Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms: Exercises

4.9.1 · D4 · Maths › Probability Theory & Statistics › Probability space — sample space Ω, sigma-algebra F, measure

Shuru karne se pehle quick symbol reminder (taaki kuch bhi unseen na lage):

$Ω$ = sample space — wo set jisme sab kuch jo ho sakta hai, shamil hai.
$ω$ = $Ω$ ke andar ek single outcome.
$F$ = sigma-algebra — un subsets ki list jinhe hum measure karne ki permission rakhte hain ("events").
$A^{c}$ = $A$ ka complement — $Ω$ mein sab kuch jo $A$ mein nahi hai.
$A \cup B$ = " $A$ ya $B$ "; $A \cap B$ = " $A$ aur $B$ "; $A ∖ B = A \cap B^{c}$ = " $A$ mein hai par $B$ mein nahi".
$A \subseteq B$ = " $A$ , $B$ ke andar hai" — $A$ ka har outcome $B$ ka bhi outcome hai (ek subset).
$⊔$ = un pieces ka union jo kabhi overlap nahi karte (disjoint).
$∣ A ∣$ = $A$ ke andar outcomes ki count (sirf finite $Ω$ ke liye meaningful).

Level 1 — Recognition

Recall Solution L1.1

Har baar teen rules check karo: $Ω$ contain karta hai; complement ke under closed hai; (countable) union ke under closed hai.

(i) $Ω$ contain karta hai ✓. Complements: $Ω^{c} = \emptyset$ ✓, $\emptyset^{c} = Ω$ ✓. Inke unions andar hi rehte hain ✓. Haan, ye sigma-algebra hai (sabse chota possible, "trivial" wala).
(ii) ${a}$ contain karta hai lekin ${a}^{c} = {b, c}$ missing hai. Complement rule fail hoti hai. Sigma-algebra nahi hai.
(iii) $Ω$ contain karta hai ✓. Complements: ${a} \leftrightarrow {b, c}$ dono present hain ✓, $\emptyset \leftrightarrow Ω$ ✓. Unions: ${a} \cup {b, c} = Ω$ ✓. Haan, ye sigma-algebra hai.

Recall Solution L1.2

${1}$ aur ${2}$ disjoint hain aur ${1} \cup {2} = Ω$ . Finite additivity (A3 se) demand karti hai ki $P (Ω) = P ({1}) + P ({2}) = 0.7 + 0.7 = 1.4 \neq = 1.$ Lekin A2 kehta hai $P (Ω) = 1$ . Contradiction. Valid nahi — ye additivity / normalization consistency violate karta hai (numbers $1$ se zyada ho jaate hain). ✗

Axiom pakadna

A3 (additivity) force karta hai

P ({1}) + P ({2}) = P (Ω)

, jise A2

1

par fix karta hai;

0.7 + 0.7 = 1.4

ise tod deta hai.

Level 2 — Application

Recall Solution L2.1

(a) Complement rule $P (A^{c}) = 1 - P (A)$ ke saath $A = rain$ : $P (no rain) = 1 - 0.4 = 0.6.$
(b) Note karo ki $rain \cap wind \subseteq rain$ (har rainy-and-windy din khaskar ek rainy din bhi hota hai). Pehle hum axioms se monotonicity prove karte hain (ye ek theorem hai, axiom nahi). Bade set ko boundary par ek disjoint split ki tarah likhte hain: $rain = (rain \cap wind) ⊔ (rain ∖ wind) .$ Finite additivity (A3 se) deta hai $P (rain) = P (rain \cap wind) + P (rain ∖ wind)$ . Aakhri term A1 se $\geq 0$ hai, isliye $P (rain \cap wind) \leq P (rain), i.e. 0.25 \leq 0.40. ✓$ Generally isi argument se monotonicity prove hoti hai: $A \subseteq B \Rightarrow P (A) \leq P (B)$ . Ek joint event kabhi bhi apne kisi ek part se zyada likely nahi ho sakta.

Recall Solution L2.2

$A = {2, 4, 6}$ , $B = {4, 5, 6}$ , $A \cap B = {4, 6}$ .

Counting: $A \cup B = {2, 4, 5, 6}$ , isliye $P (A \cup B) = \frac{4}{6} = \frac{2}{3}$ .
Inclusion–exclusion $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ : $\frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} . ✓$ $P (A \cap B) = 2/6$ ka subtraction double-counted overlap ${4, 6}$ ko remove karta hai.

Figure — Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms

Level 3 — Analysis

Recall Solution L3.1

Proof. Inclusion–exclusion deta hai $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . A1 (non-negativity) se, $P (A \cap B) \geq 0$ , isliye ise subtract karne se right side sirf ghati hai: $P (A \cup B) = P (A) + P (B) - \geq 0 P (A \cap B) \leq P (A) + P (B) . ■$ Application. $A = overheat$ , $B = power cut$ ke saath: $P (A \cup B) \leq 0.03 + 0.02 = 0.05.$ To kisi ek se bhi fail hone ki probability at most $0.05$ hai. (Equality tabhi hogi jab $A, B$ disjoint hon; overlap hone par strictly kam hogi.)

Recall Solution L3.2

Nested intervals define karo jo point par squeeze karte hain: $C_{n} = [0.5, 0.5 + \frac{1}{n}], C_{1} \supseteq C_{2} \supseteq \dots, {0.5} = ⋂_{n = 1}^{\infty} C_{n} .$ Har interval ki length $P (C_{n}) = \frac{1}{n}$ hai. Jaise-jaise $n \to \infty$ (arrow $\to$ matlab "arbitrarily close ho jaata hai"), $\frac{1}{n} \to 0$ .

Limit tak jaane ki permission kyun hai — continuity from above, yahan prove ki gayi. Hume $P (⋂_{n} C_{n}) = lim_{n} P (C_{n})$ justify karna hai; ye ek theorem hai, axiom nahi. $D_{n} = C_{1} ∖ C_{n}$ set karo. Kyunki $C_{n}$ shrink karte hain, $D_{n}$ grow karte hain ( $D_{1} \subseteq D_{2} \subseteq \dots$ ) jinka union $C_{1} ∖ ⋂_{n} C_{n}$ hai. Continuity from below (neeche L5.1 mein A3 se proved) se, $P (C_{1} ∖ ⋂_{n} C_{n}) = lim_{n \to \infty} P (D_{n}) .$ Disjoint split $C_{1} = C_{n} ⊔ D_{n}$ par finite additivity (valid kyunki $C_{n} \subseteq C_{1}$ ) se $P (D_{n}) = P (C_{1}) - P (C_{n})$ milta hai, aur isi tarah $P (C_{1} ∖ ⋂_{n} C_{n}) = P (C_{1}) - P (⋂_{n} C_{n})$ . Substitute karke aur $P (C_{1})$ cancel karke: $P (⋂_{n} C_{n}) = lim_{n \to \infty} P (C_{n}) .$ Yahi continuity from above hai. Ise yahan apply karne par, $P ({0.5}) = lim_{n \to \infty} \frac{1}{n} = 0.$ Impossible kyun nahi: point $0.5$ $Ω$ ka ek genuine element hai, isliye ye outcome ho sakta hai; "probability 0" yahan matlab hai "almost never", na ki "never" — sirf $\emptyset$ truly impossible hota hai.

Level 4 — Synthesis

Recall Solution L4.1

Derivation. $B \cup C$ ko ek event treat karo aur $A$ aur $(B \cup C)$ par two-event rule apply karo: $P (A \cup (B \cup C)) = P (A) + P (B \cup C) - P (A \cap (B \cup C)) .$ $P (B \cup C) = P (B) + P (C) - P (B \cap C)$ expand karo. Aakhri term ke liye distributive law $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ use karo, phir dobara two-event: $P ((A \cap B) \cup (A \cap C)) = P (A \cap B) + P (A \cap C) - P (A \cap B) \cap (A \cap C) (A \cap B \cap C) .$ Saare pieces wapas substitute karo: $P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C) . ■$ Ye Inclusion–exclusion principle se match karta hai: singles add karo, pairs subtract karo, triple add karo. Numbers. $0.6 + 0.5 + 0.4 - 0.3 - 0.2 - 0.2 + 0.1 = 0.9.$ To $P (at least one) = 0.9$ .

Recall Solution L4.2

$A = {1, 2}$ se shuru karo aur closure rules force karo:

$Ω = {1, 2, 3, 4}$ contain karna chahiye (rule 1).
$A^{c} = {3, 4}$ contain karna chahiye (complement rule).
$\emptyset$ contain karna chahiye ( $= Ω^{c}$ ). Ab unions/intersections check karo: $A \cup A^{c} = Ω$ (present), $A \cap A^{c} = \emptyset$ (present). Kuch naya generate nahi hota. Isliye $σ (A) = {\emptyset, {1, 2}, {3, 4}, Ω} .$ Koi choti kyun nahi: in charon mein se koi bhi hata do aur ek closure rule toot jaati hai — e.g. ${3, 4}$ hatane se $A$ ka complement chala jaata hai. Isme exactly $4 = 2^{2}$ elements hain, jo do "atoms" ${1, 2}$ aur ${3, 4}$ se match karta hai.

Level 5 — Mastery

Recall Solution L5.1

Idea: nested sets ko disjoint rings mein badlo taaki A3 (countable additivity) apply ho sake. "Step $n$ par add hone wala naya material" define karo: $B_{1} = A_{1}, B_{n} = A_{n} ∖ A_{n - 1} (n \geq 2) .$ Ye $B_{n}$ pairwise disjoint hain (har ek sirf wahi capture karta hai jo pehle nahi tha), aur $⋃_{n = 1}^{\infty} B_{n} = A, ⋃_{k = 1}^{n} B_{k} = A_{n} .$ Disjoint $B_{n}$ par A3 apply karo: $P (A) = \sum_{n = 1}^{\infty} P (B_{n}) = lim_{N \to \infty} \sum_{n = 1}^{N} P (B_{n}) .$ Lekin finite additivity se partial sum exactly $P (A_{N})$ hai: $\sum_{n = 1}^{N} P (B_{n}) = P (⋃_{k = 1}^{N} B_{k}) = P (A_{N}) .$ Isliye $P (A) = lim_{N \to \infty} P (A_{N}) . ■$ Ye L3.2 ke shrinking-interval trick ke peeche ka engine hai (uska "from above" twin, jo humne wahan isi result se derive kiya tha) aur Random variables aur Expectation and Lebesgue integral mein use hone wale limits ke peeche bhi.

Recall Solution L5.2

Setup (translation invariant ka matlab). Circle par kaam karo: $[0, 1)$ ko ek loop par points ki tarah identify karo, to koi number $t$ add karne ka matlab hai har point ko $t$ se rotate karna aur $1$ se aage wale hisse ko wapas shuru mein laana (ye addition modulo 1 hai, $x \oplus t$ likha jaata hai). "Translation invariant" ka matlab hai ki ek set aur uski rotated copy ko same probability milti hai: $P (S \oplus t) = P (S)$ — ye ek uniform law ke liye natural demand hai, kyunki loop par koi jagah special nahi hai. Construction (is step mein axiom of choice chahiye). Do points ko equivalent kaho agar unka difference rational ho. Ye $[0, 1)$ ko disjoint equivalence classes mein kaat deta hai. Ek set $V$ banao har class se exactly ek representative chun ke — uncountably many classes mein se ek-ek point ek saath select karna precisely axiom of choice hai (iske bina aisa $V$ exist karna zaroori nahi; yahi reason hai ki $V$ "non-constructive" aur slippery hai). Contradiction. $V$ ko har rational $q \in [0, 1)$ se rotate karne par countably many disjoint copies $V_{q} = V \oplus q$ milte hain jinka union poora $[0, 1)$ hai. Translation invariance se har copy ki same probability $p = P (V)$ hoti hai. Countable additivity (A3) tab force karta hai: $1 = P ([0, 1)) = \sum_{q \in Q \cap [0, 1)} P (V_{q}) = \sum_{q} p .$ Agar $p = 0$ to sum $0 \neq = 1$ hai; agar $p > 0$ to sum $+ \infty \neq = 1$ hai. $p$ ki koi bhi value kaam nahi karti. Isliye $V$ ko consistent probability assign nahi ki ja sakti — ise exclude karna padega. Isliye $F$ , $2^{Ω}$ se chota hona chahiye; hum Borel sets use karte hain. (Ye Measure theory & Lebesgue integration ke neeche ka rigorous backbone hai.)

Wrap-up recall

Recall Har level ke liye one-line answers

L1: saari teen closure rules / teeno axioms check karo — akele $\emptyset, Ω$ ka hona kaafi nahi hai.
L2: complement $= 1 - P$ ; union $=$ sum $-$ overlap; monotonicity ek disjoint split $+$ A1 se aati hai.
L3: $P (A \cup B) \leq P (A) + P (B)$ ; ek point ka $P = 0$ ho sakta hai phir bhi occur kar sakta hai (continuity from above).
L4: teen-event I–E = singles $-$ pairs $+$ triple; sabse chota $σ (A) = {\emptyset, A, A^{c}, Ω}$ .
L5: continuity from below disjoint rings $B_{n}$ se; Vitali set (axiom of choice) force karta hai $F ⊊ 2^{Ω}$ .

Related: Conditional probability & independence in events ko ek step aage le jaata hai ye poochhkar ki "diya hua $B$ hua hai, to $P (A)$ kaise badlega?".