4.9.11 · D3Independence of random variables — formal definition
Intuition Yeh page kis kaam ki haiParent note Independence — formal definition ne tumhe rules diye. Yeh page har tarah ke case phenk'ta hai tumhare saamne: discrete tables, continuous densities, honest traps jahan support maayine rakhta hai, degenerate constants, limiting behaviour, ek word problem, aur ek exam twist. Iske baad tumhe koi bhi independence question surprise nahi karna chahiye.
Shuru karne se pehle, tests ko carefully state karte hain — including events version jo hume baad mein chahiye — taaki is page par koi cheez bina define kiye use na ho.
Definition Formal test (general → concrete)General (events form). X Y sabhi sets A , B ⊆ R P ( X ∈ A , Y ∈ B ) = P ( X ∈ A ) P ( Y ∈ B ) . Concrete forms. X ⊥ Y har point par :
Discrete: p X , Y ( x , y ) = p X ( x ) p Y ( y ) x , y
Continuous: f X , Y ( x , y ) = f X ( x ) f Y ( y )
Definition Is page par use hone wale symbols (simple alfaaz mein)
E [ Z ] Z expectation ("long-run average value"): density ke liye, E [ Z ] = ∫ z f Z ( z ) d z ∑ z p Z ( z ) Covariance and correlation .Cov ( X , Y ) = E [ X Y ] − E [ X ] E [ Y ] covariance , ek single number jo X Y linear co-movement measure karta hai. Formal home: Covariance and correlation .U ( a , b ) [ a , b ] Uniform distribution : constant density f ( t ) = b − a 1 a ≤ t ≤ b 0 U ( − 1 , 1 ) 2 1 [ − 1 , 1 ] U ( 0 , 1 ) 1 [ 0 , 1 ]
Common mistake "Rectangle" rule ko zyada mat badhaoIndependence khud sirf factorisation f X , Y = f X f Y purely define hoti hai (ya events form se) — yeh alag se rectangular support require nahi karti. f X f Y factorisation shortcut mein aata hai: agar tumhe f X , Y = g ( x ) h ( y ) product-shaped support par dikhe, toh tum marginals compute kiye bina independence conclude kar sakte ho. Agar support product set nahi hai, toh shortcut available nahi hai aur tumhe honest test f X , Y = f X f Y
Do simple-alfaaz reminders taaki yahan koi undefined symbol use na ho:
Marginal of X X Y poori row sum karke milta hai; density mein tumhe yeh doosre variable ko integrate out karke milta hai.Support bas "wo set of points jahan probability zero nahi hai." Ise plane mein ek shaded shape ki tarah socho.
Har independence question in cells mein se kisi ek mein aata hai. Neeche ke examples cell ke hisaab se label kiye gaye hain taaki tum dekh sako ki poori grid cover ho gayi.
Cell
Case class
Kya mushkil banata hai
Example
A Discrete table, genuinely independent
sabhi cells check karne padenge, sirf ek nahi
Ex 1
B Discrete table, dependent
ek bura cell sab barbad kar deta hai
Ex 2
C Continuous, separable density on a rectangle
density factors → independent
Ex 3
D Continuous, support variables ko couple karta hai
density "separable lagti" hai par hai nahi
Ex 4
E Degenerate / constant variable
Y Ex 5
F Zero-covariance trap
uncorrelated but NOT independent
Ex 6
G Limiting behaviour
ek parameter → boundary answer flip kar deta hai
Ex 7
H Real-world word problem
words ko translate karo → factorise karo
Ex 8
I Exam twist
ek independent pair ka function
Ex 9
Hum geometric cases (C, D) ke liye ek figure banate hain taaki tum dekh sako ki support ki shape kyun maayine rakhti hai. Figure ko aise padho: x y x = 0.35 y poore [ 0 , 1 ] x = y x = 0.55 y cut down hokar [ 0.55 , 1 ] x y 0 1
Left panel ek rectangle support hai: har fixed x y x y triangle hai (x ≤ y x x y shrink kar deta hai. Woh shrinking information transfer hai → dependence.
Worked example Ex 1 · Do fair-ish dice talliesJoint PMF:
X \ Y 0
1
2
0
0.06
0.15
0.09
1
0.14
0.35
0.21
Kya X Y
Forecast: aage padhne se pehle yes ya no guess karo. (Rows suspiciously proportional lag rahi hain...)
Step 1 — X
Yeh step kyun? Independence marginals ke relative define hoti hai, isliye pehle unhe chahiye.
p X ( 0 ) = 0.06 + 0.15 + 0.09 = 0.30 p X ( 1 ) = 0.14 + 0.35 + 0.21 = 0.70
Step 2 — Y
Kyun? Same reason, ab doosre variable ke liye.
p Y ( 0 ) = 0.06 + 0.14 = 0.20 p Y ( 1 ) = 0.15 + 0.35 = 0.50 p Y ( 2 ) = 0.09 + 0.21 = 0.30
Step 3 — HAR cell ko p X p Y
Kyun har cell? Ek passing cell luck hai; independence har jagah hold karni chahiye (matrix cell A ka poora point yahi hai).
( 0 , 0 ) 0.30 ⋅ 0.20 = 0.06 ( 0 , 1 ) 0.30 ⋅ 0.50 = 0.15 ( 0 , 2 ) 0.30 ⋅ 0.30 = 0.09 ( 1 , 0 ) 0.70 ⋅ 0.20 = 0.14 ( 1 , 1 ) 0.70 ⋅ 0.50 = 0.35 ( 1 , 2 ) 0.70 ⋅ 0.30 = 0.21
Conclusion: saatein cells match kar rahe hain → X ⊥ Y
Verify: marginals ko har ek 1 sum karna chahiye: 0.30 + 0.70 = 1 0.20 + 0.50 + 0.30 = 1 0.06 + 0.15 + 0.09 + 0.14 + 0.35 + 0.21 = 1
Joint distribution and marginals dekho yeh samajhne ke liye ki row/column sums kyun marginals hain.
Worked example Ex 2 · Ek rebellious cell
X \ Y 0
1
0
0.10
0.30
1
0.30
0.30
Independent?
Forecast: numbers "round" hain, isliye yes kehna tempting hai.
Step 1 — marginals. Kyun? Hamesha ki tarah.
p X ( 0 ) = 0.40 , p X ( 1 ) = 0.60 p Y ( 0 ) = 0.40 , p Y ( 1 ) = 0.60
Step 2 — corner ( 0 , 0 ) Yahan se kyun shuru karein? Agar ek bhi cell fail ho toh kaam khatam — baaki check karne ki zaroorat nahi.
p X ( 0 ) p Y ( 0 ) = 0.40 ⋅ 0.40 = 0.16 0.10 0.16 = 0.10
Conclusion: ek bhi mismatch fatal hai → NOT independent .
Verify: kya table 1 sum karta hai? 0.10 + 0.30 + 0.30 + 0.30 = 1 hai ek valid PMF — bas dependent hai). Hum "gossip" bhi measure kar sakte hain: kyunki p X , Y ( 0 , 0 ) < p X ( 0 ) p Y ( 0 ) X = 0 Y = 0 kam likely ho jaata hai. Dekho Covariance and correlation .
Worked example Ex 3 · Ek clean productf X , Y ( x , y ) = 6 x 2 y [ 0 , 1 ] × [ 0 , 1 ] 0
Forecast: ( stuff in x ) × ( stuff in y )
Step 1 — confirm karo ki yeh valid density hai (1 tak integrate hoti hai).
Kyun? Agar yeh 1 tak integrate nahi hoti toh yeh density nahi hai aur aage kuch bhi matter nahi karta.
∫ 0 1 ∫ 0 1 6 x 2 y d x d y = 6 ( ∫ 0 1 x 2 d x ) ( ∫ 0 1 y d y ) = 6 ⋅ 3 1 ⋅ 2 1 = 1
Step 2 — marginals compute karo aur check karo ki woh joint reproduce karte hain.
Yeh step kyun (koi "rectangle" appeal nahi)? Honest, hamesha valid test f X , Y = f X f Y [ 0 , 1 ] 0 0 f X ( x ) = ∫ 0 1 6 x 2 y d y = 6 x 2 ⋅ 2 1 = 3 x 2 0 ≤ x ≤ 1 0 f Y ( y ) = ∫ 0 1 6 x 2 y d x = 6 y ⋅ 3 1 = 2 y 0 ≤ y ≤ 1 0 f X ( x ) f Y ( y ) = 3 x 2 ⋅ 2 y = 6 x 2 y = f X , Y ( x , y ) 0 independent .
Step 3 — shortcut, cross-check ke liye.
Iska zikr kyun? Kyunki 6 x 2 y = ( 6 x 2 ) ( y ) aur support product set [ 0 , 1 ] 2
Verify: har marginal 1 tak integrate hoti hai: ∫ 0 1 3 x 2 d x = 1 ∫ 0 1 2 y d y = 1 f X f Y f X , Y
Worked example Ex 4 · Ek density jo jhooth bolti haif X , Y ( x , y ) = 8 x y 0 ≤ x ≤ y ≤ 1 0
Forecast: 8 x y = ( ?) ( ?) Trap dekho.
Step 1 — valid density?
Kyun? Pehle wali jaisi hi sanity gate. Triangle par integrate karo (har y x 0 y ∫ 0 1 ∫ 0 y 8 x y d x d y = ∫ 0 1 8 y ⋅ 2 y 2 d y = ∫ 0 1 4 y 3 d y = 1
Step 2 — support dekho.
Yeh step pehle kyun? Figure ka right panel ek triangle dikhata hai: x y [ x , 1 ] product set nahi hai , factorisation shortcut available nahi hai — hum "8 x y
Step 3 — true marginals compute karo (integrate out).
Kyun? Product rule honestly test karne ke liye humein real marginals chahiye, fake split nahi. Dono marginals [ 0 , 1 ] 0 0 f X ( x ) = ∫ x 1 8 x y d y = 8 x ⋅ 2 1 − x 2 = 4 x ( 1 − x 2 ) 0 ≤ x ≤ 1 0 f Y ( y ) = ∫ 0 y 8 x y d x = 8 y ⋅ 2 y 2 = 4 y 3 0 ≤ y ≤ 1 0
Step 4 — test karo.
Kyun? Independence ke liye f X , Y = f X f Y f X ( x ) f Y ( y ) = 4 x ( 1 − x 2 ) ⋅ 4 y 3 = 16 x y 3 ( 1 − x 2 ) nahi hai 8 x y NOT independent.
Verify: har marginal 1 tak integrate honi chahiye: ∫ 0 1 4 x ( 1 − x 2 ) d x = 4 ( 2 1 − 4 1 ) = 1 ∫ 0 1 4 y 3 d y = 1 ( x , y ) = ( 0.9 , 0.2 ) 0 0.9 ≤ 0.2 f X ( 0.9 ) f Y ( 0.2 ) = 4 ( 0.9 ) ( 1 − 0.81 ) ⋅ 4 ( 0.008 ) = 0.6840 ⋅ 0.032 = 0
Common mistake Cell D ka sabak8 x y 6 x 2 y sirf farq support ki shape mein hai. Shortcut off-limits hai jab support product set nahi hai, isliye hamesha support draw karo aur, doubt mein, honest test f X , Y = f X f Y
Worked example Ex 5 · Ek variable jisme koi randomness nahiX ∼ U ( 0 , 1 ) Y = 3 X Y
Forecast: Y X
Step 1 — Y Kyun? Y events form use karte hain. Ek constant ka P ( Y ∈ B ) = 1 3 ∈ B 0
Step 2 — events definition directly test karo.
Events form kyun? Yeh master definition hai; dono concrete forms iske special cases hain, aur yahan na PDF form na PMF form cleanly apply hota hai, isliye hum source par jaate hain: P ( X ∈ A , Y ∈ B ) = P ( X ∈ A ) P ( Y ∈ B ) A , B
Agar 3 ∈ B = P ( X ∈ A , Y ∈ B ) = P ( X ∈ A ) Y ∈ B = P ( X ∈ A ) ⋅ 1 = P ( X ∈ A )
Agar 3 ∈ / B = 0 = P ( X ∈ A ) ⋅ 0 = 0
Conclusion: factorisation sabhi A , B independent .
Verify (sanity): ek constant kisi bhi cheez ke baare mein zero information carry karta hai, isliye independence exactly wahi hai jo intuition demand karta hai. Har degenerate (a.s.-constant) variable har doosre variable se independent hota hai.
Worked example Ex 6 · Famous trapX ∼ U ( − 1 , 1 ) 2 1 [ − 1 , 1 ] Y = X 2 Cov ( X , Y ) = 0 X , Y
Forecast: Y literally X , isliye woh independent nahi ho sakte — lekin covariance sensor zero read kar sakta hai.
Step 1 — means. Kyun? Covariance (upar E [ X Y ] − E [ X ] E [ Y ] E [ X ] , E [ Y ] , E [ X Y ] E [ X ] = ∫ − 1 1 x ⋅ 2 1 d x = 0 E [ Y ] = E [ X 2 ] = ∫ − 1 1 x 2 ⋅ 2 1 d x = 3 1
Step 2 — cross term. Kyun? Covariance ko E [ X Y ] E [ X Y ] = E [ X ⋅ X 2 ] = E [ X 3 ] = ∫ − 1 1 x 3 ⋅ 2 1 d x = 0
Step 3 — assemble karo. Yeh step kyun? Ab jo teeno pieces haath mein hain, unhe covariance formula Cov = E [ X Y ] − E [ X ] E [ Y ] Cov = 0 − 0 ⋅ 3 1 = 0
Step 4 — lekin dependent. Kyun? Independence ke liye P ( ∣ X ∣ < 0.1 , Y > 0.5 ) = P ( ∣ X ∣ < 0.1 ) P ( Y > 0.5 ) ∣ X ∣ < 0.1 ⇒ Y = X 2 < 0.01 P ( ∣ X ∣ < 0.1 , Y > 0.5 ) = 0 0 not independent.
Verify: Cov = 0 Covariance and correlation dekho): covariance sirf linear coupling sense karta hai; ∪ Y = X 2 ⇒ kabhi reverse nahi (siwa Jointly Gaussian random variables ke).
Worked example Ex 7 · Ek knob jo tum ghuma sakte hof X , Y ( x , y ) = 1 + θ ( 2 x − 1 ) ( 2 y − 1 ) [ 0 , 1 ] 2 θ ∈ [ − 1 , 1 ] θ θ X , Y
Forecast: "θ ( 2 x − 1 ) ( 2 y − 1 ) x y θ
Step 1 — marginals. Kyun? f X , f Y [ 0 , 1 ] 0 f X ( x ) = ∫ 0 1 [ 1 + θ ( 2 x − 1 ) ( 2 y − 1 ) ] d y = 1 + θ ( 2 x − 1 ) = 0 ∫ 0 1 ( 2 y − 1 ) d y = 1 f Y ( y ) = 1 dono marginals θ ( 0 , 1 )
Step 2 — joint ko product se compare karo. Kyun? Independence ke liye f X , Y = f X f Y = 1 ⋅ 1 = 1 sabhi x , y 1 + θ ( 2 x − 1 ) ( 2 y − 1 ) = 1 x , y iff θ = 0
Step 3 — limiting cases. Kyun? Matrix ka Cell G boundary values ke baare mein hai.
θ → 0 → 1 θ = + 1 X Y θ = − 1 X Y
Conclusion: independent sirf θ = 0 ; har doosra allowed θ
Verify: θ = 0 f X , Y = 1 = f X f Y θ Cov ( X , Y ) = θ /36 0 θ = 0 do coincide karte hain kyunki coupling purely linear hai.
Worked example Ex 8 · Factory floor par do machinesMachine A ek given day mein probability 0.02 0.05 X = 1 Y = 1 P ( exactly one machine aaj fail hoti hai )
Forecast: "exactly one" = A fails & B doesn't, YA B fails & A doesn't. Compute karne se pehle estimate karo (roughly 0.02 + 0.05 = 0.07
Step 1 — independence use karke joint PMF mein translate karo.
Kyun? Independence humein marginals cell-by-cell multiply karne deta hai — yeh assumption ka poora payoff hai.
p X , Y ( x , y ) = p X ( x ) p Y ( y ) p X ( 1 ) = 0.02 , p X ( 0 ) = 0.98 p Y ( 1 ) = 0.05 , p Y ( 0 ) = 0.95
Step 2 — do "exactly one" cells.
Yeh do kyun? "Exactly one" matlab ( X , Y ) = ( 1 , 0 ) ( 0 , 1 ) p ( 1 , 0 ) = 0.02 ⋅ 0.95 = 0.019 p ( 0 , 1 ) = 0.98 ⋅ 0.05 = 0.049
Step 3 — add karo. Kyun add? Do cells mutually exclusive outcomes hain, isliye "ek ya doosre" ki probability unka sum hai.
P = 0.019 + 0.049 = 0.068
Verify: saare chaar cells 1 sum karte hain: p ( 1 , 1 ) = 0.02 ⋅ 0.05 = 0.001 p ( 0 , 0 ) = 0.98 ⋅ 0.95 = 0.931 0.001 + 0.019 + 0.049 + 0.931 = 1 0.068 0.07 Independence of events version se compare karo: identical arithmetic, kyunki indicator variables ARE events.
Worked example Ex 9 · Sum aur differenceX , Y f ( t ) = e − t t > 0 S = X + Y S summands S
Forecast: do independent lifetimes jodte hain — result positive value par peak karna chahiye, 0
Step 1 — convolution kyun? Yeh tool kyun, "multiply" kyun nahi? Hum independent variables jod rahe hain. Independent variables ke sum ki density unki densities ka convolution hai (dekho Sums of independent random variables — convolution ) — multiplication joint ke liye hai, convolution sum ke liye hai.
Step 2 — convolution set up karo.
Yeh limits kyun? f S ( s ) = ∫ 0 s f X ( t ) f Y ( s − t ) d t s t t > 0 s − t > 0 t ∈ ( 0 , s )
Step 3 — evaluate karo.
Kyun? Exponential densities substitute karo aur integrand simplify karo.
f S ( s ) = ∫ 0 s e − t e − ( s − t ) d t = ∫ 0 s e − s d t = s e − s s > 0 ( 2 , 1 ) 0
Step 4 — twist. Summands ke baare mein kyun pucha? Ek baar jab tum S = X + Y S S = 2 X ∈ ( 0 , 2 ) X S not independent hain, chahe X Y
Conclusion: f S ( s ) = s e − s s > 0 ( 2 , 1 ) S X not independent hai.
Verify: f S ∫ 0 ∞ s e − s d s = 1 E [ S ] = E [ X ] + E [ Y ] = 1 + 1 = 2 ∫ 0 ∞ s ⋅ s e − s d s = 2 s e − s s = 1 e − s ( 1 − s ) = 0
Recall Kaun sa phrase kaun sa cell trigger karta hai?"Support ek triangle hai" ::: Cell D — dependent; support unhe couple karta hai aur factorisation shortcut off-limits hai, isliye honest test chalao.
"Zero covariance" ::: Cell F — phir bhi dependent ho sakta hai; covariance sirf linear ties dekhta hai.
"Probability 1 se ek value" ::: Cell E — constant har cheez se independent hai (events form use karo).
"Independent variables ka sum" ::: Cell I — convolution use karo, aur sum apne parts se independent nahi hota.
Mnemonic KISI BHI independence problem ke liye safe checklist(1) Marginals compute karo (ya read off karo). (2) Har point/cell par f X , Y = f X f Y p X , Y = p X p Y shortcut sirf tab step 1 skip karne deta hai jab support product set ho — yeh ek convenience hai, kabhi extra requirement nahi.
