Quiz 5B for MATH 105 SECTION 205
March 20, 2015
Given Name
Family Name
Student Number
1. Let f (x) =
ae−2x , if x ≥ 1,
for some constant a such that f (x) is a PDF for some continuous random
0,
otherwise
variable X.
(a) (0.5 points) Find a.
2e2
(a)
Z
Answer.
∞
f (x) dx = 1, that is,
Since f (x) is a PDF, then
−∞
∞
Z
f (x) dx
1 =
Z−∞
∞
=
1
Z
ae−2x dx
∞
e−2x dx
1
1 −2x ∞
a· − e
2
1
a −2
e .
2
= a
=
=
Then
a = 2e2 .
(b) (0.5 points) Find the cumulative distribution function F (x) of X.
(b)
Z
Answer.
F (x) =
x
Since F (x) =
f (t) dt, then
−∞
a. If x < 1, then
Z
x
F (x) =
Z
x
f (t) dt =
−∞
0 dt = 0.
−∞
0,
if x ≤ 1,
1 − e2−2x , if x > 1.
b. If x ≥ 1, then
x
Z
f (t) dt
F (x) =
−∞
Z 1
Z
=
x
f (t) dt +
−∞
Z 1
x
Z
0 dt +
=
−∞
2
f (t) dt
1
2e2 e−2t dt
1
Z
x
= 2e
e−2t dt
1
=
1 −2t x
2e · − e
2
1
2
= e2 (−e−2x + e−2 )
= −e2−2x + e2−2
= 1 − e2−2x .
In summary, we get
F (x) =
0,
if x ≤ 1,
2−2x
1−e
, if x > 1.
(c) (0.5 points) Compute Pr(X ≤ 2).
(c)
Answer.
1 − e−2
In fact, we have
Pr(X ≤ 2) = F (2) = 1 − e2−2·2 = 1 − e−2 .
(d) (0.5 points) Compute the mean of X.
(d)
Answer.
In fact, we have
Z
∞
E(X) =
xf (x) dx
Z−∞
∞
x · 2e2 e−2x dx
1
Z ∞
2
= 2e
xe−2x dx.
=
1
Z
For
xe−2x dx, use the integration by parts, by ’ILATE’ rule, we have
u = x =⇒ du = dx
Z
e−2x
−2x
dv = e
dx =⇒ v = e−2x dx = −
.
2
3
2
Then we have
Z
xe−2x dx =
=
=
=
=
Z
u dv
Z
uv − v du
−2x Z −2x
e
e
+
dx
x· −
2
2
Z
xe−2x 1
−
+
e−2x dx
2
2
xe−2x e−2x
−
+ C.
−
2
4
Notice that
x
e2x
1
= lim
x→∞ 2e2x
= 0.
lim xe−2x =
x→∞
lim
x→∞
By l’Hospital’s rule
Hence, we get
E(X) =
=
=
=
∞
xe−2x e−2x 2e −
−
2
4 1
e−2
e2 · e−2 +
2
−2
3e
e2 ·
2
3
.
2
2
(e) (0.5 points) Compute the variance of X.
(e)
Answer.
Notice that Var(X) = E(X 2 ) − [E(X)]2 . Since
Z ∞
x2 f (x) dx
E(X 2 ) =
Z−∞
∞
=
x2 · 2e2 e−2x dx
1
Z ∞
2
= 2e
x2 e−2x dx.
1
Z
For
x2 e−2x dx, use the integration by parts, by ’ILATE’ rule, we have
u = x2 =⇒ du = 2xdx
Z
e−2x
−2x
dv = e
dx =⇒ v = e−2x dx = −
.
2
4
Then we have
Z
Z
x2 e−2x dx =
u dv
Z
= uv − v du
−2x Z −2x
e
e
2
+
· 2x dx
= x · −
2
2
Z
x2 e−2x
= −
+ xe−2x dx
2
x2 e−2x xe−2x e−2x
−
−
+ C By the computations in part d
= −
2
2
4
Notice that
lim xe−2x =
x→∞
=
=
lim x2 e−2x =
x→∞
=
=
=
x
e2x
1
lim
x→∞ 2e2x
0
x2
lim 2x
x→∞ e
2x
lim
x→∞ 2e2x
2
lim
x→∞ 4e2x
0.
lim
x→∞
By l’Hospital’s rule
By l’Hospital’s rule
By l’Hospital’s rule
Hence, we get
2
E(X ) =
=
=
=
∞
x2 e−2x xe−2x e−2x 2e −
−
−
2
2
4 1
−2
e
e−2 e−2
2e2 ·
+
+
2
2
4
−2
5e
2e2 ·
4
5
.
2
2
Therefore, we get
2 2
5
3
25 9
16
−
=
− =
= 4.
Var(X) = E(X ) − [E(X)] =
2
2
4
4
4
2
2
(f) (0.5 points) Compute the standard deviation of X.
(f)
Answer.
In fact, we have
σ(X) =
2. For a sequence
{an }∞
n=1
Z
such that an =
0
1
√
p
√
Var(X) = 4 = 2.
dx
.
+ x2
n2
2
(a) (0.5 points) Find the explicit formula of an .
(a)
Z
Answer.
For
√
dx
, let x = n tan(θ), then
+ x2
Z
Z
dx
√
=
n2 + x2
Z
=
n2
√
n2 + 1 1 ln + n
n
dx = n sec2 (θ) dθ and n2 + x2 = n2 sec2 (θ). So
n sec2 (θ)
dθ
n sec(θ)
sec(θ) dθ
= ln | sec(θ) + tan(θ)| + C.
Since tan(θ) =
x
, then
n
√
sec(θ) =
So we get
Z
n2 + x2
.
n
√
n2 + x2
x
dx
√
= ln + + C.
n
n
n2 + x2
Hence we have
Z
1
an =
√
0
=
=
=
=
dx
n2 + x2
√
1
n2 + x2
x
ln + n
n 0 √
√
n2
n2 + 1 1 0 + − ln + ln n
n
n
n
√
n2 + 1 1 + − ln 1
ln n
n
√
n2 + 1 1 ln + .
n
n
(b) (0.5 points) Find a100 .
√
(b)
Answer.
ln
10001 + 1
100
By the result of part a, we have
√
√
1002 + 1
1 10001 + 1
a100 = ln +
.
= ln
100
100
100
(c) (0.5 points) If {an }∞
n=1 has a limit, find this limit.
(c)
Answer.
In fact, since
"√
lim
n→∞
#
n2 + 1 1
+
= 1.
n
n
0
Then
lim an = ln 1 = 0.
n→∞
3. For a sequence {an }∞
k=1 such that a1 = 1 and an+1 = 4an for all n ≥ 1.
(a) (0.5 points) Compute a2 , a3 , a4 , a5 .
(a)
Answer.
4, 6, 64, 256
Since a1 = 1 and an+1 = 4an , then
a2 = 4a1 = 4 · 1 = 4
a3 = 4a2 = 4 · 4 = 16
a4 = 4a3 = 4 · 16 = 64
a5 = 4a4 = 4 · 64 = 256
(b) (0.5 points) Find the explicit formula of an .
(b)
4n−1
Answer.
Since an+1 = 4an for all n ≥ 1, then {an } is a geometric sequence with ratio r = 4, which
implies that
an = a1 rn−1 = 1 · 4n−1 = 4n−1 .
√
cos( n) 3 tan−1 (n)
4. (a) (0.5 points) Compute lim 2 · √
.
+
n→∞
n3 + 3
n
(a)
√
π
π
Notice that −1 ≤ cos( n) ≤ 1 and − ≤ tan−1 (n) ≤ , then
2
2
√
1
tan−1 (n)
π
cos( n)
π
π
−√ ≤ √
≤
≤
,
≤ √ , and −
3
3
3
2(n + 3)
n +3
2(n + 3)
n
n
n
0
Answer.
for all n ≥ 1.
π
π
Since lim √ = lim
= 0, by the squeeze theorem, we have
3
n→∞
n n→∞ 2(n + 3)
√
cos( n)
tan−1 (n)
√
lim
= lim
= 0.
n→∞
n→∞
n3 + 3
n
Hence we get
√
cos( n) 3 tan−1 (n)
√
lim 2 ·
+
= 0.
n→∞
n3 + 3
n
n3 + 2n
23n−1
−1
(b) (0.5 points) Compute lim tan
+ n−10 .
n→∞
n2 + 2n + 1
9
(b)
π
2
n3 + 2n
Answer.
Since lim
= ∞, then lim tan−1
n→∞ n2 + 2n + 1
n→∞
geometric sequence with ratio r:
r =
=
=
=
Then
lim
n→∞
n3 + 2n
n2 + 2n + 1
π
= . For
2
23·2−1 91−10
·
92−10 23·1−1
25 1
·
22 9
23
9
8
< 1.
9
23n−1
= 0.
9n−10
Therefore, we get
lim
(c) (0.5 points) Evaluate
k=5
tan
n→∞
∞
X
−1
n3 + 2n
n2 + 2n + 1
23n−1
π
+ n−10 = .
9
2
1
.
(3k + 1)(3k + 4)
(c)
Answer.
1
48
By partial fraction decomposition, we have
1
1
1
1
1
=
−
=
−
.
(3k + 1)(3k + 4)
3(3k + 1) 3(3k + 4)
3(3k + 1) 3[3(k + 1) + 1]
So we have
Sn =
n+4
X
k=5
1
(3k + 1)(3k + 4)
1
1
−
3(3 · 5 + 1) 3(3 · 6 + 1)
1
1
+
−
3(3 · 6 + 1) 3(3 · 7 + 1)
1
1
+
−
3(3 · 7 + 1) 3(3 · 8 + 1)
..
.
1
1
−
+
3(3 · (n + 3) + 1) 3(3 · (n + 4) + 1)
1
1
+
−
3(3 · (n + 4) + 1) 3(3 · (n + 5) + 1)
1
1
=
−
3(3 · 5 + 1) 3(3 · (n + 5) + 1)
1
1
1
=
−
→ , as n → ∞.
48 3(3 · (n + 5) + 1)
48
=
So we get
∞
X
k=5
1
1
= lim Sn = .
(3k + 1)(3k + 4) n→∞
48
23n−1
, it’s a
9n−10
(d) (0.5 points) Evaluate
∞
X
ln
k=2
k+1
.
k
∞ or divergent
(d)
Answer.
Let Sn =
n+1
X
ln
k=2
k+1
k+1
, notice that ln
= ln(k + 1) − ln k, then
k
k
Sn = ln 3 − ln 2
+ ln 4 − ln 3
+ ln 5 − ln 4
..
.
+ ln(n + 1) − ln(n)
+ ln(n + 2) − ln(n + 1)
= ln(n + 2) − ln 2 → ∞,
Then
∞
X
k=2
ln
as n → ∞.
k+1
is divergent.
k
(e) (0.5 points) Evaluate
∞
X
3 · 42k−5
74k+5
k=8
.
3·411
737
3
1 − 474
(e)
Answer.
Notice that
∞
X
3 · 42k−5
k=8
74k+5
is a geometric series with initial term a:
3 · 42·8−5
74·8+5
3 · 411
.
737
a =
=
And the ratio r:
r =
=
=
3 · 42·9−5
74·8+5
·
74·9+5
3 · 42·8−5
13
37
4
7
· 41
11
4
7
43
< 1.
74
So we get
∞
X
3 · 42k−5
k=8
(f) (0.5 points) Is the series
∞
X
k=100
√
74k+5
11
=
3·4
a
37
= 7 43 .
1−r
1 − 74
k+1
convergent or divergent?
k
(f)
divergent
√
Answer.
Approach I: Let f (x) =
x+1
=
x
r
x+1
=
x2
r
1
1
+ , then f (x) is positive and decreasx x2
ing. Notice that
∞
Z
Z
∞
f (x) dx =
1
Z √
For
1
√
x+1
dx.
x
x+1
dx, then
x
Z √
Z
√
u
x+1
dx =
· 2u du let u = x + 1, then x = u2 − 1 and dx = 2udu
2
x
u −1
Z
u2
= 2
du
u2 − 1
Z 1
du
= 2
1+ 2
u −1
Z
1
= 2u + 2
du.
2
u −1
By the partial fraction decomposition, we have
u2
1
1
1
=
−
.
−1
2(u − 1) 2(u + 2)
Then
Z √
Z x+1
1
1
dx = 2u + 2 ·
−
du
x
2(u − 1) 2(u + 2)
Z 1
1
= 2u +
−
du
u−1 u+1
= 2u + ln |u − 1| − ln |u + 1| + C
√
√
√
= 2 x + 1 + ln | x + 1 − 1| − ln | x + 1 + 1| + C
√
x + 1 − 1
√
+ C.
= 2 x + 1 + ln √
x + 1 + 1
Notice that
√
lim 2 x + 1 = ∞,
x→∞
and
√
x + 1 − 1
= ln 1 = 0.
lim ln √
x→∞
x + 1 + 1
Then
Z
∞
1
√
x + 1 − 1 ∞
√
= ∞.
f (x) dx = 2 x + 1 + ln √
x + 1 + 1 1
So by the integral test, we know that the series
∞
X
k=100
Approach II: Notice that
√
k+1
1
≥ ,
k
k
√
k+1
is divergent.
k
for all k ≥ 1.
∞
X1
X
Since
diverges, by the comparison test, then the series
k
k=100
√
k+1
is divergent.
k
(g) (0.5 points) Is the series
∞
X
4
convergent or divergent?
2
k
ln
k
k=9
convergent
(g)
Answer.
Z
For
Let f (x) =
4
, then f (x) is positive and decreasing for x ≥ 9. Notice that
x ln2 x
Z ∞
Z ∞
4
f (x) dx =
dx
x ln2 x
9
9
Z ∞
1
= 4
dx.
x ln2 x
9
1
dx, then
x ln2 x
Z
Z
du
1
Let u = ln x, then du = dx
u2
x
1
= − +C
u
1
+ C Since u = ln x.
= −
ln x
1
dx =
x ln2 x
So we get
Z
∞
f (x) dx =
9
=
So by the integral test, we know that the series
1 ∞
4 −
ln x 9
4
.
ln 9
∞
X
4
is convergent.
k ln2 k
k=9
ax2 + b, if 0 ≤ x ≤ 1,
for some constants a and b such that f (x) is a probability density
0,
otherwise
function for some continuous random variable X.
5. Let f (x) =
(a) (1 point) Find conditions for a and b.
Answer.
Since f (x) is a PDF for some random variable, then f (x) ≥ 0 for all x. For 0 ≤ x ≤ 1,
since f (x) = ax2 + b is a non-negative quadratic polynomial and symmetric
with respect to x = 0, then
Z
∞
f (1) = a + b ≥ 0 and f (0) = b ≥ 0. On the other hand, we must have
f (x) dx = 1, then
∞
Z
∞
1 =
f (x) dx
−∞
Z 1
=
=
=
So we get
(ax2 + b) dx
0
ha
i1
x3 + bx 3
0
a
+ b.
3
a
+ b = 1, that is, a + 3b = 3. In summary, we must have the following conditions:
3
a + b ≥ 0,
b ≥ 0,
and a + 3b = 3.
(b) (1 point) Compute a and b such that E(X) = 1.
Answer.
Since E(X) = 1, then
1 = E(X)
Z ∞
xf (x) dx
=
−∞
1
Z
x(ax2 + b) dx
=
0
Z
=
1
(ax3 + bx) dx
a 4 b 2 1
x + x 4
2
0
a b
+ .
4 2
0
=
=
So we get
a b
+ = 1, that is, a + 2b = 2. By the result of part a, we have a + 3b = 3, then
4 2
a = 0,
and b = 1.
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