, but it is not.
So the problem is ill-posed.
#77(b), p. 385. We need the individual marginal pdf’s. Let us start with Y :
Z ∞
2
1
fY (y) = √
e−y e−(x−y) /2 dx = e−y y > 0.
2π −∞
Therefrom we have
Z
MY (t) =
=
∞
ty −y
e e
Z
∞
dy =
e−y(1−t) dy
0

1
1−t
0
if −∞ < t < 1,

∞
otherwise.
2
To find the mgf of X, we will
R R not find fX first; it is too complicated. Instead we use
the formula, E[g(X, Y )] =
g(x, y)f (x, y) dx dy. In this way, we obtain
Z ∞Z ∞
tX 2
1
etx e−y e−(x−y) /2 dx dy
MX (t) = E e
=√
2π 0
−∞
Z ∞
Z ∞
1
−y
tx −(x−y)2 /2
√
e
=
e e
dx dy.
2π −∞
0
Now the term inside (· · ·) is the moment generating function of a N (y, 1). Therefore,
Z ∞
2
MX (t) =
e−y ety+t /2 dy
0
Z ∞
t2 /2
=e
e−y+ty dy
0
 2
t /2
e
if −∞ < t < 1,
= 1−t

∞
otherwise.
#3, p. 427. Let X1 , . . . , Xn denote the grades of n students. We know that the Xi ’s are independent and idedntically distributed with E[Xi ] = 75. If n is large, then by the CLT,
the class average (X1 + · · · + Xn )/n is approximately normally distributed with mean
µ = 75 and variance σ 2 /n, where σ 2 is the variance of her test. We wish to find (a
large) n such that
X1 + · · · + Xn
P 70 ≤
≤ 80 ≥ 0.9.
n
By the CLT, the left-hand side is approximately
√ √ σ2
5 n
5 n
≤ 80 = Φ
−Φ −
P 70 ≤ N 75,
n
σ
σ
√ 5 n
− 1.
= 2Φ
σ
Set this equal to 0.9 to find that n must satisfy
√ 5 n
Φ
≈ 0.95.
σ
Now use the table on page 203 to find that the preceding implies that
√
5 n
≈ 1.65.
σ
Therefore, the minimum n is
n ≈ 0.1089σ 2 ,
provided that n (equivalently, σ) is large enough that the appeal to the CLT is valid.
For example, suppose σ = 25. Then, 0.1089σ 2 ≈ 68, which is large enough. This
would then imply that if the standard deviation of the grades is 25, then 69 students
would suffice.
3