Visual walkthrough — Confidence intervals

1.3.21 · D2 · AI-ML › Probability & Statistics › Confidence intervals

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Step 1 — Ek sample se ek dot milta hai

KYA HAI. Hamare paas cheezein ka ek bada pile hai (log, test images, jo bhi). Har cheez ka ek number hai. Poore pile ka ek sach wala average hai jo hum dekh nahi sakte — use $\mu$ (Greek letter "mu", sirf "sab cheezoin ka asli average" ka naam) kehte hain. Hum unme se randomly $n$ uthate hain aur unka average nikalte hain:

$$\bar{x}=\underset{\text{divide by how many}}{\underbrace{\frac{1}{n}}}\underset{\text{add them all up}}{\underbrace{\sum _{i=1}^n{x}_i}}$$

• $x_i$ — $i$-va measured number ($i$, $1$ se $n$ tak jaata hai).
• $n$ — humne kitni cheezein uthayin (sample size).
• $\bar{x}$ — "x-bar" padho, is ek mutthi ka average.

KYUN. $\bar{x}$ invisible $\mu$ ke liye humara sabse accha single guess hai. Lekin yeh sirf ek mutthi se ek guess hai.

PICTURE. Figure dekho: faint cloud poori population hai. Char colored dots char alag-alag mutthiyaan hain jo hum draw kar sakte the — har ek alag $\bar{x}$ deti hai. Sachcha $\mu$ (yellow line) fixed baitha hai jabki hamare guesses uske around scatter karte hain.

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Step 2 — Grab hazaron baar repeat karo: averages ki ek distribution

KYA HAI. Socho ek mutthi uthao, $\bar{x}$ likh lo, aur hazaron baar repeat karo. Woh $\bar{x}$ values apna ek bell-shaped histogram banate hain, jise sampling distribution of the mean kehte hain.

KYUN. Hum ek $\bar{x}$ ko control nahi kar sakte, lekin hum describe kar sakte hain ki $\bar{x}$ ki poori family kaisi behave karti hai. Family ka prediction karna hume kisi bhi ek member ke baare mein reason karne deta hai.

Central Limit Theorem promise karta hai ki decent-sized $n$ ke liye yeh histogram ek normal (bell) curve hai:

$$\bar{X}\sim N\left(\underset{\text{centred on truth}}{\underbrace{\mu }},\underset{\text{spread shrinks with }n}{\underbrace{\frac{{\sigma }^2}{n}}}\right)$$

• $N(\text{centre},\text{spread})$ — us centre aur us spread wali bell curve.
• ${\sigma }^2$ — ek raw measurement ka variance (spread${}^2$).
• ${\sigma }^2/n$ — average ka variance; bada $n$ ⇒ narrow bell.

PICTURE. Bell exactly $\mu$ par centred hai (shifted nahi!) aur iski width Standard Error $\sigma /\sqrt{n}$ hai. Dekho kaise $n$ double karne se bell narrow ho jaati hai.

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Step 3 — Ek universal ruler par slide aur stretch karo (standardize)

KYA HAI. Har normal bell ka alag centre $\mu$ aur width $SE$ hoti hai. Hum ek master bell mein convert karte hain — $0$ par centred, width $1$ — centre subtract karke aur width se divide karke:

$$Z=\frac{\stackrel{\text{how far from truth}}{\overbrace{\bar{X}-\mu }}}{\underset{\text{measured in SE units}}{\underbrace{\sigma /\sqrt{n}}}}\sim N(0,1)$$

• $Z$ — wohi quantity $\bar{X}$, lekin "$\mu$ se kitne standard errors door" mein relabel ki gayi.
• Neeche ka $\sigma /\sqrt{n}$ Step 2 ka $SE$ hai.

KYUN. Master bell par hum exactly jaante hain ki kaunse points kisi bhi tail area ko cut off karte hain — woh numbers fixed constants hain, kuch aisa nahi jo hum har problem mein recompute karein. Standardize karna "har problem alag hai" ko "ek ruler sab ke liye fit" mein badal deta hai.

PICTURE. Left: teen alag real bells. Right: shift-and-squeeze ke baad woh sab ek standard bell par collapse ho jaati hain. $x$-axis ab $SE$ units mein label hai.

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Step 4 — Master bell ka beech wala $(1-\alpha )$ hissa nikalo

KYA HAI. Ek confidence level choose karo, maano $95\%$. Bacha hua $\alpha =0.05$ rakho (woh $5\%$ jo hum miss karne ke liye ready hain). Ise $\alpha /2=0.025$ ki do barabar tails mein baanto. Master bell par woh point jo right tail mein $2.5\%$ chhodta hai use $z_{\alpha /2}$ kehte hain.

$$P\left(-z_{\alpha /2}\le Z\le z_{\alpha /2}\right)=1-\alpha$$

• $\alpha$ — total tail probability jo hum sacrifice karte hain (miss rate).
• $\alpha /2$ — uska aadha, ek har side ke liye.
• $z_{\alpha /2}$ — woh ruler mark jo us tail ko cut off karta hai; $95\%$ ke liye yeh $1.96$ hai.

Barabar kyun baanto? Hamare paas over-guessing se zyada under-guessing se darne ka koi reason nahi hai, isliye hum dono tails ko symmetrically treat karte hain. Yahi two-tailed logic Hypothesis Testing mein use hoti hai.

PICTURE. Shaded green beech mein saare $Z$ values ka $95\%$ hai; do red tails mein $2.5\%$ each hai. $\pm 1.96$ par vertical marks $\pm z_{\alpha /2}$ hain.

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Step 5 — Un-standardize: statement ko andar-bahar palto

KYA HAI. Ab hum $\mu$ ko free karne ke liye Step 3 ko algebraically undo karte hain. Probability ke andar ki inequality se shuru karo aur har part ko $SE=\sigma /\sqrt{n}$ se multiply karo:

$$-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le \bar{X}-\mu \le z_{\alpha /2}\frac{\sigma }{\sqrt{n}}$$

$\bar{X}$ subtract karo, phir $-1$ se multiply karo (jo dono $\le$ signs ko flip karta hai):

$$\boxed{\bar{X}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le \mu \le \bar{X}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}}$$

• Left/right edges — random hain, kyunki inke saath $\bar{X}$ hai.
• Beech mein $\mu$ — fixed hai, sirf do hilti hui walls ke beech trap hua hai.

KYUN. Rearrange karne se $\mu$ (jo hum chahte hain) akela beech mein aa jaata hai, toh picture ban jaati hai "ek moving bracket jo ya toh ek ठुका हुआ target ko pakad leti hai ya miss kar deti hai."

PICTURE. Yellow $\mu$ ek fixed vertical line hai. Blue brackets das repeated experiments hain; har bracket $\bar{x}\pm z_{\alpha /2}SE$ wide hai. Lagbhag das mein se nau brackets yellow line ko straddle karte hain — daswan (red) miss kar deta hai. Brackets hilte hain; line nahi hilti.

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Step 6 — Edge case: hum kabhi actually $\sigma$ jaante nahi

KYA HAI. Real life mein population spread $\sigma$ unknown hota hai. Hum ise usi sample se estimate karte hain:

$$s=\sqrt{\frac{1}{n-1}\sum _{i=1}^n(x_i-\bar{x}{)}^2}$$

• $n-1$ ($n$ nahi) — Bessel's correction; ek degree of freedom $\bar{x}$ compute karne mein kharch ho gayi, isliye hum ek kam se divide karte hain.

$s$ ko $\sigma$ ki jagah rakhne se extra randomness aati hai, isliye master bell ab kaafi nahi hai. Hum ek thodi si moti cousin use karte hain, t-distribution with $df=n-1$:

$$T=\frac{\bar{X}-\mu }{s/\sqrt{n}}\sim t_{n-1}$$

KYUN moti curve? Kyunki $s$ true spread ko under-guess kar sakta hai, isliye extreme $T$ values zyada baar hote hain. Mote tails ⇒ wider critical marks ⇒ honestly wider intervals. Jaise $n\to \infty$, t-curve wapas normal mein melt ho jaati hai.

PICTURE. Overlay: normal (blue) aur $df=3$ wali $t$ (red). Same peak, lekin red ke tails thicker hain aur uska $2.5\%$ mark zyada door hai — woh extra distance $\sigma$ na jaanne ki price hai.

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Step 7 — Degenerate limits: picture kya tod deta hai

KYA AUR KYUN. Knobs ko extremes par push karo aur dekho interval kaise respond karta hai — aise tum kisi bhi use ko sanity-check karte ho.

• $n=1$: $SE=\sigma /\sqrt{1}=\sigma$ aur $s$ undefined hai ($n-1=0$ se divide ho raha hai). Ek data point spread ke baare mein kuch nahi batata → koi interval nahi.
• $n\to \infty$: $SE=\sigma /\sqrt{n}\to 0$, bracket $\bar{x}=\mu$ par collapse ho jaata hai. Infinite data ⇒ zero uncertainty.
• Confidence $\to 100\%$ ($\alpha \to 0$): $z_{\alpha /2}\to \infty$, bracket $(-\infty ,\infty )$ tak wide ho jaata hai. "Certain" tabhi jab tum sab kuch claim karo.
• Skewed data / heavy outliers: Step 2 mein CLT bell ek jhooth hai; normal shortcut fail hoti hai aur tumhe Bootstrap Methods ya Bayesian Credible Intervals ki taraf jaana chahiye.

PICTURE. Margin of error $E$ ko $n$ ke against plot kiya — ek $1/\sqrt{n}$ curve. Pehle steep drop, phir flat: zyada data ke "diminishing returns." Dashed asymptote $E=0$ par baitha hai.

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Ek-picture summary

Sab ek saath: raw scatter → averages ki bell → standard ruler → carved middle → $\mu$ ke around un-flipped bracket, aur $t$ use karke jab $\sigma$ guess kiya ho.

Recall Feynman retelling — ek 12-saal ke bachche ko batao

Mujhe ek giant pile ka asli average $\mu$ chahiye, lekin mujhe sirf ek chhoti si mutthi milti hai, toh main us mutthi ka average $\bar{x}$ leta hoon. Agar main sochu ki mutthiyaan baar baar uthata rehta hoon, toh unke averages ek bell mein pile ho jaate hain jo sache $\mu$ par centred hai, aur bell ki width $\sigma /\sqrt{n}$ hai — zyada data, patli bell. Main us bell ko ek universal bell par slide-aur-squeeze karta hoon jo zero par centred hai, jahaan mujhe woh do marks ($\pm 1.96$ par) pata hain jo beech ke $95\%$ ko fence karte hain. Main un marks ko real units mein stretch karta hoon aur rearrange karta hoon jab tak $\mu$ akela beech mein na aa jaaye, ek bracket $\bar{x}\pm 1.96\sigma /\sqrt{n}$ se trap hua. Bracket sample se sample tak hilta hai; $\mu$ fixed rehta hai; har $100$ brackets mein se lagbhag $95$ use pakad lete hain. Kyunki main sachchi $\sigma$ kabhi nahi jaanta, main ise $s$ se estimate karta hoon ($n-1$ se divide karke), aur us extra guessing ki price chukane ke liye main thode mote $t$-marks use karta hoon — jo quietly wapas normal marks ban jaate hain jab mere paas bahut saara data ho.

Recall Quick self-test

$SE$ ko $n$ se nahi balki $\sqrt{n}$ se kyun divide karte hain? ::: Averaging variance ko $n$ se cut karta hai; spread ($SE$) variance ka square root hai, isliye yeh $\sqrt{n}$ se drop karta hai. $[167,173]$ compute karne ke baad, kya "random" hai? ::: Dono endpoints (unke saath $\bar{x}$ hai); sach wala $\mu$ fixed hai. $z\to t$ kab switch karna zaroori hai? ::: Jab $\sigma$ unknown ho aur $n$ chhhota ho ($<30$); $t$ ke fat tails $\sigma$ ko $s$ se estimate karne ki price hain. $\alpha \to 0$ hone par interval ka kya hota hai? ::: $z_{\alpha /2}\to \infty$, toh interval $(-\infty ,\infty )$ tak blow up ho jaata hai.

Related tools jo yahi machinery reuse karte hain: A/B Testing, Cross-validation, Hypothesis Testing.