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Welcome students, the MOOC's online course
on Statistical Inference. This is lecture
00:00:25.480 --> 00:00:37.960
number 11 in this week. We are studying Order
Statistics, in the last two lectures we have
00:00:37.960 --> 00:00:47.940
developed the theory and we have seen some
problems involving order statistics. In this
00:00:47.940 --> 00:00:58.609
class I will also solve quite a few problems
which will help you in solving the problems
00:00:58.609 --> 00:01:05.320
given in the tutorial sheets and some more
difficult problems in general.
00:01:05.320 --> 00:01:35.399
Consider exponential distribution let x 1
and x 2 be, two independent samples from exponential
00:01:35.399 --> 00:02:04.720
1. Therefore, f x is equal to e to the power
minus x. What is the expected value, of the
00:02:04.720 --> 00:02:25.640
maximum of x and y? Note that I have changed
the notation from x 1, x 2 to x this is just
00:02:25.640 --> 00:02:52.700
for convenience.
So, solution
00:02:52.700 --> 00:03:27.060
so what is the Pdf of x 2? Pdf of x 2 is 2
1 minus e to the power minus x. This is the
00:03:27.060 --> 00:03:34.330
capital f x or the cumulative distribution
function and e to the power minus x is the
00:03:34.330 --> 00:03:51.450
small f x or the Pdf. Therefore, expected
value of x 2 is equal to integration 0 to
00:03:51.450 --> 00:04:05.280
infinity x 2 1 minus e to the power minus
x into e to the power minus x dx.
00:04:05.280 --> 00:04:22.330
Now, we can split it into two parts, 0 to
infinity 2 x e to the power minus x dx minus
00:04:22.330 --> 00:04:40.300
0 to infinity 2 x minus e to the power 2 x
into e to the power minus 2 x dx.
00:04:40.300 --> 00:04:55.430
So, part 1 is equal to 0 to infinity 2 x e
to the power minus x dx is equal to 2 into
00:04:55.430 --> 00:05:10.810
0 to infinity x e to the power minus x dx
is equal to 2 into expected value of x where
00:05:10.810 --> 00:05:20.290
x follows exponential with 1 and we know that
if a random variable follows exponential with
00:05:20.290 --> 00:05:32.291
lambda. Then it is expectation is 1 upon lambda
therefore, this comes out to be 2 upon 1 is
00:05:32.291 --> 00:05:47.470
equal to 2.
But 2 is 0 to infinity two x e to the power
00:05:47.470 --> 00:06:08.070
minus 2 x dx is equal to 0 to infinity x into
2 e to the power minus 2 x dx and this part
00:06:08.070 --> 00:06:23.640
is expected value of x where x is exponential
with 2.
00:06:23.640 --> 00:06:34.600
Therefore this expected value is going to
be 1 upon 2 because if the parameter is lambda
00:06:34.600 --> 00:06:51.401
then expected value is 1 upon lambda. Therefore,
the answer is 2 minus 1 upon 2 is equal to
00:06:51.401 --> 00:07:13.190
3 by 2. That is expected value of max of x
comma y is equal to 3 by two that is the answer.
00:07:13.190 --> 00:07:25.810
Now, let me consider slightly more challenging
problem the setup is same. We are working
00:07:25.810 --> 00:08:01.900
on X, Y there independent samples from exponential
lambda. Our aim is to find out the Pdf of
00:08:01.900 --> 00:08:17.030
minimum of X, Y upon maximum of X, Y . And
note that these two are not quite independent.
00:08:17.030 --> 00:08:25.320
So, one way of doing it is we can look at
that joint distribution of this to minimum
00:08:25.320 --> 00:08:36.890
of X, Y and maximum of X, Y and from there
we may like to compute the Pdf of their ratios.
00:08:36.890 --> 00:08:50.149
I do it in a slightly tricky way. So, let
us proceed note that
00:08:50.149 --> 00:09:08.649
this value is always between
00:09:08.649 --> 00:09:26.310
between 0 and 1.
Let z equal to minimum of X, Y upon maximum
00:09:26.310 --> 00:09:42.629
of X,Y. So, we are looking at probability
Z less than equal to z that is we are looking
00:09:42.629 --> 00:10:02.470
at the cdf of Z. Where 0 less than Z less
than 1. Now this event we can have in two
00:10:02.470 --> 00:10:27.830
possible ways, one is x by y less than equal
to z and x is less than y plus probability
00:10:27.830 --> 00:10:37.130
y by x is less than equal to z and y less
than x .
00:10:37.130 --> 00:10:56.690
So, consider x by y less than equal to z and
x less than y. This is equivalent to x less
00:10:56.690 --> 00:11:09.209
than equal to y z and since z is less than
one this is automatically satisfied.
00:11:09.209 --> 00:11:23.250
Similarly, a
event y by x less than equal to z and y less
00:11:23.250 --> 00:11:35.490
than x is equivalent to, y less than equal
to x z.
00:11:35.490 --> 00:11:56.870
Therefore F z at z that is probability Z less
than equal to z is equal to probability x
00:11:56.870 --> 00:12:13.670
less than equal to y z plus probability y
less than equal to xz. Is equal to, In the
00:12:13.670 --> 00:12:29.089
first case; x will have to be less than equal
to yz and y can be anything between 0 to infinity.
00:12:29.089 --> 00:12:43.499
Therefore, range of y is 0 to infinity , but
for a given y x can go only from 0 to y z.
00:12:43.499 --> 00:13:02.950
And here we are writing the joint density
of X, Y dx dy plus a very symmetric term which
00:13:02.950 --> 00:13:20.139
is 0 to infinity integration 0 to x z f xy
xy dy dx.
00:13:20.139 --> 00:13:28.070
So, this is clear this is the joint density
of X, Y both are independent. So, you can
00:13:28.070 --> 00:13:37.260
write it as a product of these two and then
x for a given y x ranges between 0 to y z.
00:13:37.260 --> 00:13:44.249
Note that the z is fixed here and then y,
I am integrating it over y from 0 to infinity
00:13:44.249 --> 00:13:53.380
and a very similar thing we are doing in the
second case.
00:13:53.380 --> 00:14:25.529
Therefore the Pdf of z is equal to f z z can
be obtained by differentiating
00:14:25.529 --> 00:14:48.959
the cdf of z with respect to z.
So, let us consider the first term 0 to infinity
00:14:48.959 --> 00:15:13.069
0 to y z f of X, Y x comma y dx dy. So, when
we differentiate with respect to z look that
00:15:13.069 --> 00:15:28.860
the outer term outer integration does not
involve any z. Therefore, dz of this term
00:15:28.860 --> 00:15:54.690
is equal to 0 to infinity 0 to y z f X, Y
X, Y dx dy is equal to ok let me write it
00:15:54.690 --> 00:16:05.119
again.
0 to infinity d dz of integration 0 to y z
00:16:05.119 --> 00:16:21.459
f xy at xy dx the whole thing is dy
is equal to 0 to infinity.
00:16:21.459 --> 00:16:37.339
Now, we are cutting out this differentiation
therefore, it is going to be f xy yz .the
00:16:37.339 --> 00:16:50.259
value of x has to be replaced with y z, that
is y z comma y multiplied by dyz dz which
00:16:50.259 --> 00:17:18.020
is is equal to y dy is equal to 0 to infinity.
what is fx y? x is exponential with lambda
00:17:18.020 --> 00:17:23.230
and y is also exponential with lambda and
they are independent.
00:17:23.230 --> 00:17:39.700
So, we can write it as fx at yz multiplied
by f y at y into y dy is equal to integration
00:17:39.700 --> 00:17:54.100
0 to infinity lambda e to the power minus
lambda y z multiplied by lambda e to the power
00:17:54.100 --> 00:18:12.260
minus lambda y y dy which is is equal to integration
0 to infinity y lambda square e to the power
00:18:12.260 --> 00:18:30.299
minus lambda y into z plus 1 dy.
Is equal to lambda square integration 0 to
00:18:30.299 --> 00:18:45.409
infinity y e to the power minus lambda z plus
one into y dy.
00:18:45.409 --> 00:19:03.120
Now, let us look at this term, this resembles
exponential distribution where for a fix z
00:19:03.120 --> 00:19:14.120
the parameter is lambda into z plus 1 therefore,
suppose we write it as lambda square upon
00:19:14.120 --> 00:19:27.720
lambda into z plus 1 into 0 to infinity lambda
into z plus 1 y e to the power minus lambda
00:19:27.720 --> 00:19:44.840
into z plus 1 y dy. Then basically this is
expected value of a random variable which
00:19:44.840 --> 00:19:55.049
is distributed as exponential with lambda
z plus 1. If you accept that part then we
00:19:55.049 --> 00:20:05.769
know that this expectation is going to be
1 upon lambda z plus 1.
00:20:05.769 --> 00:20:22.169
Therefore what we get is
lambda square into lambda z plus 1 , multiplied
00:20:22.169 --> 00:20:41.309
by 1 upon lambda z plus 1 is equal to 1 upon
z plus 1 square. Therefore, from the first
00:20:41.309 --> 00:21:03.500
part we get 1 upon z plus 1 square. If we
look at the holder expression we had this
00:21:03.500 --> 00:21:17.590
as the first term and this has the second
term. And if we watch carefully we find that
00:21:17.590 --> 00:21:25.139
these two are basically the same thing because
this is independent . So, in the second case
00:21:25.139 --> 00:21:32.820
when I am differentiating with respect to
z it is going to be fx y at x comma x z and
00:21:32.820 --> 00:21:47.620
multiplied by x.
So, let me write it down d dz of 0 to infinity
00:21:47.620 --> 00:22:08.110
0 to x z fxy x comma y dy dx. And in a very
similar way what we did for that first part
00:22:08.110 --> 00:22:24.919
this is going to be 0 to infinity.
d dz of 0 to infinity 0 to x z f xy X, Y dy
00:22:24.919 --> 00:22:39.020
dx. This is is equal to 0 to infinity f xy
at x coma. Now y is going to get this value
00:22:39.020 --> 00:22:53.510
x z, now we differentiate with xz with respect
to z that gives us an x dx is equal to 0 to
00:22:53.510 --> 00:23:13.350
infinity. fx at x if y at xz x dx is equal
to 0 to infinity.
00:23:13.350 --> 00:23:26.190
lambda e to the power minus lambda x multiplied
by lambda e to the power minus lambda x z
00:23:26.190 --> 00:23:38.740
into x into dx is equal to 0 to infinity lambda
square, e to the power minus lambda z plus
00:23:38.740 --> 00:23:50.090
one x x dx.
And now we see that this becomes same as what
00:23:50.090 --> 00:23:59.120
we get in the first term. Therefore, without
evaluating it any further we can say that
00:23:59.120 --> 00:24:18.350
this is going to give us one upon one plus
z whole square. Therefore, the Pdf of minimum
00:24:18.350 --> 00:24:35.970
over x comma y upon maximum over x comma y
is equal to 2 into 1 plus z whole square 0
00:24:35.970 --> 00:24:51.230
less than z less than 1.
Now, many of you may wonder why all the examples
00:24:51.230 --> 00:25:01.669
that I am giving are from uniform distribution
or from exponential distribution.
00:25:01.669 --> 00:25:37.190
Why? The advantage of uniform and exponential
distributions or both of them have closed
00:25:37.190 --> 00:25:57.659
form for there cdf and we know that in the
distribution of order statistics cdf or capital
00:25:57.659 --> 00:26:08.940
fx plays very important role because it comes
in the Pdf of the order statistic. And therefore,
00:26:08.940 --> 00:26:16.080
the advantage of these two distributions is
that we can easily handle them under order
00:26:16.080 --> 00:26:22.720
statistics.
So, no wonder that most of our examples for
00:26:22.720 --> 00:26:34.500
this topic have come from uniform and exponential
distribution. Now now I give you a problem
00:26:34.500 --> 00:26:42.950
where I will go beyond exponential and uniform.
In particular I look at normal distribution
00:26:42.950 --> 00:27:04.910
and we shall see how complicated it becomes.
X and Y are independent samples from normal
00:27:04.910 --> 00:27:31.100
0 sigma square. What is the expected value
of the minimum of x and y? So, that is the
00:27:31.100 --> 00:28:01.860
question. 2 into 1 minus F x into small f
x.
00:28:01.860 --> 00:28:17.059
Therefore expected value of minimum of xy
is equal to integration minus infinity to
00:28:17.059 --> 00:28:40.600
infinity 2 x 1 minus f x into small f x d
x is equal to minus infinity to infinity 2
00:28:40.600 --> 00:29:01.090
x to 1 minus minus infinity to x 1 over root
over 2 pi sigma e to the power minus y square
00:29:01.090 --> 00:29:15.919
upon 2 sigma square d y multiplied by 1 over
root over 2 pi sigma e to the power minus
00:29:15.919 --> 00:29:36.970
x square upon 2 sigma square d x.
So, this I write as a sum of two parts 2 x
00:29:36.970 --> 00:29:46.750
into 1 over root over 2 pi sigma e to the
power minus x square upon 2 sigma square dx
00:29:46.750 --> 00:29:59.950
I have taken out the 1 and this gives this
minus minus infinity to infinity 2 x into
00:29:59.950 --> 00:30:12.929
1 over root over 2 pi sigma e to the power
minus x square 2 sigma square. This I have
00:30:12.929 --> 00:30:23.270
written first multiplied by the integration
of minus infinity to x 1 over root over 2
00:30:23.270 --> 00:30:36.889
pi sigma e to the power minus y square upon
2 sigma square dy and then whole thing is
00:30:36.889 --> 00:30:43.179
dx.
So, now you understand how complicated this
00:30:43.179 --> 00:30:51.820
expression becomes. Although this part is
very very simple because this part gives us
00:30:51.820 --> 00:31:08.240
0, why because this is 2 times expectation
of x where x is normal with 0 sigma square.
00:31:08.240 --> 00:31:22.629
Therefore, whole concentration goes now to
this term and that is complicated I argue
00:31:22.629 --> 00:31:29.009
to work up on this one and see your mathematics
is strong enough to handle this thing. But
00:31:29.009 --> 00:31:44.330
what I will show you today are shortcut or
a tricky way to solve the same problem.
00:31:44.330 --> 00:32:16.120
So, we were at this stage 2 times minus infinity
to infinity x into 1 minus fx into fx dx.
00:32:16.120 --> 00:32:36.759
Which we write as follows 2 times minus infinity
to infinity. 1 minus fx into x times fx dx
00:32:36.759 --> 00:32:54.330
where f is 1 over root over 2 pi sigma e to
the power minus x squared upon 2 sigma square.
00:32:54.330 --> 00:33:11.200
Therefore log of fx is equal to log of 1 upon
root over 2 pi sigma minus x squared upon
00:33:11.200 --> 00:33:31.419
2 sigma square therefore, d log f x dx is
equal to we are differentiating this with
00:33:31.419 --> 00:33:38.919
respect to x this becomes 0 because it is
a constant and what we are getting is minus
00:33:38.919 --> 00:33:53.399
2 x upon 2 sigma square is equal to minus
x upon sigma square.
00:33:53.399 --> 00:34:15.000
Now, on the left hand side we get d log fx
dx is equal to 1 over f x into f prime x.
00:34:15.000 --> 00:34:33.089
Therefore, f prime x upon fx is equal to minus
x upon sigma square or x fx is equal to minus
00:34:33.089 --> 00:34:46.250
sigma square f prime x . So, we got an expression
involving x fx which we can now replace with
00:34:46.250 --> 00:34:59.430
minus sigma square f prime x. So, let us go
back to our expression. We had 2 into this
00:34:59.430 --> 00:35:08.930
which involves x time fx and we know that
we can replace it with minus sigma square
00:35:08.930 --> 00:36:13.059
into f prime x .
So, let us focus on minus infinity to infinity
00:36:13.059 --> 00:36:31.309
1 minus F x into a prime x dx this is is equal
to first function into integration of second
00:36:31.309 --> 00:36:43.930
function and these we put the limit minus
infinity to infinity minus minus infinity
00:36:43.930 --> 00:37:07.400
to infinity fx. Now derivative of the first
function is minus fx dx. Is equal to now the
00:37:07.400 --> 00:37:16.230
first part that x is equal to infinity capital
Fx is equal to one therefore, it is 0 and
00:37:16.230 --> 00:37:23.819
at x is equal to minus infinity all small
fx is equal to 0 as the normal distribution
00:37:23.819 --> 00:37:33.980
tappers on both sides therefore, the first
part is 0. Therefore, what we are getting
00:37:33.980 --> 00:37:52.170
is integration minus infinity to infinity
f x into fx dx.
00:37:52.170 --> 00:38:05.230
Is equal to integration minus infinity to
infinity 1 over root over 2 pi sigma e to
00:38:05.230 --> 00:38:13.599
the power minus x square, to sigma square
multiplied by 1 over root over 2 pi sigma
00:38:13.599 --> 00:38:31.270
dx is equal to 1 over 2 pi sigma square minus
infinity to infinity e to the power minus
00:38:31.270 --> 00:38:48.020
2 x square 2 sigma square dx is equal to 1
over 2 pi sigma square into minus infinity
00:38:48.020 --> 00:38:58.080
to infinity e to the power minus x square
upon sigma square d x and remember that we
00:38:58.080 --> 00:39:11.170
started with
minus to sigma square into minus infinity
00:39:11.170 --> 00:39:32.430
to infinity 1 minus fx into a prime x dx.
Therefore our answer is minus 2 sigma square
00:39:32.430 --> 00:39:45.550
into 1 over 2 pi sigma square into minus infinity
to infinity, e to the power minus x square
00:39:45.550 --> 00:39:59.569
upon sigma square dx. Is equal to minus 1
upon pi into minus infinity to infinity e
00:39:59.569 --> 00:40:15.500
to the power minus x square upon sigma square
dx. Is equal to minus 1 upon pi minus infinity
00:40:15.500 --> 00:40:30.270
to infinity e to the power minus x square
upon 2 into sigma over root 2 whole square
00:40:30.270 --> 00:40:41.599
dx. Why I did it? I wanted to bring it into
2 sigma square form and therefore, I had multiplied
00:40:41.599 --> 00:40:47.339
with 2 and divided it by 2 and when I put
it inside the square it becomes sigma over
00:40:47.339 --> 00:40:55.099
root 2.
What is this? If we multiply it with root
00:40:55.099 --> 00:41:03.770
over 1 over root over 2 pi sigma over root
2. Then that is going to give me the density
00:41:03.770 --> 00:41:13.460
function of a normal distribution. So, let
me write it as 1 over pi minus infinity to
00:41:13.460 --> 00:41:29.190
infinity 1 over root over 2 pi sigma over
root 2 e to the power minus x square into
00:41:29.190 --> 00:41:45.430
2 sigma over root 2 whole square multiplied
by to compensate for this i write sigma root
00:41:45.430 --> 00:41:53.400
pi because these cancels and this gives me
sigma root pi dx.
00:41:53.400 --> 00:42:06.799
Is equal to minus 1 over pi sigma root pi
integration minus infinity to infinity 1 over
00:42:06.799 --> 00:42:15.099
root over 2 pi sigma over root 2 into e to
the power minus x square upon 2 sigma over
00:42:15.099 --> 00:42:24.490
root 2 whole square dx and this part is equal
to 1 because I am integrating a Pdf from minus
00:42:24.490 --> 00:42:35.500
infinity to infinity therefore, this is equal
to minus sigma over root pi . So, that is
00:42:35.500 --> 00:43:03.210
the answer.
Now, you can see that since we were dealing
00:43:03.210 --> 00:43:10.990
with a normal distribution things have not
been as straightforward as we had with respective
00:43:10.990 --> 00:43:22.270
uniform or exponential distribution, but we
have solved the problem in a tricky way. And
00:43:22.270 --> 00:43:29.829
there is a longer way of doing it which I
have done half way and left it as a exercise
00:43:29.829 --> 00:43:37.840
for you to solve it at home. Now let me give
you another tricky problem.
00:43:37.840 --> 00:44:09.069
Show that if x is a symmetric distribution
around mean is equal to mu, then f r at mu
00:44:09.069 --> 00:44:30.990
minus x is equal to f of n minus r plus 1
at mu plus x that is x is a distribution which
00:44:30.990 --> 00:44:43.090
is symmetric around the mean mu. So, let us
assume some arbitrary just Pdf and suppose
00:44:43.090 --> 00:44:51.869
we have taken n samples x 1 let me order them
x 2 x n.
00:44:51.869 --> 00:45:02.369
So, if the Pdf of the rth order statistics
from the beginning at mu minus x will be same
00:45:02.369 --> 00:45:14.359
as the n minus r plus 1 at statistics or the
rth order statistics from this side at mu
00:45:14.359 --> 00:45:24.480
plus x if sorry if this is mu minus x this
is mu plus x. These two will have the same
00:45:24.480 --> 00:45:45.300
density function. How do you prove it mathematically.
Now f r at mu minus x is equal to n factorial
00:45:45.300 --> 00:46:02.700
upon r minus one factorial into n minus r
factorial f at nu minus x whole to the power
00:46:02.700 --> 00:46:23.980
r minus 1 1 minus f nu minus x at n minus
r multiplied by f at nu minus x . This we
00:46:23.980 --> 00:46:33.030
know from our earlier classes that first a
r minus 1 has to be less than nu minus x and
00:46:33.030 --> 00:46:40.049
that probabilities f mu minus x multiplied
by r minus 1 the first r minus 1 have been
00:46:40.049 --> 00:46:49.650
taken care of the rth1 at mu minus x. That
we know that will be given by f mu minus x
00:46:49.650 --> 00:46:58.040
and the remaining n minus r all of them should
be above mu minus x. So, why my 1 minus f
00:46:58.040 --> 00:47:15.280
mu minus x whole to the power n minus r.
Also f n minus r plus 1 at mu plus x is equal
00:47:15.280 --> 00:47:24.740
to n factorial. We are imitating this instead
of what now I am putting n minus r plus 1.
00:47:24.740 --> 00:47:34.770
So, this is going to be n minus r factorial
instead of or I am putting n minus r plus
00:47:34.770 --> 00:47:54.799
1 here . So, that is going to be r minus 1
factorial f at mu plus x whole to the power
00:47:54.799 --> 00:48:05.230
n minus r because n minus r plus 1. So, one
will be less so it is n minus r 1 minus f
00:48:05.230 --> 00:48:18.589
at mu plus x how many will be on this side
there will be only r minus 1 of them because
00:48:18.589 --> 00:48:26.730
I am looking at the rth from the end multiplied
by f at nu plus x.
00:48:26.730 --> 00:48:56.210
So, we got the two expressions here.
Since f is symmetric around mu f of mu plus
00:48:56.210 --> 00:49:18.190
x is equal to f of mu minus x. Also
f at mu minus x is equal to 1 minus f of mu
00:49:18.190 --> 00:49:28.630
plus x. That is very simple because it is
symmetric f of mu minus x is this and f of
00:49:28.630 --> 00:49:37.609
mu plus x is this. Therefore, 1 minus therefore,
f of mu minus x is equal to 1 minus f of mu
00:49:37.609 --> 00:49:48.130
plus x therefore, if I put instead of f of
mu minus x 1 minus f of mu plus x and instead
00:49:48.130 --> 00:49:58.340
of f of mu plus x 1 minus f nu minus x. We
can see that these two terms are same that
00:49:58.340 --> 00:50:16.390
is therefore, f r at mu minus x is equal to
f of n minus r plus 1 at nu plus x.
00:50:16.390 --> 00:50:32.599
I will stop with one more problem.
And this time will be looking at Order Statistics
00:50:32.599 --> 00:50:52.490
from a finite population. Suppose the population
is
00:50:52.490 --> 00:51:04.140
x one ;
that means, we have a population of n items
00:51:04.140 --> 00:51:20.750
and they put them in increasing order x 1
x 2 xn. We take a random sample
00:51:20.750 --> 00:51:43.650
of size n x 1 x 2 xn let the corresponding
order statistic
00:51:43.650 --> 00:52:13.410
be x 1 x 2 x n. And then we want to compute
some of the probabilities.
00:52:13.410 --> 00:52:29.780
So, what we are looking at, that in the sample
the rth order statistic is actually the dth
00:52:29.780 --> 00:52:46.150
element in the sequence. Since this part is
sorted
00:52:46.150 --> 00:52:57.160
and I am looking at that the rth order statistic
is same as this what is going to be the probability.
00:52:57.160 --> 00:53:06.720
Since rth order statistics is equal to xd.
We have exactly d minus 1 element on this
00:53:06.720 --> 00:53:24.450
side and in the sample, we have r minus 1
element below this. Therefore, note that d
00:53:24.450 --> 00:53:41.760
cannot be less than r or in other words suppose
I am looking at ten samples from 1 2 3 up
00:53:41.760 --> 00:53:51.460
to 100. Then the fifth order statistic cannot
be less than 5. it has to be at least 5 right.
00:53:51.460 --> 00:53:58.369
So, that is what I am saying here that the
rth order statistic is is equal to xd means
00:53:58.369 --> 00:54:19.150
they cannot be less than r.
So, what is the probability? So, among the
00:54:19.150 --> 00:54:36.381
first d minus 1 I have to choose r minus 1
in d minus 1 c r minus 1 ways right. So, x
00:54:36.381 --> 00:54:54.820
1 less than x 2 . So, from here I have to
choose r minus one and from here I have to
00:54:54.820 --> 00:55:10.860
choose n minus r and 1 will get from xd. So,
this can be chosen in d minus 1 c r minus
00:55:10.860 --> 00:55:21.299
1 ways, Here how many elements are there N
minus d elements and from here I have to choose
00:55:21.299 --> 00:55:33.690
n minus r elements and the number of possible
ways of selection is N cn.
00:55:33.690 --> 00:55:43.690
Therefore this is going to be the probability
that the rth order statistics of the sample
00:55:43.690 --> 00:56:05.130
is actually get one of the population.
In a similar way probability x or is equal
00:56:05.130 --> 00:56:28.230
to xd and x s is equal to x k is d minus one
c r minus 1 multiplied by k minus d minus
00:56:28.230 --> 00:56:47.079
1 c s minus r minus 1 multiplied by N minus
k c n minus s upon N c n
00:56:47.079 --> 00:56:56.119
This is very simple a very similar logic will
tell you that if this is the d and this is
00:56:56.119 --> 00:57:04.599
the kth 1. Then I need r minus one of them
to be here the rth one to be here, the sth
00:57:04.599 --> 00:57:16.480
one to be here n minus s once to be here and
s minus r minus 1 of them to be from here.
00:57:16.480 --> 00:57:24.560
So, the number of ways a possible selection
is coming out to be this. I want you to verify
00:57:24.560 --> 00:57:33.529
this one and I want you to attempt the problems
that will be giving you in the tutorial sheet.
00:57:33.529 --> 00:57:43.660
With that I close the chapter on Order Statistics,
from the next class I will go into properties
00:57:43.660 --> 00:58:00.590
of estimators.
Thank you so much