4.9.1
Intuition Bada picture (YEH kyun exist karta hai)Pehle hum probabilities compute kar sakein, humein agree karna hoga ki hum actually measure kya kar rahe hain .
Ek probability space woh rigorous "stage" hai jis par poori probability perform hoti hai.
Yeh teen sawaalon ke jawaab deta hai:
Kya ho sakta hai? → sample space Ω Outcomes ke kaun se collections ko hum probability assign karne ki permission rakhte hain? → sigma-algebra F Har allowed collection ko kitni probability milti hai? → measure P
F [ 0 , 1 ] har subset ko consistently probability assign nahi kar sakte — pathological sets rules tod deti hain. Isliye hum ek well-behaved family F
Definition Probability spaceEk probability space ek triple ( Ω , F , P )
Ω sample space , sabhi possible outcomes ω F sigma-algebra (Ω events kehte hain).P probability measure , ek function P : F → [ 0 , 1 ]
Yeh teen kyun? Yeh minimum closure rules hain taaki events ke kisi bhi logical combination ("not A", "A or B", "A and B", "infinitely mein se kam se kam ek") ek aisa event rahe jise hum measure kar sakein. 1–3 se aapko baaki sab milta hai:
∅ = Ω c ∈ F Countable intersections : ⋂ i A i = ( ⋃ i A i c ) c ∈ F
F F un yes/no sawaalon ki list ke roop mein socho jo tum poochh sakte ho. Agar tum poochh sakte ho "kya A hua?", toh tum yeh bhi poochh sakne chahiye "kya A nahi hua?" aur "kya A ya B hua?". Ek sigma-algebra bas yeh rule hai ki yeh sawaalon ki list logically complete hai.
Har axiom kyun?
A1 — probability ek "mass"/"size" hai; sizes kabhi negative nahi hoti.A2 — Ω kuch na kuch zaroor hoga, isliye total mass 1 par normalize hoti hai.A3 — agar ek event ko disjoint pieces mein tod do, toh probabilities add hoti hain. "Countable" (sirf finite nahi) version hi hai jo limits, integrals, aur continuous distributions ko kaam karne deta hai.
Neeche kuch bhi extra rule nahi hai — sab theorems hain jo A1–A3 se forced hain.
P ( ∅ ) = 0 A 1 = Ω , A 2 = A 3 = ⋯ = ∅ ⋃ A i = Ω P ( Ω ) = P ( Ω ) + ∑ i ≥ 2 P ( ∅ ) Yeh step kyun? A3 mujhe union ko P ( ∅ ) ∑ i ≥ 2 P ( ∅ ) = 0 ≥ 0 0 ■
A , B finite additivity P ( A ∪ B ) = P ( A ) + P ( B ) A 1 = A , A 2 = B , A 3 = A 4 = ⋯ = ∅ P ( ∅ ) = 0 P ( A ∪ B ) = P ( A ) + P ( B ) + 0 + 0 + … ■ Yeh step kyun? A3 infinite unions ke liye stated hai; ∅
Complement rule P ( A c ) = 1 − P ( A ) A A c A ∪ A c = Ω P ( Ω ) = P ( A ) + P ( A c ) P ( Ω ) = 1 P ( A c ) = 1 − P ( A ) ■ Consequence: P ( A ) = 1 − P ( A c ) ≤ 1 0 ≤ P ( A ) ≤ 1 [ 0 , 1 ] derived hai, assume nahi ki gayi.
Monotonicity A ⊆ B ⇒ P ( A ) ≤ P ( B ) B = A ∪ ( B ∖ A ) B ∖ A = B ∩ A c P ( B ) = P ( A ) + P ( B ∖ A ) Yeh step kyun? B A P ( B ∖ A ) ≥ 0 P ( B ) ≥ P ( A ) ■
Inclusion–exclusion P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) A ∪ B = ( A ∖ B ) ⊔ ( A ∩ B ) ⊔ ( B ∖ A ) . P ( A ∪ B ) = P ( A ∖ B ) + P ( A ∩ B ) + P ( B ∖ A ) P ( A ) = P ( A ∖ B ) + P ( A ∩ B ) P ( B ) = P ( B ∖ A ) + P ( A ∩ B ) Yeh step kyun? A B A ∩ B P ( A ) + P ( B ) = P ( A ∖ B ) + P ( B ∖ A ) + 2 P ( A ∩ B ) P ( A ∩ B ) P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) . ■
Worked example Finite (die) spaceEk fair die roll karo. Ω = { 1 , 2 , 3 , 4 , 5 , 6 } F = 2 Ω P ( A ) = ∣ A ∣/6 A = { even } = { 2 , 4 , 6 } B = { >3 } = { 4 , 5 , 6 }
P ( A ) = 3/6 P ( B ) = 3/6 A ∩ B = { 4 , 6 } P ( A ∩ B ) = 2/6 P ( A ∪ B ) = 6 3 + 6 3 − 6 2 = 6 4 = 3 2 Check: A ∪ B = { 2 , 4 , 5 , 6 } 4/6 Yahaan F = 2 Ω Ω har subset measurable hai — koi pathology nahi.
Worked example Continuous (uniform) spaceΩ = [ 0 , 1 ] F = P ([ a , b ]) = b − a
P ([ 0 , 0.3 ]) = 0.3 P ({ 0.5 }) = 0 Single point ki probability 0 kyun hai: { 0.5 } = ⋂ n [ 0.5 , 0.5 + n 1 ] P ([ 0.5 , 0.5 + n 1 ]) = n 1 → 0 0.5 ho sakta hai — probability 0 ≠ impossible .
Ω Yeh sahi kyun lagta hai: Finite die par, yeh hai — F = 2 Ω Fix: Uncountable Ω [ 0 , 1 ] F 2 Ω chhota ho sakta hai.
Common mistake "Additivity kisi bhi union ke liye kaam karti hai:P ( A ∪ B ) = P ( A ) + P ( B ) Yeh sahi kyun lagta hai: Yeh sach hai jab A , B Fix: Agar woh overlap karte hain toh tum A ∩ B double-count karte ho. Inclusion–exclusion use karo: P ( A ∩ B )
P ( A ) = 0 A Yeh sahi kyun lagta hai: Impossible events (∅ Fix: Converse fail hota hai. Continuous spaces mein individual points ki probability 0 hoti hai par woh phir bhi ho sakte hain. "Probability 0" = "almost never", "never" nahi.
Common mistake "Finite additivity kaafi hai; countable toh sirf fancy hai."Yeh sahi kyun lagta hai: Saare real coin/die experiments finite hain.
Fix: Countable additivity (A3) strictly stronger hai aur yahi guarantee karta hai P
Recall Quick self-test (answers hide karo)
Sigma-algebra ke teen closure rules kya hain? → Ω
P [ 0 , 1 ] ≥ 0 ≤ 1 Kolmogorov A3 state karo. → disjoint events ke liye countable additivity.
P ( A c ) P ( A ) + P ( A c ) = P ( Ω ) = 1 1 − P ( A )
Recall Feynman: ek 12-saal ke bachche ko samjhaoEk bada bag socho (woh Ω F P exactly 1 liter paint saare sawaalon par baantna: kisi sawaal ko negative paint nahi milti (A1), pura bag saara 1 liter use karta hai (A2), aur agar do sawaal kabhi overlap nahi karte, unka paint bas add ho jaata hai (A3). Probability mein baaki sab bas is paint ko bina girae idhar-udhar daalte rehna hai.
Mnemonic Trio yaad rakhho"O-F-P: Outcomes, Filter, Paint."
Ω O utcomes; F F ilter; P P aint (total = 1).
Aur axioms: "NNA" = N on-negative, N ormalized, A dditive.
Probability space mein kaun se teen objects hote hain? ( Ω , F , P )
Sample space Ω Experiment ke sabhi possible outcomes ka set.
Sigma-algebra ke teen axioms list karo. Ω
F Taaki events ka koi bhi logical combination (not, or, and) ek measurable event rahe.
Kolmogorov ke teen axioms state karo. A1 P ( A ) ≥ 0 P ( Ω ) = 1
P ( ∅ ) = 0 Disjoint Ω , ∅ , ∅ , … P ( Ω ) = P ( Ω ) + ∑ P ( ∅ ) P ( ∅ ) = 0
Complement rule derive karo. A ∪ A c = Ω P ( A ) + P ( A c ) = 1 P ( A c ) = 1 − P ( A )
0 ≤ P ( A ) ≤ 1 ≥ 0 ≤ 1
Do events ke liye inclusion–exclusion state aur derive karo. P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) A ∖ B , A ∩ B , B ∖ A
[ 0 , 1 ] F = 2 Ω Non-measurable (Vitali) sets exist karti hain jo countable additivity violate karti hain, isliye hum Borel sets use karte hain.
Kya P ( A ) = 0 A Nahi. Continuous spaces mein single points ki probability 0 hoti hai par woh phir bhi ho sakte hain ("almost never").
Monotonicity: agar A ⊆ B P ( A ) ≤ P ( B ) P ( B ) = P ( A ) + P ( B ∖ A ) P ( B ∖ A ) ≥ 0
Countable (sirf finite nahi) additivity kyun? Yeh P
Closed under countable union
Countable intersection derived
