Page 237 - DMTH401_REAL ANALYSIS

Unit 19: Arzelà’s Theorem and Weierstrass Approximation Theorem

Definition 1: Define Notes
æ n ö
k
p (x) = ç ÷ x (1 – x) , " n Î  and 0  k  n.
n–k
kn è k ø
Note that, p (x) becomes a probability mass function of binomial distribution with probability
kn
of successful trial equal to x if x Î [0, 1].
n
Lemma: (a) " n Î : å n k 0 p (x) = 1, (b) " n Î : å k 0 k pkn (x) = nx, (c) " n Î : å n k 0 (k – nx) 2
=
=
=
kn
p (x) = nx(1 – x).
kn
Proof: (a) If x Î [0, 1] the equality follows from normalisation of probability distribution. In
general
n æ n ö
-
k
n
(a + b) = å ç ÷ a b n k ,
k 0 k ø
= è
therefore
n
n n æ ö
k
n
å p (x) = å ç ÷ x (1 – x) n – k = (x + 1 – x) = 1.
= è
=
k 0 kn k 0 kø
(b) If x Î [0, 1], we can define a random variable Y the number of heads observed on unfair
n,x
x-coin tossed n-times. Then
æ n ö
k
(Y = k) = ç ÷ x (1 – x) n – k = p (x).
n,x è k ø kn
Moreover, we find the following relation with (b)
n
[Y ] = å k (x) = nx.
n,x p kn
k 0
=
In general
-
-
æ n ö n! (n 1)! æ n 1ö
k ç ÷ = k = n = n ç ÷ ,
-
è k ø k!(n k)! (k - 1)!((n 1) (k - 1))! è k - 1 ø
-
-
so
n n æ nö n æ n - 1ö
k
k
å k (x) = å k x (1 – x) n – k = å n x (1 – x) n – k =
k
1
p kn ç ÷ k 0 è ç k - ø ÷
k 0 k 0 è ø =
=
=
n æ n - 1ö n æ n - 1ö
k
= å n ç ÷ x (1 – x) n – k = nx å ç ÷ x k – 1 (1 – x) n – k =
1
1
= è
k 1 è k - ø k 1 k - ø
=
1
-
æ
n 1 n - ö
i
= nx å ç ÷ x (1 – x) n – 1 – i = nx(x + 1 – x) n – 1 = nx.
i 0 è i ø
=
(c) If x Î [0, 1], we can rewrite the formula as
n
2
Var(Y ) = å (k – nx) p (x) = nx (1 – x).
n,x kn
k 0
=
In general
æ n ö (n - 2)! æ n - 2ö
k(k – 1) ç ÷ = n(n – 1) = n(n – 1) ,
è k ø (k - 2)!((n - 2) (k - 2))! ç è k - ø ÷
2
-
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