Problem Set #3
1. Max and Min – 4 pts
a) 2 pts
By observation, we know that W ≤ Z.
Two cases: on [0;1] and on [1;2].
On the interval [0;1]:
1
w2
FW ,Z (w,z) = [ zw + w(z " w)] = zw "
2
2
2
# FW ,Z (w,z)
fW ,Z (w,z) =
=1
#w#z
!
!
On the interval [1;2]:
1
w + wz " w 2
FW ,Z (w,z) = [ zw + w(1" w)] =
2
2
2
# FW ,Z (w,z) 1
fW ,Z (w,z) =
=
#w#z
2
b) 2 pts
By observation, we see that T = W + Z = X + Y. X and Y are independently distributed.
We have three cases: on [0;1], on [1;2] and on [2;3].
On the interval [0;1]:
1 t2 t2
t
FT = " = # fT =
2 2 4
2
!
!
On the interval [0;1]:
t + [1# (2 # t)] t 1
1
1
FT = "1"
= # $ fT =
2
2
2 4
2
On the interval [0;1]:
2
$ (3 # t) 2 '
1
(3 # t ) * f = 3 # t
FT = "1" &2 #
=
1#
)
T
2
2 (
4
2
%
!
2. Building a car – 5 pts
a) 2 pts
Mean time between arrivals of fidgets:
1
E[TF ] = = 10 _ min
"
Mean time between arrivals of whoosies:
)
1&$
E[TW ] = ( % t " (0.1" e#0.1"t + 0.02 " e#0.02"t ) dt+ = 30 _ min
2' 0
*
!
b.1) 1 pt
The Poisson process being memoryless, the mean time from our return to the workstation until
the first fidget arrives is 10 min.
!
b.2) 1 pt
Mean time to arrival of first whoosy:
2
# " W &2
,
1
1
E[TW ] ) E 2 [TW ]
2/
1+ %
1+
(
2 +
2 ) 30 1
.
2
E [TW ]
+
+2
$ E[TW ] '
1
E[VW ] = E 2 [TW ]
= 30 2
= 15 * .1+ 1
2E[TW ]
60
30 2
.
1
.10
, 2500 + 100 ) 900 /
E[VW ] = 15 * .1+
10 = 43.33_ min
900
!
b.3) 1pt
We can have two cases here: either the fidget arrives first, or it’s the whoosy.
43.3
P{Fidget _ first} =
$ ".e
0
!
43.3
# ".t
.dt =
$ 0.1% e
#0.1%t
= 0.987
0
If the fidget arrives first, then the total waiting time is the mean waiting time for the first whoosy to arrive: 43.33 min.
If the whoosy is first, however, we will have to wait 10 more minutes on average, since the
Poisson process is memoryless.
Therefore: E{Total _ waiting _ time} = 0.987 " 43.33 + (1# 0.987) " (43.33 + 10) = 43.46 _ min
!
3. Spatial Poisson on a line – 4 pts
a) 2 pts
The distance to first ambulance obeys to a Poisson process with a pdf d1 (t) = " .e#" .t
D1 is the corresponding cumulative distribution function.
1
Therefore: E[D1 ] = _ mile is the average length we have to check before finding an ambulance.
"
!
E[D1]
!
Road
Accident
This distance is centered on the accident: the average distance from the accident to the nearest
ambulance (E[D’1]) is half this value. Since the ambulance is driven at 1 mile per minute,
1
E[T'1 ] =
_ min
2."
b) 2 pts
!
1
mile on average to find the next ambulance, starting on
"
the border of the area previously searched (memoryless property of the Poisson pdf). However,
since we have already determined on which side was the previous ambulance, we know that the
second ambulance is necessarily on the other direction.
!
Similarly, we will again have to cover
Therefore, the distance to the second nearest ambulance is: E[D'2 ] =
E[T'2 ] =
1 1 3
+ =
_ min
2." " 2."
!
4. Covered rectangle – 4 pts
!
Coverage rectangle has sides of length L1 = [Max(X) – Min(X)] and L2 = [Max(Y) – Min(Y)].
Its area is A = L1 x L2.
10
$ x '10
FM (x) = P{Max(X) " x} = # P[X i " x] = & ) , x * [0;2]
% 2(
i=1
$ x '9
f M (x) = 5& )
% 2(
2
E[Max(X)] =
,
0
!
1 1 3
+ =
_ miles and
2." " 2."
2
$ x '9
5 - x11 0 20
x + 5 + & ) dx =
/ 2 =
%2(
512 . 11 10 11
& 2 # x )10
Fm (x) = P{Min(X) " x} = 1# P{Min(X) $ x} = 1# % P[X i $ x] = 1# (
+ , x , [0;2]
' 2 *
i=1
10
& 2 # x )9
f M (x) = 5(
+
' 2 *
2
E[Min(X)] =
.
0
2
2
&
2
/(2 # x)11 2
& 2 # x )9
& 2 # x )10 )
5 (/ x(2 # x)10 2
5
2
x - 5 -(
+ dx =
+ dx ++ =
1
4 # .(
1
4 =
(
' 2 *
512 '0 #10 30 0 ' #10 *
* 512 0 #110 30 11
Therefore, E[L1 ] =
20 2 18
" =
11 11 11
$ y '10
FM (y) = P{Max(Y ) " y} = # P[Yi " y] = & ) , y * [0;10]
% 10 (
i=1
10
!
!
$ y '9
Similarly, f M (y) = & )
%2(
10
E[Max(Y )] =
,
0
10
$ y '9
1 - y11 0
100
y + & ) dy = 9 / 2 =
%10 (
10 . 11 10
11
& 10 # y )10
Fm (y) = P{Min(Y ) " y} = 1# P{Min(Y ) $ y} = 1# % P[Yi $ y] = 1# (
+ , y , [0;10]
' 2 *
i=1
!
&10 # y ) 9
f M (x) = (
+
' 10 *
10
10
&
10
10
& 10 # y ) 9
&10 # y )10 ) 1 /(10 # y)11 2
1 (/ y(10 # y)10 2
10
+
E[Min(X)] = . y - (
+ dy = 9 (1
+ dy + = 9 1
4 # .(
4 =
' 10 *
10 '0 #10 30 0 ' #10 *
0
* 10 0 110 30 11
100 10 90
Therefore,
E[L2 ] =
" =
11 11 11
18 90 1620
As a consequence,
E[A] = E[L1 ] " E[L2 ] = "
=
# 13.39 _ miles2 # 33.67 _ km 2
11 11 121
10
!
! 5. Cookies, etc – 3pts
No standardized answer for this question, obviously.
!