Subject - Statistics Paper - Probability Module - Set Theory - few combinatorial results

Probability theory is built upon set theory. In this module we shall discuss few important set function that are essential and useful to derive few important results in probability theory. These proofs are sometimes much easier than the conventional proofs that we are aware of.

1 Indicator function

1.1 Definition Let pΩ, A,P q be a probability space where Ω is the sample space, A is the σ-field generated by the class of subsets of Ω, and P is the probability function defined on pΩ, Aq. Now for any set A P A, we define an indicator function as:

1 if ω P A I ω ap q “ c #0 if ω R A This means that an indicator function takes only two values 0 or 1. This looks very simple, but has immense application not only in probability theory but also in entire mathematics.

1.2 Properties Now we discuss few properties of an indicator function, Although looks very obvious, but we shall provide proofs to these properties, designated as ‘Results’.

2 Result 1 IA “ IA for every A P A.

Proof 1 Take any A P A. Then we have,

2 If ω P A, IApωq “ 1 “ IApωq. 2 If ω R A, IApωq “ 0 “ IApωq. Hence the proof.

1 Result 2 Iφ “ 0 and IΩ “ 1.

Proof 2 Proofs are obvious and we omit.

Result 3 For any A, B P A, A Ă B ðñ IA “ IB.

Proof 3 Note that B “ pA X Bq Y pAc X Bq. Therefore,

If ω P A X B,IA “ IB “ 1. c If ω P A X B,IA “ 0 ă 1 “ IB.

Hence IA ď IB whenever A Ă B.

Result 4 For any A P A, IAc “ 1 ´ IA.

Proof 4

c If ω P A ,IAc “ 1 “ 1 ´ 0 “ 1 ´ IA rsince, IA “ 1s c If ω R A , i.e. ω P A, IAc “ 0 “ 1 ´ 1 “ 1 ´ IA. Hence the proof.

Result 5 For any A, B P A, IAXB “ IA.IB.

Proof 5

IAXB “ 1 ðñ ω P A X B ðñ ω P A and ω P B

ðñ IA.IB “ 1. Similarly we can show this for the value 0, that complete the proof.

Result 6 For any A, B P A such that A X B “ φ, IpAYBq “ IA ` IB.

Proof 6 Since A X B “ φ, then we have,

IAXBpωq “ 1 ðñω P A Y B ðñω P A or ω P B

ðñtIApωq “ 1,IBpωq “ 0u or tIApωq “ 0,IBpωq “ 1u but not both

ðñIAXBpωq “ IApωq ` IBpωq.

2 Result 7 For any A, B P A, IpA Y Bq “ IA ` IB ´ IA.IB. Proof 7 Note that A Y B “ A Y pAc X Bq and B “ pA X Bq Y pAc X Bq. Therefore, from Result 6, ` ˘ IBpωq “ IpAXBqpωq ` IpAcXBqpωq

ùñ IpAcXBqpωq “ IBpωq ´ IpAXBqpωq. (1) Again, since A Y pAc X Bq “ φ, we have, from equation (1),

IAYBpωq “ IpAYpAcXBqqpωq “ IApωq ` IpAcXBqpωq

“ IApωq ` IBpωq ´ IpAXBqpωq. Hence the proof is complete.

2 Indicator function and probability theory

Let pΩ, A,P q be a probability space. Then for any A P A, we can always treat IApωq for ω P A as a random variable. Clearly its expectation exists. Then we have, c EpIAq “ 1.P pAq ` 0.P pA q “ P pAq Thus probability of a set (event) can be looked as expectation of an indicator variable (function) defined on that set. Thus simple observation leads us to vey simple proofs of some wellknown results or theorems in probability theory.

2.1 Few results: use of indicator variable To prove few results below, we always consider the underlying probability space as pΩ, A,P q. Result 8 For any A, B P A, prove that P pA Y Bq “ P pAq ` P pBq ´ P pA X Bq. Proof 8 Note that from Result 7,

IAYB “ IA ` IB ´ IAXB. (2) Taking expectations on both sides of (2), we have,

E IAYB “ E IA ` IB ´ IAXB ùñ P p´A Y b¯q ` P p´Aq `¯ P p´Bq ´¯ P p´A X B¯q. Hence the proof is complete.

3 Result 9 For any A P A, prove that P pAcq “ 1 ´ P pAq.

Proof 9 Taking expectations on both sides of Result 4, proof follows immediately.

Result 10 (Poincare Theorem) Let Ai P A for i “ 1, 2, . . . , n. Then prove that,

n n n n´1 P “ P pAiq ´ P p Ai X Aj ` ¨ ¨ ¨ ` p´1q P A1 X A2 X ¨ ¨ ¨ An . i“1 i“1 iăj“1 ´ č ¯ ÿ ÿ ` ˘ ` ˘ n Proof 10 First note that A is a σ-field. Hence for Ai P A, i “ 1, . . . , n, i“1 P A and Ai1 Y Ai2 Y ¨ ¨ ¨ Air P A for 1 ď r ď n. c Ť ` ˘ n n c Again, by De Morgan’s law, we know that, i“1 “ i“1 Ai . ˜ ¸ We also know from Result 4 and Result 5, IŤAc “ 1 ´ IŞA and IAXB “ IAIB. Then,

I n 1 I n c 1 I c I c I c i“1 Ai “ ´ i“1 Ai “ ´ A1 A2 ¨ ¨ ¨ An Ť Ş “ 1 ´ p1 ´ IA1 qp1 ´ IA2 q ¨ ¨ ¨ p1 ´ IAn q n n n “ IAi ´ IAi IAj ` ¨ ¨ ¨ ` p´1q IA1XA2X...XAn (3) i“1 iăj1 ÿ ÿ Now take expectations on both sides of (3), we have,

n n n E I n E I I I 1 I i“1 Ai “ Ai ´ Ai Aj ` ¨ ¨ ¨ ` p´ q A1XA2X...XAn ˜ ¸ ˜ i“1 iăj1 ¸ Ş n ÿ ÿn n “ Ep IAi ´ E IAi IAj ` ¨ ¨ ¨ ` p´1q E IA1XA2X...XAn i“1 iăj1 ÿn ` ˘ ÿn ` ˘ ` ˘ n P I n P I P I I 1 P I ùñ i“1 Ai “ p Ai ´ Ai Aj ` ¨ ¨ ¨ ` p´ q A1XA2X...XAn ˜ ¸ i“1 iăj1 Ş ÿ ` ˘ ÿ ` ˘ ` ˘

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