Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Visual walkthrough

4.9.6 · D2 · Maths › Probability Theory & Statistics › Common discrete distributions — Bernoulli, Binomial, Poisson

Step 1 se pehle, aao is kahaani ke chaar characters par agree kar lein. Agar koi unfamiliar lage, theek hai — har ek ko ek picture milegi.

Step 1 — Minute ko $n$ tiny slots mein kaato

KYA. Hum ek continuous minute lete hain (call kisi bhi instant par aa sakti hai) aur pretend karte hain ki yeh actually $n$ alag, barabar chhote time-slots hain. Har slot mein, zyada se zyada ek call hoti hai: ya toh call (success, chance $p$ ) ya khamoshi (failure, chance $q = 1 - p$ ). Yeh exactly ek Bernoulli trial hai har slot ke liye.

KYUN. Ek Bernoulli flip hi woh atom hai jiske saath hum compute kar sakte hain. Time continuous aur daraauna hai; $n$ independent flips ki ek row friendly hai. Toh hum minute ko $n$ flips se approximate karte hain aur promise karte hain ki error baad mein $n \to \infty$ lekar theek karenge.

PICTURE. Red band hamara single minute hai. Neeche, wohi minute $n = 12$ slots mein kata hua — har ek ek coin flip hai.

Figure — Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial

Agar exactly $k$ slots mein call hai, toh humein $n$ flips mein $k$ successes mile hain — yeh parent note ka Binomial hai.

Step 2 — Binomial probability likho

KYA. $n$ independent flips mein exactly $k$ successes ki chance Binomial formula hai: $P (X = k) = which slots (k n) k calls p^{k} n - k silences (1 - p)^{n - k} .$

Term by term, wahan jo baithe hain:

$(k n)$ — count karta hai ki $n$ slots mein se kaun se $k$ slots mein calls hain (order matter nahi karta, toh hum "choose" karte hain).
$p^{k}$ — $k$ successful slots mein se har ek ka factor $p$ hai; multiply isliye kyunki slots independent hain.
$(1 - p)^{n - k}$ — baaki $n - k$ silent slots mein se har ek ka factor $q = 1 - p$ hai.

KYUN. Yeh finite $n$ ka exact answer hai. Baki page par hamara kaam yeh dekhna hai ki yeh kya converge karta hai.

PICTURE. Ek specific arrangement: 3 red (call) slots, 9 black (silent) slots. Aisi $(3 12)$ arrangements hain.

Step 3 — Average ko fixed raho: $p = λ / n$ substitute karo

KYA. Hum ab woh ek rule impose karte hain jo limit ko meaningful banata hai: calls ki average number per minute, $n p$ , ek fixed number $λ$ ke barabar honi chahiye. $n p = λ$ solve karne par $p = λ / n$ milta hai. Ise substitute karo: $P (X = k) = (k n) (\frac{λ}{n})^{k} (1 - \frac{λ}{n})^{n - k} .$

KYUN. Agar hum sirf $p \to 0$ jaane dein bina $n p$ pin kiye, toh saari calls gayab ho jaati hain aur answer boringly $P (X = 0) = 1$ ho jaata hai. Agar $p$ fixed rahe jab $n \to \infty$ , toh calls infinity tak blast ho jaati hain. $n p = λ$ hold karna Goldilocks path hai: bahut saare slots, har ek mein call ki tiny chance, lekin ek steady average.

PICTURE. Jab hum $n$ double karte hain (finer slots), har red slot chhota ho jaata hai (smaller $p$ ), lekin total red length — expected number of calls $n p$ — exactly wohi $λ$ rehti hai.

Step 4 — "Choose" term ko kholo

KYA. $(k n) = \frac{n !}{k ! ( n - k )!}$ expand karo aur $(\frac{λ}{n})^{k} = \frac{λ ^{k}}{n ^{k}}$ ko alag karo. Teen tidy chunks mein regroup karo: $P (X = k) = \frac{λ ^{k}}{k !} \cdot chunk A \frac{n !}{( n - k )! n ^{k}} \cdot chunk B (1 - \frac{λ}{n})^{n} \cdot chunk C (1 - \frac{λ}{n})^{- k} .$

Har piece kahan se aayi:

$\frac{λ ^{k}}{k !}$ — $λ^{k}$ $p^{k}$ se aaya hai; $\frac{1}{k !}$ $(k n)$ ka leftover hai $n$ -parts nikalne ke baad. Is chunk mein koi $n$ nahi hai — yeh limit se guzar kar bachi rehti hai.
Chunk $A = \frac{n !}{( n - k )! n ^{k}}$ — binomial coefficient ka " $n$ -flavoured" leftover.
Chunk $B = (1 - \frac{λ}{n})^{n}$ — silence term ka bulk.
Chunk $C = (1 - \frac{λ}{n})^{- k}$ — chhoti correction kyunki exponent $n - k$ tha, $n$ nahi.

KYUN. Hum algebra alag karte hain taki har chunk ka ek clean, known limit ho. Divide and conquer: har chunk ko apne step mein solve karo.

PICTURE. Teen chunks as teen dials, har ek $n \to \infty$ hone par turn hone wali. Sirf red dial ( $B$ ) ek non-trivial jagah jaati hai.

Step 5 — Chunk $A \to 1$ : the falling factorial

KYA. Chunk $A$ ko dhyan se dekho: $\frac{n !}{( n - k )! n ^{k}} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - k + 1 )}{n ^{k}} = \frac{n}{n} \cdot \frac{n - 1}{n} \dots \frac{n - k + 1}{n} .$ Exactly $k$ fractions hain, aur $n \to \infty$ par har ek $\to 1$ jaata hai ( $- 1, - 2, \dots$ huge $n$ ke saamne negligible ho jaate hain). $k$ ones ka product $1$ hai: $\frac{n !}{( n - k )! n ^{k}} ⟶ 1.$

Term by term:

$n! / (n - k)!$ — yeh top $k$ integers $n, n - 1, \dots, n - k + 1$ ka product hai (baaki sab cancel ho jaate hain).
$n^{k}$ — un $k$ terms mein se har ek ke liye ek factor $n$ , neeche baithe hain.
Har ratio $\frac{n - j}{n} = 1 - \frac{j}{n} \to 1$ kyunki fixed $j$ ke liye $\frac{j}{n} \to 0$ .

KYUN. $k$ ek fixed chhoti number hai (jaise 5), jabki $n$ infinity ki taraf race karta hai. Ek enormous number se fixed amount subtract karna proportionally kuch nahi badalta.

PICTURE. $k$ ratios plotted jab $n$ badhta hai — sab height $1$ par red horizontal line ki taraf climb karte hain.

Step 6 — Chunk $B \to e^{- λ}$ : exponential aata hai

KYA. Chunk $B = (1 - \frac{λ}{n})^{n}$ show ka star hai. Iska limit exponential ki definition hai: $(1 - \frac{λ}{n})^{n} ⟶ e^{- λ} .$ Yeh standard limit $\left(1+\frac{x}{n}\right)^n \to e^{x}$ hai $x = - λ$ ke saath.

KYUN yahi tool aur koi nahi? Hum ek number jo $1$ se thoda sa kam hai use $n$ times khud se multiply kar rahe hain, jabki $n \to \infty$ . Do runaway effects ladte hain: base $\to 1$ (jo $1$ deta) aur exponent $\to \infty$ (jo $0$ deta). Koi bhi outright nahi jeetta — unke beech ka tension ek precise finite number par land karta hai, aur woh number $e^{- λ}$ define hota hai. Koi bhi aur elementary function "shrinking base to the power of exploding exponent" capture nahi karta; exponential exactly is sawaal ka jawab dene ke liye exist karta hai.

PICTURE. $(1 - \frac{λ}{n})^{n}$ ki value (red curve) $n$ ke against $λ = 3$ ke liye plotted, dashed line $e^{- 3} \approx 0.0498$ par flat ho rahi hai.

Step 7 — Chunk $C \to 1$ , aur answer assemble karo

KYA. Chunk $C = (1 - \frac{λ}{n})^{- k}$ . Andar, $\frac{λ}{n} \to 0$ , toh base $\to 1$ ; ek fixed power $- k$ tak raise hone par, yeh $\to 1^{- k} = 1$ rehta hai. Ab Steps 4–6 ke surviving pieces multiply karo: $P (X = k) = \frac{λ ^{k}}{k !} \cdot A 1 \cdot B e^{- λ} \cdot C 1 = \frac{λ ^{k} e ^{- λ}}{k !} .$

Final formula term by term padhna:

$λ^{k}$ — expected count raise to the number jo humne poocha; zyada calls poocha $\Rightarrow$ higher power.
$e^{- λ}$ — "khaali slots mein kuch nahi hua" factor, limit ka fingerprint.
$\frac{1}{k !}$ — is fact ko divide out karta hai ki $k$ calls time mein indistinguishable hain ( $(k n)$ ka survivor).

KYUN yeh legitimate hai. Limits ka finite product limit of product ke barabar hota hai (har chunk converge hota hai), toh hum chunk-by-chunk limits le sakte hain aur multiply kar sakte hain.

PICTURE. $λ = 3$ ke liye finished Poisson bars: poora probability mass, $k$ ki har value ke liye ek red bar.

Step 8 — Edge aur degenerate cases (koi gap mat chhodho)

KYA & KYUN & PICTURE, teen cases:

Worked example Teen scenarios jinhe formula survive karna chahiye

(i) $k = 0$ (kuch nahi hota). $P (X = 0) = \frac{λ ^{0} e ^{- λ}}{0 !} = e^{- λ}$ . Yeh pure chunk $B$ hai: koi calls nahi, toh sirf "saare slots silent" factor bachta hai. Sanity check: $λ = 0$ par (impossible event), $P (X = 0) = e^{0} = 1$ — certainly kuch nahi hota. ✓

(ii) $λ \to 0$ (vanishingly rare). $k \geq 1$ wala har term ek $λ^{k} \to 0$ carry karta hai, toh saara mass $k = 0$ par pile ho jaata hai. Distribution collapse ho jaata hai "always zero" par — correct hai aise event ke liye jo kabhi hota hi nahi.

(iii) Kya probabilities 1 tak sum hoti hain? Honi chahiye, ya phir yeh distribution nahi hai: $\sum_{k = 0}^{\infty} \frac{λ ^{k} e ^{- λ}}{k !} = e^{- λ} \sum_{k = 0}^{\infty} \frac{λ ^{k}}{k !} = e^{- λ} e^{λ} = 1.$ Yahan $\sum_{k \geq 0} \frac{λ ^{k}}{k !} = e^{λ}$ exponential ki Taylor series hai — wohi $e$ jo Step 6 mein aaya tha, ab total probability $1$ guarantee karta hai. ✓

Ek-picture summary

Poori derivation ek single flow mein: time kaato → Binomial → $n p = λ$ pin karo → teen chunks → $1 \times e^{- λ} \times 1$ → Poisson. Ek non-trivial limit (red) exponential hai.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Ek minute imagine karo jahan phone ring kar sakta hai. Main "kisi bhi instant" handle nahi kar sakta, toh minute ko $n$ tiny slots mein kaatta hoon aur kehta hoon har slot ek coin flip hai — ring ya khamoshi. Ek slot mein ring ki chance $p$ hai, aur main insist karta hoon ki rings ki average number, $n p$ , ek fixed $λ$ ke barabar ho (maano 3). Jab main finer aur finer kaatta hoon, har slot ki ring-chance $p = λ / n$ ghatti hai, lekin total expected rings kabhi nahi badliti. Exact Binomial answer teen pieces mein split hota hai. Unme se do lazy hain — woh bas $1$ ki taraf drift karte hain jab slots multiply hote hain. Teesra piece, "saare silent slots ek saath multiply," interesting wala hai: $1$ se thoda sa kam number khud se infinitely baar multiply hota hai, exactly $e^{- λ}$ par land karta hai. Surviving pieces multiply karo — $\frac{λ ^{k}}{k !}$ , times $1$ , times $e^{- λ}$ , times $1$ — aur nikal aata hai $P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !}$ . $k = 0$ par check karo: sirf $e^{- λ}$ , kuch ring na hone ki chance. Saare $k$ add karo: exponential series $e^{λ}$ rebuild karta hai, cancel ho jaata hai, $1$ deta hai. Perfect. Yahi Poisson hai: infinitely many microscopic coin flips, sirf ek number $λ$ se yaad rahe.

Recall Quick self-test

Hum $n p = λ$ kyun fixed rakhein, sirf $p \to 0$ kyun na jaane dein? ::: Warna saari rings gayab ho jaati hain aur $P (X = 0) \to 1$ ; $n p = λ$ fix karna ek steady average rakhta hai toh limit meaningful hoti hai. Teen chunks mein se kaun sa exponential deta hai, aur uska limit kya hai? ::: Chunk $B = (1 - λ / n)^{n} \to e^{- λ}$ . Poisson ke liye $P (X = 0)$ kya hai? ::: $e^{- λ}$ . Chunk $A = \frac{n !}{( n - k )! n ^{k}}$ ko 1 par kyun bheja ja sakta hai? ::: Yeh $k$ ratios $\frac{n - j}{n} \to 1$ ka product hai fixed $j$ ke liye, aur $k$ fixed hai.