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

4.9.9 · D4 · Maths › Probability Theory & Statistics › Chi-squared, t, F distributions — definition, degrees of fre

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Level 1 — Recognition

Goal: distribution ka naam bolo aur uske degrees of freedom pehchano. Counting ke alaawa koi arithmetic nahi.

Exercise 1.1. Tum $7$ independent standard normals ke squares aapas mein jod dete ho: $W = Z_{1}^{2} + \dots + Z_{7}^{2}$ . $W$ kaunsi distribution follow karta hai, aur uske degrees of freedom (df) kya hain?

Recall Solution 1.1

Yeh object kya hai: yeh ek sum of squared independent standard normals hai. Yahi chi-squared ki exact definition hai. Df count karo: df = kitne independent squared normals hum ne sum kiye. Humne $7$ sum kiye. $W \sim χ_{7}^{2}, df = 7.$ Sanity check mean se: $E [W] = k = 7$ — har $Z_{i}^{2}$ mean $1$ contribute karta hai, aur $7 \times 1 = 7$ . ✓

Exercise 1.2. Ek statistic $\frac{Z}{V /9}$ likha gaya hai jahan $Z \sim N (0, 1)$ aur $V \sim χ_{9}^{2}$ independent hain. Distribution aur uske df batao.

Recall Solution 1.2

Template se match karo. $t$ ki definition hai $t_{k} = \frac{Z}{V / k}$ : upar ek normal, aur neeche ek chi-squared apne df se divided, phir square-root liya. Yahan neeche $V /9$ hai jahan $V \sim χ_{9}^{2}$ , isliye $k = 9$ . $\frac{Z}{V /9} \sim t_{9}, df = 9.$

Exercise 1.3. Identify karo: $\frac{U /3}{V /8}$ jahan $U \sim χ_{3}^{2}$ , $V \sim χ_{8}^{2}$ independent hain. Yeh kya hai, aur numerator df kaunsa hai?

Recall Solution 1.3

Template se match karo. $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ — har chi-squared apne df se divided, phir ratio. Upar $U \sim χ_{3}^{2}$ use ho raha hai isliye $d_{1} = 3$ ; neeche $V \sim χ_{8}^{2}$ isliye $d_{2} = 8$ . $\frac{U /3}{V /8} \sim F_{3, 8}, numerator df d_{1} = 3, denominator df d_{2} = 8.$

Level 2 — Application

Goal: $E$ aur $Var$ formulas mein substitute karo.

Exercise 2.1. $E [χ_{11}^{2}]$ aur $Var (χ_{11}^{2})$ nikalo.

Recall Solution 2.1

Formulas: $E [χ_{k}^{2}] = k$ aur $Var (χ_{k}^{2}) = 2 k$ . Yeh kyun hold karta hai: har ek $Z^{2}$ ka $E [Z^{2}] = 1$ aur $Var (Z^{2}) = 2$ hota hai; independence ki wajah se means aur variances $k$ terms par add ho jaate hain. $E [χ_{11}^{2}] = 11, Var (χ_{11}^{2}) = 2 \cdot 11 = 22.$

Exercise 2.2. $t_{6}$ ke liye, $E [t_{6}]$ aur $Var (t_{6})$ compute karo.

Recall Solution 2.2

Mean. $t$ $0$ ke baare mein symmetric hai (numerator $Z$ symmetric hai, denominator ek positive number hai), isliye $k > 1$ ke liye, $E [t_{k}] = 0$ . Yahan $k = 6 > 1$ : $E [t_{6}] = 0$ . Variance. $Var (t_{k}) = \frac{k}{k - 2}$ for $k > 2$ . $k = 6$ rakhne par: $Var (t_{6}) = \frac{6}{6 - 2} = \frac{6}{4} = 1.5.$ Note karo $1.5 > 1$ : $N (0, 1)$ se zyada spread (jiska variance $1$ hai), kyunki random denominator extra uncertainty add karta hai.

Exercise 2.3. $F_{4, 10}$ ke liye, $E [F_{4, 10}]$ compute karo.

Recall Solution 2.3

Formula: $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2}$ for $d_{2} > 2$ . Mean ke liye sirf denominator df matter karta hai. Yahan $d_{2} = 10$ : $E [F_{4, 10}] = \frac{10}{10 - 2} = \frac{10}{8} = 1.25.$ Interpretation: $1$ ke paas, jaisa expect kiya tha — $F$ do per-df variances compare karta hai, aur equal variances ke under har part ka mean $\approx 1$ hota hai.

Level 3 — Analysis

Goal: constraints aur estimated parameters track karke khud sahi df chuno.

Exercise 3.1. $N (μ, σ^{2})$ se size $n = 13$ ka ek sample liya gaya hai. True $σ^{2} = 5$ hai aur sample se $s^{2} = 6$ milta hai, jahan $s^{2} = \frac{1}{n - 1} \sum (x_{i} - \overset{x}{ˉ})^{2}$ . $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ ki exact distribution batao aur uski observed value compute karo.

Recall Solution 3.1

Distribution. Key statistics result: normal distribution se liye gaye sample ke liye, $\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n - 1}^{2} .$ Df $= n - 1$ kyun, $n$ nahi: $\overset{x}{ˉ}$ compute karne se deviations par ek constraint lag jaata hai $\sum (x_{i} - \overset{x}{ˉ}) = 0$ . Yeh ek constraint hai, isliye $n$ deviations mein se ek baaki $n - 1$ se determine ho jaata hai — ek df lost ho jaata hai. Dekho Sample Variance and Bessel's Correction. $n = 13$ ke saath: df $= 12$ , isliye distribution $χ_{12}^{2}$ hai. Observed value. $\frac{( n - 1 ) s ^{2}}{σ ^{2}} = \frac{12 \cdot 6}{5} = \frac{72}{5} = 14.4.$

Exercise 3.2. Ek one-sample $t$ -test mein $n = 25$ , $\overset{x}{ˉ} = 103$ , $s = 10$ hai, $μ_{0} = 100$ test ho raha hai. $t$ -statistic compute karo aur uski reference distribution batao. $Z$ use karna chahiye ya $t$ , aur kyun?

Recall Solution 3.2

Statistic. $t = \frac{x ˉ - μ _{0}}{s / n} = \frac{103 - 100}{10/ 25} = \frac{3}{10/5} = \frac{3}{2} = 1.5.$ Reference distribution. $t_{n - 1} = t_{24}$ . Df $s^{2}$ ke andar chhupe chi-squared se aata hai, jo $n - 1 = 24$ df carry karta hai (wahi constraint jaisa 3.1 mein). $Z$ ya $t$ ? $t$ use karo. Humne true $σ$ nahi jaana tha; hum ne ise random estimate $s$ se replace kiya. Denominator mein woh extra randomness fatter-tailed $t_{24}$ maangti hai, $Z$ nahi. Dekho Hypothesis Testing.

Exercise 3.3. Do independent samples: sample A mein $n_{1} = 7$ , sample B mein $n_{2} = 16$ hain. Equal true variances ki assumption ke under, tum $F = s_{1}^{2} / s_{2}^{2}$ form karte ho. Numerator aur denominator df kya hain, aur $F$ kaunsi distribution follow karta hai?

Recall Solution 3.3

Har sample ki spread ek chi-squared hai. 3.1 ki logic se, $\frac{( n _{1} - 1 ) s _{1}^{2}}{σ ^{2}} \sim χ_{n_{1} - 1}^{2}$ aur $\frac{( n _{2} - 1 ) s _{2}^{2}}{σ ^{2}} \sim χ_{n_{2} - 1}^{2}$ . Per df ratio banao. Equal $σ^{2}$ ke under, common $σ^{2}$ cancel ho jaata hai, aur $\frac{s _{1}^{2}}{s _{2}^{2}} = \frac{[ ( n _{1} - 1 ) s _{1}^{2} / σ ^{2} ] / ( n _{1} - 1 )}{[ ( n _{2} - 1 ) s _{2}^{2} / σ ^{2} ] / ( n _{2} - 1 )} = \frac{U / d _{1}}{V / d _{2}} \sim F_{d_{1}, d_{2}} .$ Df. $d_{1} = n_{1} - 1 = 6$ (numerator, sample A se), $d_{2} = n_{2} - 1 = 15$ (denominator, sample B se). $F = \frac{s _{1}^{2}}{s _{2}^{2}} \sim F_{6, 15} .$

Level 4 — Synthesis

Goal: do facts combine karo — additivity, $t^{2} = F$ link, reciprocal symmetry.

Exercise 4.1. $A \sim χ_{4}^{2}$ aur $B \sim χ_{9}^{2}$ independent hain. $A + B$ ki distribution kya hai, aur uska mean aur variance kya hain?

Recall Solution 4.1

Additivity. $χ_{4}^{2}$ matlab $4$ squared normals; $χ_{9}^{2}$ matlab $9$ aur independent squared normals. Inhe append karne par $4 + 9 = 13$ squared independent normals milte hain: $A + B \sim χ_{13}^{2} .$ Mean aur variance. $E [A + B] = 13, Var (A + B) = 2 \cdot 13 = 26.$ Additivity ke bina cross-check: means add hote hain ( $4 + 9 = 13$ ✓) aur independence se variances bhi add hote hain ( $8 + 18 = 26$ ✓).

Exercise 4.2. Ek random variable $T \sim t_{20}$ hai. $T^{2}$ ki distribution nikalo, aur phir $E [T^{2}]$ do alag tareekon se compute karo yeh confirm karne ke liye ki dono agree karte hain.

Recall Solution 4.2

$t^{2} = F$ link. $t_{k} = \frac{Z}{V / k}$ likhte hain, toh $T^{2} = \frac{Z ^{2}}{V / k} = \frac{Z ^{2} /1}{V / k} .$ Kyunki $Z^{2} \sim χ_{1}^{2}$ (ek squared normal, df $1$ ), upar ek $χ_{1}^{2}$ hai apne df $1$ se divided, aur neeche $χ_{k}^{2}$ divided by $k$ . Yeh exactly ek $F$ hai $d_{1} = 1$ , $d_{2} = k$ ke saath: $T^{2} \sim F_{1, 20} .$ $E [T^{2}]$ tarika 1 — $F$ se: $E [F_{1, 20}] = \frac{d _{2}}{d _{2} - 2} = \frac{20}{18} = \frac{10}{9} \approx 1.1111$ . $E [T^{2}]$ tarika 2 — $t$ se: $E [T^{2}] = Var (T) + (E [T])^{2} = \frac{k}{k - 2} + 0 = \frac{20}{18} = \frac{10}{9}$ . Dono $\frac{10}{9}$ dete hain. ✓ (Yeh agree karna hi chahiye — $T^{2}$ dono taraf se same random variable hai.)

Exercise 4.3. Tumne ratio $s_{A}^{2} / s_{B}^{2}$ ke liye $F_{5, 12} = 0.4$ compute kiya, lekin table sirf $F_{12, 5}$ ki upper-tail values list karta hai. Reciprocal symmetry use karke convert karo.

Recall Solution 4.3

Reciprocal symmetry. Numerator aur denominator swap karne se value invert hoti hai aur df swap ho jaate hain: $\frac{1}{F _{d_{1}, d_{2}}} \sim F_{d_{2}, d_{1}} .$ Toh agar $F_{5, 12} = 0.4$ hai, toh $\frac{1}{F _{5, 12}} = \frac{1}{0.4} = 2.5 \sim F_{12, 5} .$ Interpretation: ek chhota $F_{5, 12}$ (sample B zyada variable) ek bade $F_{12, 5} = 2.5$ ban jaata hai jab tum konsa variance upar hai woh flip karte ho — same evidence, tumhari table ke liye re-express kiya gaya.

Level 5 — Mastery

Goal: limiting behaviour aur structural links ke baare mein reason karo.

Exercise 5.1. Explain karo, phir limit compute karo: $k \to \infty$ hone par $Var (t_{k})$ ka kya hoga, aur $t_{k}$ kaunsi distribution approach karta hai? Trend dikhane ke liye $k = 3, 10, 100$ ke numerical variances do.

Recall Solution 5.1

Limit kyun hai. $t_{k} = \frac{Z}{V / k}$ mein, term $V / k$ $k$ independent $Z_{i}^{2}$ 's ka average hai (kyunki $V = \sum Z_{i}^{2}$ ). Law of large numbers se woh average apne mean $E [Z^{2}] = 1$ ki taraf converge karta hai. Toh random denominator hichkichana band kar deta hai aur $1$ par lock ho jaata hai, sirf $Z$ bacha rehta hai. Isliye $t_{k} \to N (0, 1)$ . Variance limit. $lim_{k \to \infty} Var (t_{k}) = lim_{k \to \infty} \frac{k}{k - 2} = 1,$ jo $Var (Z) = 1$ se match karta hai. Trend (figure dekho): $Var (t_{3}) = \frac{3}{1} = 3, Var (t_{10}) = \frac{10}{8} = 1.25, Var (t_{100}) = \frac{100}{98} \approx 1.0204.$ $1$ se upar ka excess shrink hota hai: $3 \to 1.25 \to 1.02$ — tails normal ki taraf patli hoti jaati hain.

Figure — Chi-squared, t, F distributions — definition, degrees of freedom

Exercise 5.2. $F_{d_{1}, d_{2}}$ consider karo jahan $d_{1}$ fixed hai aur $d_{2} \to \infty$ . Dikhao $E [F] \to 1$ aur structurally explain karo yeh object kya banta hai. Phir batao ek akela $χ_{d_{1}}^{2} / d_{1}$ kya kar raha hai.

Recall Solution 5.2

Mean limit. $lim_{d_{2} \to \infty} E [F_{d_{1}, d_{2}}] = lim_{d_{2} \to \infty} \frac{d _{2}}{d _{2} - 2} = 1.$ Structural reason. $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ . Jab $d_{2} \to \infty$ , denominator $V / d_{2}$ $d_{2}$ squared normals ka average hai → $E [Z^{2}] = 1$ ki taraf converge karta hai (phir se law of large numbers). Ratio sirf numerator tak collapse ho jaata hai: $F_{d_{1}, d_{2}} d_{2} \to \infty \frac{U}{d _{1}}, U \sim χ_{d_{1}}^{2} .$ Toh bahut bade denominator df wala $F$ essentially ek single chi-squared divided by its df hai — spread ka ek per-df measure jiska mean $1$ hai. Isliye, ek bade doosre sample ke saath, $F$ se variances compare karna sirf pehle sample ki spread taulne tak reduce ho jaata hai. ANOVA ke liye foundational.

Exercise 5.3. $n = 2$ par ek $t$ -test (toh $k = 1$ ) $t_{1}$ deta hai. Variance formula $\frac{k}{k - 2}$ se $\frac{1}{- 1} = - 1$ milta hai. Variance negative nahi ho sakta — is degenerate case ko honestly explain karo aur batao sach mein kya sahi hai.

Recall Solution 5.3

Formula apply nahi hota. $Var (t_{k}) = \frac{k}{k - 2}$ sirf $k > 2$ ke liye stated hai. $k = 1$ ke liye formula apne domain ke bahar hai — daalne par meaningless $- 1$ milta hai, jo ek signal hai ki variance exist nahi karta, na ki koi real value. Sach mein kya sahi hai. $k = 1$ ke liye, random denominator $V /1 = ∣ Z_{2} ∣$ arbitrarily close to $0$ ho sakta hai, isliye $t_{1} = \frac{Z _{1}}{∣ Z _{2} ∣}$ ke tails itne heavy hain ki na uska mean exist karta hai na variance (dono integrals diverge karte hain). Actually $t_{1}$ standard Cauchy distribution hai. Honest boundary summary:

$t_{k}$ ka mean sirf $k > 1$ ke liye exist karta hai.
$t_{k}$ ka variance sirf $k > 2$ ke liye exist karta hai; $k = 2$ ke liye, $\frac{k}{k - 2} = \frac{2}{0}$ undefined hai (infinite variance); $k = 1$ ke liye mean bhi fail karta hai. Lesson: substitute karne se pehle hamesha moment formula ke saath attached domain condition check karo.

Recall Ek-line self-test

$F$ ka mean kaunsa df control karta hai? ::: Denominator df $d_{2}$ , $E = \frac{d _{2}}{d _{2} - 2}$ ke zariye ( $d_{2} > 2$ ). $t^{2} \sim F_{1, k}$ kyun hai? ::: Kyunki $Z^{2} \sim χ_{1}^{2}$ , isliye square karne par ek df- $1$ chi-squared $χ_{k}^{2} / k$ ke upar aa jaata hai — $F$ template. $t_{k} \to N (0, 1)$ kyun hota hai? ::: Denominator $V / k$ $k$ squared normals ko unke mean $1$ par average karta hai, ek constant par freeze ho jaata hai.