# MATH 5010–001 SUMMER 2003 SOLUTIONS TO ASSIGNMENT 6 Lemma. Remark. ```MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 6
Throughout these exercises, we will encounter
compute this once and for all.
Lemma. For any k ≥ 0,
Z
∞
R
y k e−y dy a few times. So let us
y k e−y dy = k!.
0
Remark. By now, you should recognize the integral as Γ(k + 1) and so the above lemma
should be familiar. Here is the familar proof.
R
R
Proof. Recall integration by parts: u dv = uv − v du. Apply this with u = y k and
dv = e−y dy to get du = ky k−1 dy and v = −e−y . Thus,
Z
0
∞
k −y
y e
Z
∞
k −y dy = −y e
+k
0
∞
y
0
k−1 −y
e
Z
dy = k
∞
0
y k−1 e−y dy.
R∞
That is, ak = 0 y k e−y dy satisfies the recursion, ak = kak−1 for all k ≥ 1. Apply this
recursion to itself,
ak = kak−1 = k(k − 1)ak−2 = &middot; &middot; &middot; = k!a0 .
R ∞ −y
But a0 = 0 e dy = 1. This proves the lemma.
Problems from pp. 228–231
R
2. First find C by setting
Z
1=C
f dx = 1. But
∞
0
−x/2
xe
Z
dx = 4C
∞
0
ye−y dy
(y = x/2).
Thanks to the Lemma, this equals 4C, which means that C = 14 . Consequently,
1
P {X ≥ 5} =
4
&quot;
Z
∞
5
−x/2
xe
Z
dx =
Z
∞
−y = −ye +
5/2
=
∞
5/2
∞
5/2
ye−y dy
(y = x/2)
#
e−y dy
5 −5/2
7
+ e−5/2 = e−5/2 .
e
2
2
6. (a) Change variable (y = x/2), and apply the Lemma: E(X) =
R∞
2 0 y 2 e−y dy = 4.
1
1
4
R∞
0
x2 e−x/2 dx =
R
R1
6. (b) First, find c by setting f dx = 1. That is, 1 = c −1 (1 − x2 ) dx =
R1
In particular, E(X) = 34 −1 x(1 − x2 ) dx = 0.
6. (c) E(X) =
R∞
5
5
x
4c
3 .
So, c = 34 .
dx = +∞.
21. Let X denote the height, in inches, of a randomly-selected 25-year old man. The first
part asks for P {X ≥ 74}. Since X is normally distributed with mean &micro; = 61 and
standard deviation σ = 2.5, Z = (X − &micro;)/σ is normally distributed with mean zero
and standard deviation one. So use the normal table (p. 203) to obtain
74 − 71
P {X ≥ 74} = P Z ≥
= 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
2.5
In the second part, we are asked to find:
P {X ≥ 77 | X ≥ 72} =
1 − Φ(2.4)
1 − 0.9918
P {X ≥ 77}
=
=
= 0.023796.
P {X ≥ 72}
1 − Φ(0.4)
1 − 0.6554
31. (a) If X is uniformly distributed on (0, A), then for any number a ∈ (0, A),
Z
Z
Z
1 A
1 a
1 A
E{|X − a|} =
|x − a| dx =
(a − x) dx +
(x − a) dx
A 0
A 0
A a
(A − a)2
a2
+
.
=
2A
2A
a
Call this ψ(a) and minimize ψ by letting ψ 0 = 0 and check ψ 00 . Now ψ 0 (a) = A
− A−a
=
A
2a−A
A
2
A
0
00
A . Therefore, ψ = 0 when a = 2 . Also, ψ (a) = A ≥ 0 so the value of a = 2 is
a minimum. Finally,
A
A
(A/2)2 (A/2)2
+
= .
min E{|X − a|} = ψ
=
2
2A
2A
4
a∈(0,A)
31. (b) In a similar way,
Z ∞
Z a
Z ∞
−λx
−λx
E{|X − a|} =
|x − a|λe
dx = λ
(a − x)e
dx + λ
(x − a)e−λx dx
0
0
a
Z a
Z ∞
= λe−λa
yeλy dy + λe−λa
ze−λz dz.
0
0
I have changed variables by letting y = a − x and z = a − x. Change variables once
more (w = λy, v = λz) to obtain:
Z
Z
e−λa aλ w
e−λa ∞ −v
E{|X − a|} =
we dw +
ve dv
λ
λ
0
0
&quot;Z
#
aλ
e−λa
=
wew dw + 1 .
λ
0
2
The Lemma was used in the last line. Now integrate by parts to see that
R aλ w
aλ
wew |aλ
− eaλ + 1. Therefore,
0 − 0 e dw = aλe
R aλ
0
wew dw =
e−λa aλeaλ − eaλ + 2
λ
1
2e−aλ
=a− +
.
λ
λ
E{|X − a|} =
Call this ψ(a) and minimize. Clearly, ψ 0 (a) = 1 − 2e−aλ and ψ 00 (a) = 2λe−aλ ≥ 0. So
we have a minimum where ψ 0 = 0; i.e., e−aλ = 12 , which is at a = ln(2)
λ . Thus,
min E{|X − a|} = ψ
a&gt;0
3
ln(2)
λ
=
ln 2
.
λ
```