Independence of random variables — formal definition: Question bank

4.9.11 · D5 · Maths › Probability Theory & Statistics › Independence of random variables — formal definition

Jo core slogan hum baar baar test karte hain: joint object ko har point par marginals ke product mein factor hona chahiye, AUR support ek rectangle (product set) hona chahiye. Dono mein se koi bhi half tooto aur independence fail ho jaati hai.

True ya false — justify karo

Independence ka matlab hai ki

X

aur

Y

ek saath kabhi bhi bade nahi ho sakte.

False. Independence kehta hai ki ek ko jaanne se doosre ke baare mein koi information nahi milti — woh phir bhi ittifaqan ek saath bade ho sakte hain; woh joint probability bas dono alag probabilities ka product hoti hai.

Agar

X ⊥ Y

toh

P (X \in A, Y \in B) = P (X \in A) P (Y \in B)

sirf intervals ke liye hold karna zaroori hai.

False. Definition maangti hai ki yeh saare Borel sets

A, B

ke liye hold kare; aisa hota hai ki intervals

(- \infty, x]

check karna kaafi hai kyunki woh har Borel set generate karte hain, lekin property khud har jagah required hai.

Agar

Cov (X, Y) = 0

toh

X

aur

Y

independent hain.

False. Zero covariance sirf linear association khatam karta hai;

X \sim U (- 1, 1)

ke saath

Y = X^{2}

mein

Cov = 0

hai phir bhi

Y

poori tarah

X

se determine hota hai. Dekho Covariance and correlation.

Agar

X ⊥ Y

toh

Cov (X, Y) = 0

True. Independence se milta hai

E [X Y] = E [X] E [Y]

, isliye

Cov = E [X Y] - E [X] E [Y] = 0

. Arrow sirf isi direction mein chalta hai.

Ek joint density jo

g (x) h (y)

ke roop mein factor hoti hai woh hamesha independence imply karti hai.

False. Yeh independence imply karta hai tabhi jab support ek product set (rectangle) ho; triangle

0 \leq x \leq y \leq 1

par constant density

2

2 \cdot 1

" ke roop mein "factor" hoti hai phir bhi

X, Y

dependent hain.

Jointly Gaussian

X, Y

ke liye, zero correlation independence imply karta hai.

True. Gaussians woh special family hain jahaan uncorrelated

\Leftrightarrow

independent hota hai; off-diagonal covariance ko zero set karne se joint density factor ho jaati hai. Dekho Jointly Gaussian random variables.

Agar

X ⊥ Y

toh

X^{2}

aur

Y^{3}

bhi independent hain.

True. Independent variables ke kisi bhi (measurable) functions bhi independent hote hain, kyunki

X^{2}

ke baare mein sawaal sirf

X

ke baare mein sawaal hai, aur unhe

Y

ke baare mein saare sawaalon se independent assume kiya gaya tha.

Agar

X, Y, Z

mein har pair independent hai, toh teeno mutually independent hain.

False. Pairwise independence strictly weak hoti hai; teen variables pairwise independent ho sakte hain phir bhi three-way product rule

P (X \in A, Y \in B, Z \in C) = P (X \in A) P (Y \in B) P (Z \in C)

fail kar sakte hain.

Agar

F_{X, Y} (x, y) = F_{X} (x) F_{Y} (y)

ek convenient point par hold kare, toh independence establish ho jaati hai.

False. Factorisation saare

x, y

ke liye hold hani chahiye; ek matching point (ya PMF table mein ek matching cell) coincidence se ho sakta hai.

Do independent variables ka ek hi distribution hona zaroori hai.

False. Independence unke beech ke relationship ke baare mein hai, unke individual laws ke baare mein nahi; ek

Exp (1)

aur ek Bernoulli independent ho sakte hain.

Error dhundho

f_{X, Y} (x, y) = 2

0 \leq x \leq y \leq 1

factors as

2 \cdot 1

, isliye

X ⊥ Y

Support ek rectangle nahi hai: constraint

x \leq y

variables ko baandhti hai.

X = 0.9

jaanne se

Y \geq 0.9

force hota hai, aur wakai

f_{X} f_{Y} = 4 y (1 - x) \neq = 2

"Woh uncorrelated hain, aur correlation dependence measure karta hai, isliye woh independent hain."

Correlation sirf linear dependence measure karta hai. Non-linear links (jaise

Y = sin X

) covariance ke liye invisible ho sakte hain jabki variables independent se door hote hain.

p_{X, Y} (0, 0) = p_{X} (0) p_{Y} (0)

check out karta hai, isliye PMF table independence dikhata hai."

Aapko table ke har cell mein product rule verify karna hoga; ek single matching entry tab bhi ho sakti hai jab doosre cells fail kar rahe hon.

E [X Y] = E [X] E [Y]

, isliye

X

aur

Y

independent hain."

Yeh equality zero covariance ke equivalent hai, jo independence ke liye sirf necessary hai, sufficient nahi — wahi

Y = X^{2}

waala counterexample.

"Kyunki

f_{X, Y} (x, y) = x + y

[0, 1]^{2}

ka rectangular support hai, isliye

X ⊥ Y

Support theek hai, lekin

x + y

product

g (x) h (y)

mein factor nahi hota — sum ek product nahi hota, isliye density factorisation half fail karti hai.

"Conditional density

x = 0.5

par marginal ke barabar hai, isliye

X ⊥ Y

Independence ke liye zaroori hai ki

f_{Y ∣ X} (y ∣ x) = f_{Y} (y)

saare

x

ke liye hold kare, sirf ek slice ke liye nahi. Dekho Conditional distributions and conditional independence.

X

aur

Y

independent hain, isliye

X

aur

X + Y

bhi hain."

Galat reasoning:

X + Y

mein

X

hai, isliye

X

jaanne se

X + Y

ki distribution shift ho jaati hai. Sirf original independent pieces independent hain, woh derived combinations nahi jo kisi variable ko reuse karti hain.

Why questions

Kyun intervals

(- \infty, x]

check karna saare sets ke liye independence guarantee karne ke liye kaafi hai?

Kyunki woh half-infinite intervals poori Borel

σ

-algebra generate karte hain, isliye unpar product rule har measurable

A, B

tak propagate ho jaata hai.

Kyun factorisation shortcut kaam karne ke liye support ka rectangle hona zaroori hai?

Ek non-rectangular support ka matlab hai

Y

ki allowed values

X

par depend karti hain (ya vice versa), jo khud mein information transfer hai — independence iska koi bhi formula shape ho, forbid karta hai.

Kyun hum "joint factors into marginals" kehte hain "joint equals the marginals" nahi?

Joint plane par rehta hai jabki har marginal ek line par; independence precisely woh statement hai ki 2-D object do 1-D objects ke product ke roop mein rebuild hota hai.

Independence

E [X Y] = E [X] E [Y]

kyun imply karta hai?

Kyunki

f_{X, Y} = f_{X} f_{Y}

double integral

\iint x y f_{X, Y}

(\int x f_{X}) (\int y f_{Y})

mein split karne deta hai, yaani alag expectations ka product.

Zero covariance generally independence recover karne ke liye kyun kaafi nahi hai?

Covariance ek single number hai jo average linear co-movement capture karta hai; independence constraints ki ek infinite family hai (sets ke har pair ke liye ek), isliye ek number unhe saara encode nahi kar sakta.

Independent variables add karne par convolution kyun aata hai?

Kyunki independence joint density ko factor karne deti hai, aur summing saare splits

x + (s - x) = s

par integral force karta hai, jo exactly convolution

f_{X} * f_{Y}

hai. Dekho Sums of independent random variables — convolution.

Random variables ki independence events ki independence se kyun bani hai?

Ek random variable outcomes ko numbers mein badalne wali machine hai, isliye "

X

Y

ke baare mein kuch nahi batata" ka matlab hona chahiye ki har event

{X \in A}

har event

{Y \in B}

se independent hai. Dekho Independence of events.

Edge cases

Kya ek constant random variable

X = c

har doosre variable

Y

se independent hai?

Haan.

{X \in A}

ya toh poora sample space hai ya empty, aur dono kisi bhi event se independent hain, isliye product rule trivially hold karta hai.

Kya

X

khud se independent ho sakta hai?

Sirf tab agar

X

(almost surely) constant ho. Warna

P (X \leq x, X \leq x) = P (X \leq x) \neq = P (X \leq x)^{2}

, kyunki ek non-trivial probability

p

ke liye

p \neq = p^{2}

hota hai.

Agar joint density measure zero ke set (ek curve) par zero hai, toh kya woh independence tod deta hai?

Nahi. Densities "almost everywhere" define hoti hain, isliye measure-zero set par behaviour irrelevant hai — factorisation sirf a.e. hold karna zaroori hai.

Do discrete variables mein se har ek probability 1 se ek single value leta hai. Independent?

Haan, trivially — har ek constant hai, aur constants har cheez se independent hain, kyunki unke baare mein har event poora space ya empty hota hai.

Agar

X

aur

Y

ka rectangular support hai lekin density ek corner mein vanish karti hai, toh kya woh phir bhi independent ho sakte hain?

Sirf tab agar vanishing corner ek product

g (x) h (y)

ke consistent ho; ek "hole" jo

{g = 0} \cup {h = 0}

(ek full row/column) ke roop mein nahi likha ja sakta, matlab density ab factor nahi karti, isliye woh dependent hain.

Kya do independent joint laws ka mixture still independent hai?

Generally nahi. Do product densities ka average

\frac{1}{2} f_{1} g_{1} + \frac{1}{2} f_{2} g_{2}

deta hai, jo usually factor nahi karta — mixing independent ingredients se bhi dependence induce karta hai.

Recall One-line survival kit

Independence = factor everywhere + rectangular support. Zero covariance ek symptom hai, diagnosis nahi. Independent cheezein ke functions independent rehte hain; recombinations jo kisi variable ko reuse karti hain, woh nahi.

Connections

Independence of Random Variables — Formal Definition
Joint distribution and marginals
Conditional distributions and conditional independence
Covariance and correlation
Jointly Gaussian random variables
Sums of independent random variables — convolution
Independence of events