Step 1 — "Saath chalna" encode karo.
Deviation X−μX lo. Yeh positive hota hai jab X high ho, negative jab low ho. Same Y ke liye.
Step 2 — Deviations ko multiply karo.
Product (X−μX)(Y−μY) hai:
positive jab dono high ya dono low hon (agree karte hain),
negative jab ek high ho aur doosra low (disagree karte hain).
Yeh step kyun? Multiplication sabse sasta operation hai jo agreement ke liye + aur disagreement ke liye − deta hai.
Step 3 — Distribution par average lo.
Expectation lo taaki agreements aur disagreements net out ho jaayein:
Cov(X,Y)=E[(X−μX)(Y−μY)].
Net positive ⇒ woh generally saath move karte hain.
Step 4 — Ek computational shortcut. Expand karo:
E[(X−μX)(Y−μY)]=E[XY−XμY−μXY+μXμY]=E[XY]−μYE[X]−μXE[Y]+μXμY=E[XY]−μXμY.
Maano U=X−μX, V=Y−μY. Kisi bhi real t ke liye consider karo:
E[(U−tV)2]≥0(ek square kabhi negative nahi hota).
Expand karo:
E[U2]−2tE[UV]+t2E[V2]≥0.
Yeh t mein ek quadratic hai jo hamesha ≥0 hai, isliye iska discriminant ≤0 hona chahiye:
(2E[UV])2−4E[U2]E[V2]≤0⇒(E[UV])2≤E[U2]E[V2].
Lekin E[UV]=Cov(X,Y), E[U2]=σX2, E[V2]=σY2. Isliye
Cov(X,Y)2≤σX2σY2⇒σXσYCov≤1.Yeh step kyun? Equality tabhi hoti hai jab U−tV=0 ho, yaani Y, X ka exact linear function ho — isliye ρ=±1 ka matlab perfectly linear hota hai.
Do doston ko jhoolon par imagine karo. Covariance poochhti hai: jab ek aage jhuulta hai, kya doosra bhi aage jhuulta hai? Agar haan → positive; agar ek aage jaaye aur doosra peeche → negative; agar koi pattern nahi → almost zero.
Lekin "woh kitna jhuulte hain" jhoolon ki size par depend karta hai, jo compare karna unfair hai. Toh correlation ek fixed report card par unki teamwork measure karna hai −1 (perfectly opposite) se +1 (perfectly saath) tak, jahaan 0 ka matlab "koi teamwork nahi." Yeh jhuulon ki size ignore karta hai aur sirf judge karta hai ki woh kitna match karte hain.