1
00:00:01,310 --> 00:00:05,530
Besides PMFs and PDFs, we can
also describe the distribution

2
00:00:05,530 --> 00:00:09,330
of a random variable, as
we know, using a CDF.

3
00:00:09,330 --> 00:00:12,260
A CDF is always well-defined.

4
00:00:12,260 --> 00:00:14,970
And for the case of a continuous
random variable,

5
00:00:14,970 --> 00:00:18,640
the CDF can be found by
integrating the PDF.

6
00:00:18,640 --> 00:00:23,230
And conversely, we can recover
the PDF from the CDF by

7
00:00:23,230 --> 00:00:25,040
differentiating.

8
00:00:25,040 --> 00:00:28,010
There is something similar that
happens for the case of

9
00:00:28,010 --> 00:00:30,410
multiple random variables,
as well.

10
00:00:30,410 --> 00:00:37,390
We can define the joint CDF as
the probability that X and Y,

11
00:00:37,390 --> 00:00:43,290
the pair X-Y, takes values
that are below certain

12
00:00:43,290 --> 00:00:46,360
numbers, little x
and little y.

13
00:00:46,360 --> 00:00:51,930
So we are talking about the
probability of the blue set in

14
00:00:51,930 --> 00:00:53,180
this diagram.

15
00:00:55,780 --> 00:01:05,140
This probability can be found
by integrating the joint PDF

16
00:01:05,140 --> 00:01:07,830
over the blue set.

17
00:01:07,830 --> 00:01:12,300
And, since we're using x and y
to be some specific numbers,

18
00:01:12,300 --> 00:01:16,070
let us use some different dummy
variables to carry out

19
00:01:16,070 --> 00:01:17,320
the integration.

20
00:01:22,310 --> 00:01:25,000
What is the range of
the integration?

21
00:01:25,000 --> 00:01:29,160
The first variable, which is s
in this integral, ranges from

22
00:01:29,160 --> 00:01:31,226
minus infinity up to x.

23
00:01:34,021 --> 00:01:37,520
And the other variable, which
is the one that we're

24
00:01:37,520 --> 00:01:42,150
integrating with respect to,
in the outer integral--

25
00:01:42,150 --> 00:01:43,380
the t variable--

26
00:01:43,380 --> 00:01:45,910
ranges from minus
infinity to y.

27
00:01:50,000 --> 00:01:53,870
Now, let us see what happens if
we start taking derivatives

28
00:01:53,870 --> 00:01:55,610
of this expression.

29
00:01:55,610 --> 00:01:59,960
If we take the derivative of
this expression with respect

30
00:01:59,960 --> 00:02:04,475
to y, what is left is
the inner integral.

31
00:02:07,010 --> 00:02:11,810
And if we take, now, a
derivative with respect to x

32
00:02:11,810 --> 00:02:15,550
of this inner integral,
we will be left with

33
00:02:15,550 --> 00:02:20,090
just the joint PDF.

34
00:02:20,090 --> 00:02:24,600
And it will be the joint PDF
evaluated at the particular

35
00:02:24,600 --> 00:02:26,329
limits of the integration.

36
00:02:26,329 --> 00:02:32,420
So, it's going to be f sub
xy at little x, little y.

37
00:02:32,420 --> 00:02:35,010
So, we have this particular
formula.

38
00:02:35,010 --> 00:02:38,320
By taking derivative with
respect to x, and then with

39
00:02:38,320 --> 00:02:41,075
respect to y, or maybe in
the opposite order.

40
00:02:41,075 --> 00:02:42,790
It doesn't matter.

41
00:02:42,790 --> 00:02:48,000
This particular derivative
gives us back the PDF.

42
00:02:48,000 --> 00:02:50,190
Let us look at an example.

43
00:02:50,190 --> 00:02:52,730
Suppose that we have a uniform

44
00:02:52,730 --> 00:02:56,820
distribution on the unit square.

45
00:02:56,820 --> 00:03:01,540
So the PDF is equal to 1
on this green square.

46
00:03:01,540 --> 00:03:04,610
And is equal to 0 otherwise.

47
00:03:04,610 --> 00:03:08,970
So, in this example, if we take
some x and y, so that the

48
00:03:08,970 --> 00:03:16,060
xy pair falls inside the
rectangle, the probability of

49
00:03:16,060 --> 00:03:20,400
the blue set is going to be just
the probability of that

50
00:03:20,400 --> 00:03:22,140
little rectangle here.

51
00:03:22,140 --> 00:03:25,860
Because everything outside
has zero probability.

52
00:03:25,860 --> 00:03:30,240
With a uniform joint PDF,
which is equal to 1, the

53
00:03:30,240 --> 00:03:33,329
probability is just the area
of the set that we are

54
00:03:33,329 --> 00:03:34,150
considering.

55
00:03:34,150 --> 00:03:37,390
And since this set that we are
considering is a rectangle

56
00:03:37,390 --> 00:03:38,120
with [sides]

57
00:03:38,120 --> 00:03:44,280
x and y, the joint CDF is
equal to x times y.

58
00:03:44,280 --> 00:03:47,340
Now, if we take the derivative
of this expression with

59
00:03:47,340 --> 00:03:53,180
respect to x, and then with
respect to y, then we're left

60
00:03:53,180 --> 00:03:56,095
just with a constant
equal to 1--

61
00:03:59,620 --> 00:04:04,590
which is as it should be, so
that it integrates to one.

62
00:04:04,590 --> 00:04:08,870
So, we have seen that CDFs
also apply to the case of

63
00:04:08,870 --> 00:04:12,610
multiple random variables, and
that we can recover the joint

64
00:04:12,610 --> 00:04:14,350
PDF from the joint CDF.