1
00:00:00,000 --> 00:00:00,260
2
00:00:00,260 --> 00:00:03,330
In this problem, Romeo and
Juliet are to meet up for a
3
00:00:03,330 --> 00:00:08,530
date, where Romeo arrives at
time x and Juliet at time y,
4
00:00:08,530 --> 00:00:12,490
where x and y are independent
exponential random variables,
5
00:00:12,490 --> 00:00:13,750
with parameters lambda.
6
00:00:13,750 --> 00:00:16,340
And we're interested in knowing
the difference between
7
00:00:16,340 --> 00:00:19,450
the two times of arrivals,
we'll call it z,
8
00:00:19,450 --> 00:00:21,940
written as x minus y.
9
00:00:21,940 --> 00:00:25,180
And we'll like to know what
the distribution of z is,
10
00:00:25,180 --> 00:00:29,400
expressed by the probability
density function, f of z.
11
00:00:29,400 --> 00:00:32,430
Now, we'll do so by using the
so-called convolution formula
12
00:00:32,430 --> 00:00:34,220
that we learn in the lecture.
13
00:00:34,220 --> 00:00:38,130
Recall that if we have a random
variable w that is the
14
00:00:38,130 --> 00:00:44,160
sum of two independent random
variables, x plus y, now, if
15
00:00:44,160 --> 00:00:47,400
that's the case, we can write
the probability [INAUDIBLE]
16
00:00:47,400 --> 00:00:50,320
function, fw, [INAUDIBLE]
17
00:00:50,320 --> 00:00:52,360
as the following integration--
18
00:00:52,360 --> 00:01:01,130
negative infinity to infinity
fx little x times f of y w
19
00:01:01,130 --> 00:01:05,540
minus x, integrated over x.
20
00:01:05,540 --> 00:01:08,800
And to use this expression to
calculate f of z, we need to
21
00:01:08,800 --> 00:01:10,490
do a bit more work.
22
00:01:10,490 --> 00:01:14,170
Notice w is expressed as a sum
of two random variables,
23
00:01:14,170 --> 00:01:19,120
whereas z is expressed as the
subtraction of y from x.
24
00:01:19,120 --> 00:01:21,370
But that's fairly easy to fix.
25
00:01:21,370 --> 00:01:23,090
Now, we can write z.
26
00:01:23,090 --> 00:01:28,820
Instead of a subtraction, write
it as addition of x plus
27
00:01:28,820 --> 00:01:30,250
negative y.
28
00:01:30,250 --> 00:01:33,120
So in the expression of the
convolution formula, we'll
29
00:01:33,120 --> 00:01:37,225
simply replace y by negative
y, as it will
30
00:01:37,225 --> 00:01:38,640
show on the next slide.
31
00:01:38,640 --> 00:01:44,600
Using the convolution formula,
we can write f of z little z
32
00:01:44,600 --> 00:01:51,090
as the integration of f of x
little x and f of negative y z
33
00:01:51,090 --> 00:01:54,210
minus x dx.
34
00:01:54,210 --> 00:01:58,900
Now, we will use the fact that
f of negative y, evaluated z
35
00:01:58,900 --> 00:02:05,450
minus x, is simply equal to f
of y evaluated at x minus z.
36
00:02:05,450 --> 00:02:08,530
To see why this is true, let's
consider, let's say, a
37
00:02:08,530 --> 00:02:10,759
discreet random variable, y.
38
00:02:10,759 --> 00:02:15,830
And now, the probability that
negative y is equal to
39
00:02:15,830 --> 00:02:19,440
negative 1 is simply the same
as probability that
40
00:02:19,440 --> 00:02:21,400
y is equal to 1.
41
00:02:21,400 --> 00:02:24,930
And the same is true for
probability density functions.
42
00:02:24,930 --> 00:02:28,590
With this fact in mind, we can
further write equality as the
43
00:02:28,590 --> 00:02:38,900
integration of x times
f of y x minus z dx.
44
00:02:38,900 --> 00:02:40,120
We're now ready to compute.
45
00:02:40,120 --> 00:02:44,310
We'll first look at the case
where z is less than 0.
46
00:02:44,310 --> 00:02:46,490
On the right, I'm writing out
the distribution of an
47
00:02:46,490 --> 00:02:49,710
exponential random variable
with a parameter lambda.
48
00:02:49,710 --> 00:02:52,520
In this case, using the
integration above, we could
49
00:02:52,520 --> 00:02:57,130
write it as 0 to infinity,
lambda e to the negative
50
00:02:57,130 --> 00:03:05,130
lambda x times lambda e to the
negative lambda x minus z dx.
51
00:03:05,130 --> 00:03:08,530
Now, the reason we chose a
region to integrate from 0 to
52
00:03:08,530 --> 00:03:11,880
positive infinity is because
anywhere else, as we can
53
00:03:11,880 --> 00:03:15,910
verify from the expression of
fx right here, that the
54
00:03:15,910 --> 00:03:20,300
product of fx times
fy here is 0.
55
00:03:20,300 --> 00:03:21,560
Follow this through.
56
00:03:21,560 --> 00:03:23,080
We'll pull out the constant.
57
00:03:23,080 --> 00:03:28,430
Lambda e to the lambda z, the
integral from 0 to infinity,
58
00:03:28,430 --> 00:03:34,660
lambda e to the negative
2 lambda x dx.
59
00:03:34,660 --> 00:03:41,610
This will give us lambda e to
the lambda z minus 1/2 e to
60
00:03:41,610 --> 00:03:48,970
the negative 2 lambda x infinity
minus this expression
61
00:03:48,970 --> 00:03:50,900
value at 0.
62
00:03:50,900 --> 00:03:57,250
And this will give us lamdba
over 2 e to the lambda z.
63
00:03:57,250 --> 00:04:02,640
So now, we have an expression
for f of z evaluated at little
64
00:04:02,640 --> 00:04:07,660
z when little z is
less than 0.
65
00:04:07,660 --> 00:04:10,870
Now that have the distribution
of f of z when z is less than
66
00:04:10,870 --> 00:04:13,340
0, we'd like to know what
happens when z is greater or
67
00:04:13,340 --> 00:04:14,620
equal to 0.
68
00:04:14,620 --> 00:04:16,820
In principle, we can go through
the same procedure of
69
00:04:16,820 --> 00:04:19,450
integration and calculate
that value.
70
00:04:19,450 --> 00:04:22,250
But it turns out, there's
something much simpler.
71
00:04:22,250 --> 00:04:27,470
z is the difference between x
and y, at negative z, simply
72
00:04:27,470 --> 00:04:29,850
the difference between
y and x.
73
00:04:29,850 --> 00:04:33,710
Now, x and y are independent and
identically distributed.
74
00:04:33,710 --> 00:04:36,270
And therefore, x minus
y has the same
75
00:04:36,270 --> 00:04:38,760
distribution as y minus x.
76
00:04:38,760 --> 00:04:42,350
So that tells us z and negative
z have the same
77
00:04:42,350 --> 00:04:43,300
distribution.
78
00:04:43,300 --> 00:04:46,650
What that means is, is the
distribution of z now must be
79
00:04:46,650 --> 00:04:48,610
symmetric around 0.
80
00:04:48,610 --> 00:04:53,530
In other words, if we know that
the shape of f of z below
81
00:04:53,530 --> 00:04:57,910
0 is something like that, then
the shape of it above 0 must
82
00:04:57,910 --> 00:04:59,190
be symmetric.
83
00:04:59,190 --> 00:05:01,570
So here's the origin.
84
00:05:01,570 --> 00:05:06,750
For example, if we were to
evaluate f of z at 1, well,
85
00:05:06,750 --> 00:05:12,140
this will be equal to the value
of f of z at negative 1.
86
00:05:12,140 --> 00:05:16,880
So this will equal to f
of z at negative 1.
87
00:05:16,880 --> 00:05:20,490
Well, with this information in
mind, we know that in general,
88
00:05:20,490 --> 00:05:26,310
f of z little z is equal to
f of z negative little z.
89
00:05:26,310 --> 00:05:30,750
So what this allows us to do is
to get all the information
90
00:05:30,750 --> 00:05:35,290
for z less than 0 and generalize
it to the case
91
00:05:35,290 --> 00:05:38,250
where z is greater
or equal to 0.
92
00:05:38,250 --> 00:05:42,150
In particular, by the symmetry
here, we can write, for the
93
00:05:42,150 --> 00:05:46,960
case z greater or equal to 0,
as lambda over 2 e to the
94
00:05:46,960 --> 00:05:49,410
negative lambda z.
95
00:05:49,410 --> 00:05:52,350
So the negative sign comes
from the fact that the
96
00:05:52,350 --> 00:05:55,650
distribution of f of z is
symmetric around 0.
97
00:05:55,650 --> 00:05:58,080
And simply, we can go back
to the expression
98
00:05:58,080 --> 00:06:00,370
here to get the value.
99
00:06:00,370 --> 00:06:04,220
And all in all, this implies
that f of z little z is equal
100
00:06:04,220 --> 00:06:07,780
to lambda over 2 e to
the negative lambda
101
00:06:07,780 --> 00:06:09,820
absolute value of z.
102
00:06:09,820 --> 00:06:11,070
This completes our problem.
103
00:06:11,070 --> 00:06:12,150