4.9.13 · Maths › Probability Theory & Statistics
Conditional expectation answer karta hai: "Agar mujhe pehle se kuch information pata hai, toh X ke average ke liye mera best guess kya hoga?"
Plain expectation E [ X ] tera best single-number guess hai jab kuch bhi nahi pata . Conditional expectation E [ X ∣ Y ] tera updated guess hai Y jaanne ke baad — aur sabse important baat, kyunki Y pehle se pata nahi hoti, yeh updated guess khud ek random variable hoti hai.
Definition Kisi event par conditional expectation
Ek discrete random variable X aur ek event B ke liye jahan P ( B ) > 0 :
E [ X ∣ B ] = ∑ x x P ( X = x ∣ B ) = ∑ x x P ( B ) P ( X = x , B )
Yeh bilkul ordinary expectation hai, bas X ki conditional probability distribution use karke B given compute ki gayi hai.
Definition Ek variable par conditional expectation
E [ X ∣ Y = y ] ek number hai (har value y ke liye alag):
E [ X ∣ Y = y ] = ∑ x x P ( X = x ∣ Y = y ) ( discrete )
E [ X ∣ Y = y ] = ∫ − ∞ ∞ x f X ∣ Y ( x ∣ y ) d x ( continuous )
Function g ( y ) = E [ X ∣ Y = y ] , jab random input Y diya jaaye, toh random variable E [ X ∣ Y ] = g ( Y ) milti hai.
Do flavours kyun? E [ X ∣ B ] ek fixed event par condition karta hai → ek number. E [ X ∣ Y ] "har possible y ke liye" saare answers ko ek object mein pack karta hai → Y ka ek random function.
Yeh THE key theorem hai. Isse scratch se build karte hain.
First principles se derivation (discrete):
E [ X ∣ Y = y ] ki definition se shuru karo aur Y par uski expectation lo:
E [ E [ X ∣ Y ] ] = ∑ y E [ X ∣ Y = y ] P ( Y = y )
Yeh step kyun? E [ X ∣ Y ] , Y ka ek function hai; ek function g ( Y ) ki expectation hoti hai ∑ y g ( y ) P ( Y = y ) .
Definition E [ X ∣ Y = y ] = ∑ x x P ( X = x ∣ Y = y ) substitute karo:
= ∑ y ( ∑ x x P ( X = x ∣ Y = y ) ) P ( Y = y )
Kyun? Bas plug in kar rahe hain. Ab P ( X = x ∣ Y = y ) P ( Y = y ) = P ( X = x , Y = y ) use karo (conditional probability ki definition):
= ∑ y ∑ x x P ( X = x , Y = y ) = ∑ x x P ( X = x ) y ∑ P ( X = x , Y = y )
Inner sum kyun collapse hoti hai? Saare y par joint sum karne se y marginalise ho jaata hai, sirf P ( X = x ) bachta hai.
= ∑ x x P ( X = x ) = E [ X ] ■
Intuition Tower Rule kyun sach honi chahiye
Socho exam scores average karna. Pehle har class ke andar average karo (E [ X ∣ Y ] class Y ke liye). Phir un class-averages ko average karo, class size ke hisaab se weight karke (P ( Y = y ) ). Tumhe wahi number milega jaise saare students ko ek saath average karo. Split karke phir combine karne se overall mean change nahi ho sakta.
Worked example 1. Do dice roll karo; total ki expected value given pehla even hai
Maano X = sum, Y = pehla die. Hume chahiye E [ X ∣ Y even ] .
E [ X ∣ Y = y ] = y + E [ doosra die ] = y + 3.5 .
Kyun? Doosra die independent hai, toh Y par conditioning sirf pehla term fix karti hai.
Y even given, Y ∈ { 2 , 4 , 6 } har ek prob 1/3 se: average y = 4 .
Toh E [ X ∣ Y even ] = 4 + 3.5 = 7.5 .
Worked example 2. Random number of summands (Wald-style)
Ek fair coin flip karo; heads par X = ek die roll; tails par X = do die rolls ka sum.
Maano Y =coin. E [ X ∣ Y = H ] = 3.5 , E [ X ∣ Y = T ] = 7 .
Tower: E [ X ] = 2 1 ( 3.5 ) + 2 1 ( 7 ) = 5.25 .
Yeh step kyun? Hum kabhi X ki full distribution compute nahi karte — bas do conditional answers ko unki probabilities se weight karke average karte hain. Yahi tower ki power hai.
Worked example 3. Continuous:
Y ∼ U ( 0 , 1 ) , X ∣ Y = y ∼ U ( 0 , y )
E [ X ∣ Y = y ] = 2 0 + y = 2 y ([ 0 , y ] par uniform ka mean).
Toh E [ X ∣ Y ] = Y /2 , ek random variable.
Tower: E [ X ] = E [ Y /2 ] = 2 1 E [ Y ] = 2 1 ⋅ 2 1 = 4 1 .
Kyun? Hum joint density par mushkil direct integral ko do aasaan steps se replace karte hain.
E [ X ∣ Y ] sirf ek number hai."
Kyun sahi lagta hai: E [ X ∣ Y = y ] hota hai ek number, toh log distinction drop kar dete hain.
Fix: E [ X ∣ Y = y ] fixed y ke liye ek number hai; lekin E [ X ∣ Y ] mein Y random rehti hai, toh yeh ek random variable hai — yehi reason hai ki E [ E [ X ∣ Y ] ] bana bhi sense karta hai.
E [ X Y ∣ Y ] = Y E [ X ] likhna.
Kyun sahi lagta hai: "known stuff nikaal bahar" aur "independence E [ X ] deta hai" yaad rehta hai aur dono merge ho jaate hain.
Fix: Y nikalna hamesha valid hai: E [ X Y ∣ Y ] = Y E [ X ∣ Y ] . E [ X ∣ Y ] ko E [ X ] se replace karne ke liye independence chahiye. Dono rules ko fuse mat karo.
Common mistake Tower mein
P ( Y = y ) se weight karna bhool jaana.
Kyun sahi lagta hai: conditional means ko average karna symmetric lagta hai.
Fix: E [ X ] = ∑ y E [ X ∣ Y = y ] P ( Y = y ) — bade groups zyada count karte hain. Group means ka plain average galat hai jab tak groups equally likely na hon.
Recall Feynman: 12-saal ke bacche ko samjhao
Tum guess karna chahte ho ki school ka koi random baccha kitna tall hai. Tera pehla guess school ka average hai — woh hai E [ X ] . Ab koi whisper karta hai "woh 8th grade mein hai" — yeh extra info hai Y . Tum update karte ho 8th-grade average par — woh hai E [ X ∣ Y ] . Cool trick (Tower Rule): agar tum har grade ka average lete ho aur unhe is hisaab se blend karte ho ki har grade mein kitne bacche hain, toh whole-school average wapas milta hai. Split karke recombine karna overall answer kabhi nahi badalta.
"Averages ko average karo, crowds ke hisaab se weight karke." — Tower Rule.
Aur: E andar, E bahar, beech wala gayab → E [ E [ X ∣ Y ]] = E [ X ] .
E [ X ∣ Y ] (note: capital Y ) kis tarah ka object hai?Ek random variable — random Y ka ek function g ( Y ) .
E [ X ∣ Y = y ] kis tarah ka object hai?Ek number, har fixed value y ke liye alag.
Tower / Law of Iterated Expectation state karo. E [ E [ X ∣ Y ] ] = E [ X ] .
Tower derive karne ka pehla step (discrete)? E [ E [ X ∣ Y ]] = ∑ y E [ X ∣ Y = y ] P ( Y = y ) .
Inner sum ∑ y P ( X = x , Y = y ) kyun collapse hoti hai? y par marginalise karne se P ( X = x ) milta hai.
"Take out what is known" rule kya hai? E [ g ( Y ) X ∣ Y ] = g ( Y ) E [ X ∣ Y ] .
Agar X ⊥ Y toh E [ X ∣ Y ] kya hai? Constant E [ X ] .
Y ∼ U ( 0 , 1 ) , X ∣ Y = y ∼ U ( 0 , y ) : E [ X ] nikalo.E [ X ∣ Y ] = Y /2 , toh E [ X ] = E [ Y /2 ] = 1/4 .
Coin: H→ek die, T→do dice ka sum. E [ X ] nikalo. 2 1 ( 3.5 ) + 2 1 ( 7 ) = 5.25 .
Tower groups par apply karte waqt common slip kya hai? Conditional means ko P ( Y = y ) se weight karna bhool jaana.
random Y ko g of Y mein daalo
averaging se wapas milta hai
g of Y constant ki tarah act karta hai
E X given Y is a random variable
Tower Rule E of E X given Y equals E X