Joint distributions — joint PMF - PDF, marginal, conditional: Question bank

4.9.10 · D5 · Maths › Probability Theory & Statistics › Joint distributions — joint PMF - PDF, marginal, conditional

True or false — justify

A joint PDF value

f_{X, Y} (x, y)

is a probability.

False — yeh ek density hai (probability per unit area). Yeh

1

se bhi zyada ho sakta hai; sirf

\iint_{A} f d x d y

kisi region ke upar ek probability hoti hai.

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

for all

(x, y)

, then

X

and

Y

are independent.

True — joint ka har point par marginals ke product mein factor hona exactly Independence of Random Variables ki definition hai.

X

and

Y

have identical marginal distributions, they must be independent.

False — marginals sirf har variable ko akele describe karti hain aur unke joint behaviour ke baare mein kuch nahi kehti; equal marginals strong dependence ke saath bhi ho sakti hain.

Every valid marginal comes from some joint, but a given pair of marginals determines a unique joint.

False — kai alag-alag joints ek hi do marginals share kar sakte hain (woh aapas mein is baat mein differ karte hain ki variables saath kaise move karte hain), isliye marginals ki pair joint ko pin down nahi karti.

\int_{- \infty}^{\infty} f_{X ∣ Y} (x ∣ y) d x = 1

for every fixed

y

with

f_{Y} (y) > 0

True — conditional ko

f_{Y} (y)

se divide karke renormalize kiya jaata hai, isliye

x

ke function ke roop mein yeh exactly

1

tak integrate hoti hai; yeh

x

mein ek genuine distribution hai.

\iint f_{X ∣ Y} (x ∣ y) d x d y = 1

False — conditional ko

y

par bhi integrate karna double-counting hai; yeh

1

ke barabar nahi hota. Sirf

x

par integral (jab

y

freeze ho)

1

ke barabar hoti hai.

Cov (X, Y) = 0

then

X

and

Y

are independent.

False — zero Covariance and Correlation ka matlab sirf koi linear association nahi hai; nonlinear dependence (jaise

Y = X^{2}

symmetric

X

ke saath) zero covariance de sakti hai lekin full dependence hoti hai.

X

and

Y

are independent then

Cov (X, Y) = 0

True — independence se

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

ho jaata hai, isliye covariance zero ho jaati hai. (Converse fail hoti hai, jaise upar bataya.)

p_{X ∣ Y} (x ∣ y)

can be larger than

p_{X, Y} (x, y)

True — joint entry ko

p_{Y} (y) \leq 1

se divide karne par woh inflate ho jaati hai, isliye conditional probability joint value se kam se kam utni hi badi hoti hai.

A conditional PDF

f_{X ∣ Y} (x ∣ y)

is a function of

y

as well as

x

Form mein True hai, lekin fixed

y

ke liye yeh

x

mein ek distribution hai; alag-alag conditioning values

y

se generally alag-shaped slices milti hain.

Spot the error

"To get

p_{X} (x)

, just delete the

Y

column you don't care about."

Galat — delete karne se sirf ek

Y

-value bachti hai; marginalize karne ka matlab hai saare

Y

-values par sum karna:

p_{X} (x) = \sum_{y} p_{X, Y} (x, y)

. Tum collapse karte ho, kabhi drop nahi karte.

"The conditional density is simply the joint evaluated along the line

Y = y

Galat — woh slice

f_{Y} (y)

tak integrate hoti hai,

1

tak nahi. Isse valid distribution banane ke liye tumhe

f_{Y} (y)

se divide karke renormalize karna hoga.

"For

f_{X, Y} = 3 x

on the triangle

0 < y < x < 1

f_{X} (x) = \int_{0}^{1} 3 x d y = 3 x

Galat bounds — is triangle par

y

sirf

0

x

tak jaata hai, isliye

f_{X} (x) = \int_{0}^{x} 3 x d y = 3 x^{2}

. Support ko ignore karne se galat marginal milti hai.

"Since

f_{X, Y} (x, y) = 3 x

splits as (a function of

x

) times (a function of

y

X

and

Y

are independent."

Galat — support

0 < y < x < 1

variables ko couple karta hai (allowed range of

y

depends on

x

karta hai). Factoring ek product-shaped support ke saath bhi hold karna chahiye; yahan woh nahi karta.

P (X = x ∣ Y = y) = P (Y = y ∣ X = x)

because both slice the same joint entry."

Galat — dono

p_{X, Y} (x, y)

upar use karte hain lekin alag marginals se divide karte hain (

p_{Y} (y)

p_{X} (x)

), isliye woh generally unequal hain (yahi Bayes' Theorem ka seed hai).

"A density integrates to

1

, so it must stay below

1

everywhere."

Galat — ek tall narrow density (jaise ek chote interval par spike) height mein

1

se zyada hoti hai jabki unit area enclose karti hai. Height unbounded hai; sirf total area fixed hai.

"If

f_{Y} (y) = 0

I can still condition on

Y = y

by dividing."

Galat —

0

se divide karna undefined hai; conditional

f_{X ∣ Y} (x ∣ y)

sirf wahan defined hoti hai jahan

f_{Y} (y) > 0

ho, yani jahan us value of

Y

par actually density ho.

Why questions

Why do we use a density

f_{X, Y}

instead of point probabilities in the continuous case?

Kyunki ek single point

(x, y)

ka area zero hota hai, isliye

P (X = x, Y = y) = 0

; density per unit area probability measure karti hai, aur real probability tabhi milti hai jab kisi region par integrate karo.

Why does summing/integrating out a variable give the marginal?

Kyunki

{X = x} = ⋃_{y} {X = x, Y = y}

disjoint events ka union hai; probability ki additivity us union ko sum (discrete) ya integral (continuous) mein badal deti hai — yahi Law of Total Probability hai.

Why must we divide by the marginal when forming a conditional?

Frozen slice

{Y = y}

ka total mass

p_{Y} (y)

(ya

f_{Y} (y)

) hota hai,

1

nahi; divide karne se yeh rescale ho jaata hai taaki naya sample space

1

tak integrate ho aur ek legitimate distribution ban sake.

Why does independence let us write

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

Independence ke under,

Y

jaanna tumhe

X

ke baare mein kuch nahi bataata, isliye

f_{X ∣ Y} = f_{X}

; chain rule

f_{X, Y} = f_{X ∣ Y} f_{Y}

mein substitute karne par product form milta hai.

Why is the chain rule

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

symmetric in the two orders?

Dono expressions same joint ko ek slice times uske weight se reconstruct karte hain; pehle

X

par ya pehle

Y

par condition karna ek hi surface ke do alag raaste hain — unhe equal karne par Bayes' rule milta hai.

Why can the support region alone destroy independence even when the formula factors?

Independence ke liye joint ko rectangular support par

g (x) h (y)

ke roop mein factor karna zaroori hai; triangular ya curved support ek variable ki allowed values ko doosre par dependent banata hai, jo ki dependence hai.

Why does knowing only marginals leave Covariance and Correlation undetermined?

Covariance

E [X Y]

par depend karti hai, jo ek joint quantity hai; marginals

E [X]

aur

E [Y]

fix karti hain lekin un marginals wale kai joints alag-alag

E [X Y]

dete hain.

Edge cases

What happens to

f_{X ∣ Y} (x ∣ y)

at a value

y

where

f_{Y} (y) = 0

Yeh undefined chhod di jaati hai — condition karne ke liye koi probability mass nahi hai, isliye

0/0

ratio ka wahan koi matlab nahi hai.

For

f_{X, Y} = 3 x

0 < y < x < 1

, what is the support of

f_{X ∣ Y} (\cdot ∣ y)

once

Y = y

is fixed?

Sirf

x \in (y, 1)

—

Y = y

freeze karne par

x > y

force hota hai, isliye conditional shrunken interval par rehti hai, pure

(0, 1)

par nahi.

X

takes a single value with probability

1

(a degenerate/constant variable), is it independent of any

Y

Haan — ek constant mein koi randomness nahi hoti, isliye

p_{X, Y} (x_{0}, y) = 1 \cdot p_{Y} (y) = p_{X} (x_{0}) p_{Y} (y)

; factorization trivially hold karti hai.

In a discrete joint table, what does a full row of zeros for

X = x

mean, and what is

p_{X ∣ Y} (x ∣ y)

there?

Iska matlab hai

p_{X} (x) = 0

(woh

X

-value kabhi occur nahi hoti); conditional

p_{X ∣ Y} (x ∣ y) = 0

har

y

ke liye hoti hai, aur

X = x

par condition karna undefined hai.

If the conditional

f_{X ∣ Y} (x ∣ y)

is the same function of

x

for every

y

, what does that tell you?

X

ki distribution

Y

par react nahi karti, yani

f_{X ∣ Y} = f_{X}

— independence ki definition.

What is

f_{X, Y}

at any point outside the stated support region?

Exactly

0

— "and

0

elsewhere" clause density ka hissa hai; ise bhoolne par integration bounds galat extend ho jaate hain aur total-mass-

= 1

check toot jaata hai.