HOMEWORK 8 - SOLUTIONS
Exercise 1
a
"
E[x ] =
"
# xf ( x ) dx = # xf (x ) dx + # xf (x ) dx =
x
x
!"
x
a
!"
0
0
0
# (x + a) f ( x + a) dx + # (a ! x) f (a ! x ) dx = 2a # f ( x + a ) dx =
=
x
x
!"
!"
0
= a (because 1 =
"
# f ( x + a) dx + # f (x + a) dx =
x
x
0
!"
0
0
f x ( x + a) dx +
#
x
!"
!"
#
!"
0
fx (a ! x ) dx = 2 # fx ( x + a ) dx
!"
0
therefore
# f ( x + a) dx = 0.5)
x
!"
The median b of x is such that P(x>b)=P(x<b)=0.5. We have just seen that
•
0
a
f x ( x + a) dx =
#
!"
# f ( x ) dx = 0.5
x
!"
Exercise 2
E[x2] =
#
"
0
2
x ce
!cx
dx = [ !e
! cx
x
2 "
] !#
0
"
0
!e
!cx
"
" !1
!1
!2 ! cx & " 2
$
!cx &
! cx
$
2x dx =
2xe
!
2e dx = 2 e
=
%c
' 0 #0 c
%c
' 0 c2
Exercise 3
"
"
•
!6t
!6t
!1.5
#0.256e dt = [ !e ]0.25 = e = 0.22
•
E[t] = 1/λ = 1/6
Exercise 4
T is the mean of t, which is an exponential random variable. Thus, the coefficient c if its
pdf is equal to 1/T.
•
2
2
1 ! "0 0.5e !0.5t dt = 1 ! [!e !0.5t ]0 = e!1 = 0.37
you can also compute it as
#
"
2
"
0.5e !0.5t dt = [!e !0.5t ]2 = e !1 = 0.37
•
P(any component still functions after 4 months) =
4
4
1 ! "0 0.5e !0.5t dt = 1 ! [!e !0.5t ]0 = e !2 = 0.13
P(at least 2 out of 5 are still working after 4 months) =
5 ! 5$
k
5&k
' k =2 #" k % (0.13) ( 0.87) = 0.13
Exercise 5
15
15
0.5e !0.5t dt = [ !e!0.5 ]0 = 1! e !7.5 = 0.99
•
"
•
The mean of the (exponential) random variable t is µ=1/λ = 2 minutes/train. P(t < 15)
= 1 - P(t >15) = 1 - P(t > 7.5µ) > 1- 1/7.5 = 0.87
0
Exercise 6
•
The variable t (time until next connection request) is an exponential random variable
with parameter c=λ=10/second. Therefore the solution is
"
"
!10 x
!10 x
!5
# 10e dx = [ !e ]0.2 = e = 0.007
0.2
•
•
Mean = 1/λ = 1/10 seconds. Variance = 1/λ2 = 1/100.
P(30 connections simultaneously) = P(≥30 connection requests in 1 second).
29
10 k
= 2.5#10 !7
Remember that P(n connections in T time) = 1 ! " e !10
k!
k=0
Exercise 7
•
µx = 0 (because fx(x)=fx(-x))
0
#
! = E[x ] " µ = E[ x ] =
2
x
2
2
x
2
$x
2
"#
0
0
1
1
2
fx ( x ) dx = $ x (x + 1) dx + $ x 2 (1" x ) dx =
0
"1
0
1
1
1
%x4 (
% x3 (
% x4 (
= $ x dx + $ x dx + $ x dx " $ x dx = ' * + ' * " ' * =
& 4 ) "1 & 3 ) "1 & 4 ) 0
"1
"1
0
0
3
="
2
2
3
1 1 1 1
1
+1 1
+ + " = 2, " -. =
4 3 3 4
3 4
6
fy(y)
|b|
(-1-a)/b
•
•
-a/b
(1-a)/b
y
Here is the computation of fy(y) (NOTE: this was not required by the question).
Assuming b > 0 (otherwise the inequalities invert)
# b (by + a + 1), !1 " by + a < 0
f x (by + a ) %
fy ( y) =
= $ b (1 ! by ! a), 0 " by + a " 1
=
1
%&
0,
elsewhere
b
# b (by + a + 1), !1 ! a " y < ! a
b
b
%%
a
1!a
= $ b (1 ! by ! a), ! " y "
b
b
%
0,
elsewhere
%&
x ! a$ 1
a
E[ y ] = E "
= ( µ x ! a) = !
# b % b
b
2
'# x " a % * a 2
1
a2
! y2 = E[ y 2 ] " µ 2y = E
" 2 = 2 E[ x 2 + a 2 " 2ax ] " 2 =
)($ b & ,+ b
b
b
=
E[ x 2 ]
1
1
2
2
2
E
x
+
a
"
2aE
x
"
a
=
= 2
[ ]
[
]
2
2
b
b
6b
(
)
Exercise 8
The root of y=g(x)= 2x is log2(y)=loge(x)/loge(2) for y>0. The derivative of g(x) at
x=log2(y) is (loge(y)/loge(2)) loge (2) = loge(y). Therefore,
1
!
, 2<y<4
f y ( y ) = " 2log e ( y )
0,
elsewhere
#
x
• P(y>2) = P(2 <4) = P(x<log2(4)) = P(x<2) = 0.5
"
3
0.5
4
x
x
x 3
E[ y ] = # 2 f x ( x )dx = 0.5# 2 dx =
2 ]1 =
[
log e (2)
log e 2
!"
1
•
Exercise 9
0,
x < !5
#%
f x ( x ) = $( x + 5) / 72, !5 " x < 7
%&
0
x>7
•
7
7
1 ) # x3 &
# x2 & ,
+
E[x] = " x ( x + 5) / 72 dx =
+ 5% ( . = 3
!5
72 * %$ 3 (' !5
$ 2 ' !5 -
•
7
7
1 ) # x4 &
# x3 & ,
E[x ] = " x ( x + 5) / 72 dx = + % ( + 5 % ( . = 17
!5
72 * $ 4 ' !5
$ 3 ' !5 -
•
7
2
7
2
Therefore, σ2 = E[x2]-E[x]2 = 17-9 = 8
7
7
7
1 ) # x5 &
# x4 & ,
3
3
+
E[x ] = " x ( x + 5) / 72 dx =
+ 5% ( . = 86.2
!5
72 * %$ 5 (' !5
$ 4 ' !5 -
Exercise 10
•
•
The root of y=g(x)=log2 x is 2y. Since log2 x = ln x / ln 2, we have that
g’(x) = 1 / x ln 2. Therefore, g’(2y) = 1 / 2y ln 2. When x=2, then y = 1. When x=4,
then y=2. Therefore,
"0.5! 2 y ln 2, 1 < y < 2 "2 y %1 ln 2, 1< y < 2
f y ( y ) = f x (2 y )2 y ln 2 = #
=#
0,
elsewhere $ 0,
elsewhere
$
To compute E[y] we can use two methods.
• Method 1: Apply the formula of the expectation directly on fy(y):
"
2
* $ 2 y !1 y ' 2 2 2 y !1 y!1
E[ y ] = # yfy ( y )dy = ln 2 # y2 dy = ln 2, &
)( ! # ln 2 dy /. =
ln
2
+
%
!"
1
1
1
2
•
$ 2 y !1 '
1
= 3!&
=
3
!
= 1.56
ln2
% ln 2 )( 1
Method 2: Use the fundamental theorem of expectation
"
E[ y ] =
4
# g( x ) fx (x )dx = 0.5# log 2 xdx =
!"
2
2
= (substitute z = log 2 x) = 0.5ln 2 # x2 x dx
1
•
then proceed as above.
P(y>2) = 1 – P(y ≤ 2) = 1 – P(log2 x ≤ 2) = 1 – P(x ≤ 4) = 0
(Could have obtained the same solution reasoning on the fact that fy(y)=0 for y>2.
Exercise 11
µy = aµ x + b and ! y2 = a 2! 2x (why?). Therefore, solving the system, we obtain a=0.5 and
b=-2.5.
Exercise 12
"
1=
"
"
# f ( x )dx = # (a g
x
1 µ1 ,$1
!"
( x ) + a2 gµ
2 ,$ 2
!"
"
"
x
1 µ1 ,$ 1
!"
( x ))dx = a1 # gµ ,$ ( x )dx + a2 # gµ
1
1
!"
# xf ( x )dx = # x( a g
E[x ] =
"
2 ,$ 2
( x )dx = a1 + a2
!"
( x ) + a2 gµ ,$ ( x ))dx =
2
2
!"
"
"
= a1 # xg µ1 ,$ 1 ( x )dx + a2 # xgµ 2 ,$ 2 ( x )dx = a1 µ1 + a2 µ2
!"
!"
"
E[x ] =
2
"
#x
2
!"
"
fx ( x )dx = a1 # x gµ1 ,$ 1 (x )dx + a2 # x 2 gµ2 ,$ 2 ( x )dx = a1 ($ 12 + µ12 ) + a2 ($ 22 + µ 22 )
2
!"
2
!"
(remember that ! = E[ x ] " µ )
2
x
2
2
x
2
1
! x2 = E[x 2 ] " E[ x ] = a1 (! 12 + µ ) + a2 (! 22 + µ22 ) " a12 µ12 " a22 µ22 " 2a1a2 µ1 µ2 =
= a1 µ12 (1" a1 ) + a2 µ22 (1" a2 ) " 2a1 a2 µ1µ 2 + a1! 12 + a2! 22
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
-10
-5
10
0
5
10
10
P(0 < x ! 10) = a1 # gµ1 ," 1 ( x )dx + a2 # gµ 2 ," 2 ( x )dx =
0
0
0.3 (G0, 3 (10) $ G0, 3 (0)) + 0.7 (G10,3 (10) $ G10,3 (0)) = 0.4996
15
20
Exercise 13
2
E[(x ! c ) ] = E[ x 2 + c 2 ! 2cx] = E[x 2 ] + c 2 ! 2c µx =
2
= " x2 + µ 2x ! 2cµx + c2 = ( µ x ! c) + " 2x