Chi-squared, t, F distributions — definition, degrees of freedom: Question bank

4.9.9 · D5 · Maths › Probability Theory & Statistics › Chi-squared, t, F distributions — definition, degrees of fre

Yeh ek concept-trap page hai. Koi heavy arithmetic nahi — har item ek misconception ya boundary case dhundh raha hai is baare mein ki teen sampling distributions kaise banti hain, degrees of freedom kyun utne hi aate hain, aur edges par kya hota hai (one degree of freedom, infinitely many, equal variances, zero deviation).

Shuru karne se pehle, parent note se teen builders yaad kar lo:

Recall Ek-line definitions jinke against tumhara test ho raha hai

$χ_{k}^{2} = Z_{1}^{2} + \dots + Z_{k}^{2}$ — independent standard normals ke squares ka sum.
$t_{k} = \frac{Z}{V / k}$ jahan $Z \sim N (0, 1)$ , $V \sim χ_{k}^{2}$ independent — ek $χ^{2}$ ke scaled square root par normal.
$F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ — do $χ^{2}$ 's ka ratio, jisme har ek ko pehle apne df se divide kiya gaya hai.

Answer cover karo, ek reason commit karo, phir reveal karo.

True or false — justify karo

Chi-squared variable negative ho sakta hai.

False — yeh squares

Z_{i}^{2}

ka sum hai, aur squares kabhi negative nahi hote; poori distribution

[0, \infty)

par rehti hai.

χ_{k}^{2}

normal ki tarah symmetric hai.

False — yeh right-skewed hai (right mein lambi tail), kyunki squaring se probability zero ke paas stack ho jaati hai aur bade positive values stretch ho jaate hain; sirf jab

k

badhta hai tab dheere-dheere bell-shaped lagta hai.

t

distribution

0

ke baare mein symmetric hai.

True — numerator

Z

0

ke baare mein symmetric hai aur positive denominator

V / k

kisi bhi sign ko favour nahi karta, isliye

t

aur

- t

equally likely hain.

F

distribution symmetric hai.

False — yeh do non-negative quantities ka ratio hai, isliye yeh

[0, \infty)

par rehta hai aur right-skewed hai; koi reflection symmetry nahi hai, sirf reciprocal symmetry

1/ F_{d_{1}, d_{2}} \sim F_{d_{2}, d_{1}}

hai.

t_{k}

ka variance har finite

k > 2

ke liye

N (0, 1)

se bada hai.

True —

Var (t_{k}) = \frac{k}{k - 2} > 1

har finite

k > 2

ke liye, kyunki random denominator (ek estimated

σ

) extra spread inject karta hai; yeh sirf limit

k \to \infty

mein

1

tak shrink hota hai.

Do independent chi-squareds ko add karne se ek aur chi-squared milta hai.

True —

χ_{a}^{2} + χ_{b}^{2} = χ_{a + b}^{2}

, kyunki tum bas sum mein aur independent squared normals append kar rahe ho; df simply add ho jaate hain.

Do independent

t

variables ko add karne se ek aur

t

milta hai.

False — aisa koi closure rule nahi hai; numerators normal hain lekin random denominators ek single scaled

χ^{2}

mein combine nahi hote, isliye sum ek

t

nahi hai.

k

df wale

χ^{2}

ka variance

k

hai.

False — iska mean

k

hai; iska variance

2 k

hai, kyunki har

Z_{i}^{2}

variance

2

contribute karta hai (

E [Z^{4}] - 1 = 3 - 1

se) aur

k

independent terms par variances add ho jaate hain.

F_{d_{1}, d_{2}}

aur

F_{d_{2}, d_{1}}

same distribution hain.

False — kaun sa sample upar hai yeh matter karta hai; yeh sirf reciprocal relation

F_{d_{1}, d_{2}} = 1/ F_{d_{2}, d_{1}}

ke through equal hain, identical distributions ke roop mein nahi (jab tak

d_{1} = d_{2}

na ho).

F_{d_{1}, d_{2}}

ka mean exactly

1

hai.

False — yeh

\frac{d _{2}}{d _{2} - 2}

hai (jab

d_{2} > 2

), thoda 1 se upar; yeh sirf

d_{2} \to \infty

par

1

approach karta hai kyunki tab denominator

χ^{2}

per-df exactly 1 par stabilise ho jaata hai.

Spot the error

n

size ke sample ke liye,

\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n}^{2}

Wrong df — yeh

χ_{n - 1}^{2}

hai.

\overset{x}{ˉ}

estimate karna

\sum (x_{i} - \overset{x}{ˉ}) = 0

force karta hai, ek constraint, isliye ek degree of freedom spend ho jaati hai; free directions

= n - 1

F = \frac{U}{V}

jahan

U \sim χ_{d_{1}}^{2}

V \sim χ_{d_{2}}^{2}

Div-by-df missing hai —

F = \frac{U / d _{1}}{V / d _{2}}

. Bina har ek ko apne df se divide kiye, ratio ka typical size

d_{1}, d_{2}

par depend karega aur equal variances ke under

1

ke paas nahi baithega.

"Kyunki

\overset{x}{ˉ}

normal hai, ise hamesha

Z = \frac{x ˉ - μ}{s / n}

se standardise karo."

Tumne

s

(ek estimate) likha lekin use

Z

kaha. Jab

σ

ko random

s

se replace kiya jaata hai, ratio

t_{n - 1}

follow karta hai,

N (0, 1)

nahi; dekho Sample Variance and Bessel's Correction.

t = \frac{Z}{V / k}

jahan

Z

aur

V

same data se aate hain, toh yeh dependent hain — lekin yeh theek hai."

Definition ko

Z

aur

V

ki independence require hai. Yeh isliye kaam karta hai kyunki normal sample ke liye

\overset{x}{ˉ}

(jo

Z

deta hai) aur

s^{2}

(jo

V

deta hai) provably independent hain; us fact ke bina derivation collapse ho jaata hai.

t^{2} \sim F_{k, 1}

Reversed —

t^{2} = \frac{Z ^{2} /1}{V / k} = F_{1, k}

. Squared normal

Z^{2} \sim χ_{1}^{2}

numerator hai (df

= 1

), aur

V \sim χ_{k}^{2}

denominator hai (df

= k

"Chi-squared per degree of freedom,

χ_{k}^{2} / k

, ka mean

k

hai."

Iska mean

1

hai,

k

nahi — mean-

k

variable ko

k

se divide karne par mean

1

milta hai. Yahi normalisation exactly wajah hai ki

F

ke har part ko per-df footing par rakha jaata hai.

"Degrees of freedom ek whole number hona chahiye jo sample size ke barabar ho."

Df data minus estimated parameters hai, sample size nahi; aur generally (jaise Welch's

t

, Gamma Distribution-based definitions mein) df integer bhi nahi hona chahiye.

Why questions

Hum

Z_{i}

ko sum karne ki bajaye square kyun karte hain?

Plain sum

\sum Z_{i}

phir se normal hai aur zero par average karta hai, isliye positive aur negative misses cancel ho jaate hain; squaring sign ki parwah kiye bina deviation ki total magnitude measure karta hai — ek genuine "spread" score.

t_{k} \to N (0, 1)

kyun hota hai jab

k \to \infty

Jab df badhte hain,

V / k \to 1

law of large numbers se, isliye random denominator wobbling band kar deta hai aur

1

par freeze ho jaata hai, aur

t = Z /1 = Z

reh jaata hai.

t

ke tails normal se fatter kyun hain?

Denominator

V / k

random hai; kabhi kabhi yeh chance se chhota hota hai, jo

t

ko extreme values tak inflate kar deta hai ek fixed denominator se zyada — heavier tails.

F

mein har

χ^{2}

ko apne df se divide kyun karte hain?

Ek raw

χ_{k}^{2}

ka mean

k

hai, isliye bare ratio df sizes par depend karega; per-df normalisation har part ko average

\approx 1

banata hai, toh

F \approx 1

equal variances signal karta hai — yahi interpretation ANOVA mein use hoti hai.

Sample variance ke liye df

n - 1

kyun hai,

n

nahi?

Same data se

\overset{x}{ˉ}

compute karna linear constraint

\sum (x_{i} - \overset{x}{ˉ}) = 0

impose karta hai; jab

n - 1

deviations known hoon to last forced hai, isliye sirf

n - 1

pieces truly free hain.

F

mein do

χ^{2}

's independent kyun honi chahiye?

Clean ratio distribution tabhi hold karta hai jab numerator aur denominator independently vary karein; agar unhone randomness share ki (jaise same sample), to ratio ke cancellations ka shape distort ho jaata aur

F

tables apply nahi hote.

Do variances compare karne ke liye

F

difference

s_{1}^{2} - s_{2}^{2}

se zyada natural tool kyun hai?

Ek ratio scale-free hai — equal variances ke under yeh common

σ^{2}

ki parwah kiye bina

1

ke paas centre karta hai — jabki ek difference unknown units aur scale par depend karta rehta hai.

E [Z^{2}] = 1

kyun hai,

0

nahi?

Kyunki

E [Z^{2}] = Var (Z) + (E [Z])^{2} = 1 + 0 = 1

; squared value ka mean variance pick up karta hai, chahe

E [Z] = 0

ho.

Edge cases

χ_{1}^{2}

kya hai — sabse chhota interesting case?

Yeh sirf

Z^{2}

hai, ek single standard normal ka square: heavily right-skewed with

0

ke paas ek spike (chhote

Z

's common hain) aur mean

1

, variance

2

Kya

t_{k}

ka mean hota hai jab

k = 1

Nahi —

t_{1}

Cauchy distribution hai; iske tails itne heavy hain ki mean integral converge nahi karta, isliye

E [t_{k}] = 0

sirf

k > 1

ke liye stated hai.

Kya

t_{k}

ka finite variance hai jab

k = 2

Nahi —

Var (t_{k}) = \frac{k}{k - 2}

blow up karta hai jab

k \to 2^{+}

, isliye variance sirf

k > 2

ke liye finite hai;

k = 1, 2

par tails bahut fat hain.

F_{1, 1}

extreme mein kaisa dikhta hai?

Dono df

1

ke barabar hain, isliye yeh

\frac{Z _{1}^{2}}{Z _{2}^{2}}

hai — do squared normals ka ratio; extremely heavy-tailed with koi finite mean nahi, kyunki

d_{2} = 1

d_{2} > 2

requirement fail ho jaati hai.

F_{d_{1}, d_{2}}

ka kya hota hai jab do population variances truly equal hoon?

Do normalised variance estimates mein se har ek lagbhag

1

hai, isliye

F

1

ke paas concentrate karta hai; yahi precisely null-hypothesis behaviour hai jo Hypothesis Testing aur ANOVA mein exploit hota hai.

Agar saare sample values identical hoon, toh

s^{2}

aur

χ^{2}

statistic kya hai?

Tab har

x_{i} - \overset{x}{ˉ} = 0

, isliye

s^{2} = 0

aur

\frac{( n - 1 ) s ^{2}}{σ ^{2}} = 0

—

χ^{2}

support ke bilkul left edge par ek legitimate boundary value, halanki continuous data ke liye probability zero hai.

Jab

k \to \infty

, ek standardised

χ_{k}^{2}

kaun sa shape approach karta hai?

Kyunki

χ_{k}^{2}

k

i.i.d. terms ka sum hai, Central Limit Theorem apply hota hai:

\frac{χ _{k}^{2} - k}{2 k} \to N (0, 1)

, isliye skew dheere-dheere wash out ho jaata hai aur yeh normal lagta hai.

t

ka sabse bada tail kaun sa ho sakta hai, aur kab?

Sabse heavy

k = 1

par (Cauchy) hai; tails progressively lighter hote jaate hain jab

k

badhta hai, normal thin tails mein converge hote hue jab

k \to \infty