WEBVTT
00:00:02.240 --> 00:00:05.620
We now look at an example
similar to the previous one,
00:00:05.620 --> 00:00:08.750
in which we have again two
scenarios, but in which we
00:00:08.750 --> 00:00:11.160
have both discrete
and continuous
00:00:11.160 --> 00:00:13.600
random variables involved.
00:00:13.600 --> 00:00:15.620
You have $1 and the opportunity
00:00:15.620 --> 00:00:17.200
to play in the lottery.
00:00:17.200 --> 00:00:21.730
With probability 1/2, you do
nothing and you're left with
00:00:21.730 --> 00:00:24.426
the dollar that you
started with.
00:00:24.426 --> 00:00:28.330
With probability 1/2, you decide
to play the lottery.
00:00:28.330 --> 00:00:31.780
And in that case, you get back
an amount of money which is
00:00:31.780 --> 00:00:34.120
random and uniformly
distributed
00:00:34.120 --> 00:00:37.100
between zero and two.
00:00:37.100 --> 00:00:40.170
Is the random variable,
X, discrete?
00:00:40.170 --> 00:00:44.120
The answer is no, because
it takes values on
00:00:44.120 --> 00:00:46.830
a continuous range.
00:00:46.830 --> 00:00:50.670
Is the random variable,
X, continuous?
00:00:50.670 --> 00:00:56.790
The answer is no, because the
probability that X takes the
00:00:56.790 --> 00:01:00.920
value of exactly one
is equal to 1/2.
00:01:04.209 --> 00:01:07.900
Even though X takes values in
a continuous range, this is
00:01:07.900 --> 00:01:10.530
not enough to make it a
continuous random variable.
00:01:10.530 --> 00:01:13.460
We defined continuous random
variables to be those that can
00:01:13.460 --> 00:01:15.460
be described by a PDF.
00:01:15.460 --> 00:01:18.930
And you have seen it in such a
case, any individual point
00:01:18.930 --> 00:01:20.880
should have zero probability.
00:01:20.880 --> 00:01:25.060
But this is not the case here,
and so X is not continuous.
00:01:25.060 --> 00:01:28.840
We call X a mixed
random variable.
00:01:28.840 --> 00:01:32.770
More generally, we can have a
situation where the random
00:01:32.770 --> 00:01:37.300
variable X with some probability
is the same as a
00:01:37.300 --> 00:01:41.320
particular discrete random
variable, and with some other
00:01:41.320 --> 00:01:44.140
probability it is equal
to some other
00:01:44.140 --> 00:01:46.420
continuous random variable.
00:01:46.420 --> 00:01:50.550
Such a random variable, X, does
not have a PMF because it
00:01:50.550 --> 00:01:52.240
is not discrete.
00:01:52.240 --> 00:01:56.880
Also, it does not have a PDF
because it is not continuous.
00:01:56.880 --> 00:01:59.479
How do we describe such
a random variable?
00:01:59.479 --> 00:02:02.450
Well, we can describe it in
terms of a cumulative
00:02:02.450 --> 00:02:04.070
distribution function.
00:02:04.070 --> 00:02:07.760
CDFs are always well
defined for all
00:02:07.760 --> 00:02:10.000
kinds of random variables.
00:02:10.000 --> 00:02:13.290
We have two scenarios, and
so we can use the Total
00:02:13.290 --> 00:02:18.890
Probability Theorem and write
that the CDF is equal to the
00:02:18.890 --> 00:02:23.600
probability of the first
scenario, which is p, times
00:02:23.600 --> 00:02:27.240
the probability that the random
variable Y is less than
00:02:27.240 --> 00:02:28.820
or equal to x.
00:02:28.820 --> 00:02:31.900
This is a conditional model
under the first scenario.
00:02:31.900 --> 00:02:34.870
And with some probability, we
have the second scenario.
00:02:34.870 --> 00:02:38.370
And under that scenario, X will
take a value less than
00:02:38.370 --> 00:02:42.800
little x, if and only if our
random variable Z will take a
00:02:42.800 --> 00:02:45.280
value less than little x.
00:02:45.280 --> 00:02:51.510
Or in CDF notation, this is p
times the CDF of the random
00:02:51.510 --> 00:02:57.620
variable Y evaluated at this
particular x plus another
00:02:57.620 --> 00:03:07.520
weighted term involving the CDF
of the random variable Z.
00:03:07.520 --> 00:03:11.820
We can also define the expected
value of X in a way
00:03:11.820 --> 00:03:15.940
that is consistent with the
Total Expectation Theorem,
00:03:15.940 --> 00:03:20.290
namely define the expected
value of X to be the
00:03:20.290 --> 00:03:23.850
probability of the first
scenario, in which case X is
00:03:23.850 --> 00:03:26.870
discrete times the expected
value of the associated
00:03:26.870 --> 00:03:30.120
discrete random variable, plus
the probability of the second
00:03:30.120 --> 00:03:34.740
scenario, under which X is
continuous, times the expected
00:03:34.740 --> 00:03:40.200
value of the associated
continuous random variable.
00:03:40.200 --> 00:03:42.670
Going back to our original
example, we have two
00:03:42.670 --> 00:03:49.242
scenarios, the scenarios that
we can call A1 and A2.
00:03:49.242 --> 00:03:54.110
Under the first scenario, we
have a uniform PDF, and the
00:03:54.110 --> 00:03:58.010
corresponding CDF
is as follows.
00:03:58.010 --> 00:04:01.880
It's flat until zero, then
it rises linearly.
00:04:01.880 --> 00:04:04.880
And then it stays flat,
and the value
00:04:04.880 --> 00:04:06.900
here is equal to one.
00:04:06.900 --> 00:04:11.750
So the slope here is 1/2.
00:04:11.750 --> 00:04:15.050
So the slope is equal to
the corresponding PDF.
00:04:15.050 --> 00:04:18.029
Under the second scenario, we
have a discrete, actually a
00:04:18.029 --> 00:04:19.529
constant random variable.
00:04:19.529 --> 00:04:25.020
And so the CDF is flat at zero
until this value, and at that
00:04:25.020 --> 00:04:29.800
value we have a jump
equal to one.
00:04:29.800 --> 00:04:32.690
We then use the Total
Probability Theorem, which
00:04:32.690 --> 00:04:36.700
tells us that the CDF of the
mixed random variable will be
00:04:36.700 --> 00:04:41.450
1/2 times the CDF under the
first scenario plus 1/2 times
00:04:41.450 --> 00:04:43.850
the CDF under the
second scenario.
00:04:43.850 --> 00:04:48.590
So we take 1/2 of this plot
and 1/2 of that plot
00:04:48.590 --> 00:04:49.990
and add them up.
00:04:49.990 --> 00:04:56.905
What we get is a function that
rises now at the slope of 1/4.
00:05:01.700 --> 00:05:05.970
Then we have a jump, and the
size of that to jump is going
00:05:05.970 --> 00:05:09.970
to be equal to 1/2.
00:05:09.970 --> 00:05:16.770
And then it continues at a slope
of 1/4 until it reaches
00:05:16.770 --> 00:05:17.470
this value.
00:05:17.470 --> 00:05:19.915
And after that time,
it remains flat.
00:05:22.540 --> 00:05:25.090
So this is a simple illustration
that for mixed
00:05:25.090 --> 00:05:27.720
random variables it's not
too hard to obtain the
00:05:27.720 --> 00:05:31.590
corresponding CDF even though
this random variable does not
00:05:31.590 --> 00:05:34.370
have a PDF or a PMF of its own.