Binomial distribution — PMF, mean, variance: Visual walkthrough

2.7.13 · D2 · Maths › Statistics & Probability — Intermediate › Binomial distribution — PMF, mean, variance

Step 1 — Ek trial: sabse chhota possible experiment

KYA. $n$ coins flip karne se pehle, ek ko dekho. Ek single trial ke paas exactly do raaste hain jinse woh guzar sakta hai: success (hum ise teal colour denge) ya failure (plum). Hum success ki chance ko $p$ kehte hain, jo $0$ aur $1$ ke beech ki ek number hai. Kyunki yeh do raaste hi sirf do raaste hain, failure ki chance wahi hogi jo bacha hua hai: $1 - p$ . Hum us bacha hue hisse ko apna ek naam dete hain, $q = 1 - p$ , taaki hume baar baar "one minus p" na likhna pade.

KYUN. Ek binomial experiment bas yahi ek trial hai, repeat kiya hua. Agar hum ek flip ko theek se describe nahi kar sakte, toh aage ki poori cheez ret ki neenv par bani hai. Yeh single-trial object Bernoulli distribution hai.

PICTURE. Do boxes jinki widths probabilities hain. Dhyan do ki widths zaroor poori strip bharti hain — yahi statement hai $p + q = 1$ .

Figure — Binomial distribution — PMF, mean, variance

Step 2 — Ek specific sequence: path ke saath multiply karo

KYA. Ab coin ko $n$ baar flip karo. Ek particular story likho — maano pehle $k$ flips successes ( $S$ ) hain aur baaki $n - k$ failures ( $F$ ) hain:

k successes S S \dots S n - k failures F F \dots F

Kyunki flips ek doosre ko influence nahi karte, poori story ki probability har step ki probabilities ka product hai. Har $S$ ek factor $p$ contribute karta hai; har $F$ ek factor $q$ contribute karta hai:

P (yeh exact story) = k copies of p p \cdot p \dots p \cdot n - k copies of q q \cdot q \dots q = p^{k} q^{n - k}

$p^{k}$ ::: $k$ successes, har ek $p$ ka ek factor multiply karta hua.
$q^{n - k}$ ::: baaki $n - k$ trials fail hue, har ek $q$ ka ek factor multiply karta hua.

KYUN multiply karo, aur hum allowed kyun hain? Probabilities ka multiplication exactly wahi hai jo independence ka matlab hai: "A aur B ki chance, A ki chance times B ki chance hai." Yahi BINS mein I hai. No independence → no multiplying → no clean formula.

PICTURE. Ek tree mein ek path, har edge par $p$ ya $q$ label kiya hua; path ki probability un edge labels ka product hai jinhe woh cross karta hai.

Step 3 — Stories count karo: $(k n)$ kahan se aata hai

KYA. $k$ successes wali har arrangement — chahe woh successes kahaan bhi hon — ki identical probability $p^{k} q^{n - k}$ hai (same number of $p$ 's, same number of $q$ 's, bas shuffle hua). Toh $P (X = k)$ woh single probability hai jo aisi kitni arrangements exist karti hain se multiply hoti hai. Unhe count karne ka matlab hai: $n$ slots mein se choose karo ki kaunse $k$ mein success hai. Woh count $(k n)$ likha jaata hai, padha jaata hai " $n$ choose $k$ ":

(k n) = \frac{n !}{k ! ( n - k )!}

$n!$ ::: $n$ distinct slots ki sabhi orderings.
$k!$ ::: successes ke beech mein reorderings divide out karta hai (hamein parwah nahi ki kaunsa success "pehla" hai).
$(n - k)!$ ::: usi wajah se failures ke beech reorderings divide out karta hai.

KYUN yeh tool aur koi nahi? Humne ek pure counting question poocha — "kitne tareekon se?" — jisme order matter nahi karta. Mathematical object jo precisely " $n$ mein se $k$ cheezein choose karo, order irrelevant" answer karne ke liye bana hai, woh binomial coefficient hai. Kuch aur use karna (jaise $n^{k}$ , jo order care karta hai) over- ya under-count karta.

PICTURE. $n = 3, k = 2$ ke liye hum saari stories likhte hain. Exactly $(2 3) = 3$ mein do successes hain; har ek ka weight $p^{2} q^{1}$ hai.

Step 4 — Poori distribution, aur check ki sum 1 hoti hai

KYA. $k$ ko $0$ se $n$ tak sweep karo aur successes ki har possible number ke liye ek bar milta hai. Saare bars stack karo: unki heights exactly $1$ add honi chahiye, kyunki koi na koi number of successes definitely hota hai.

KYUN yeh kaam karta hai. Sum

k = 0 \sum n (k n) p^{k} q^{n - k}

literally Binomial Theorem se $(p + q)^{n}$ ka expansion hai — woh theorem jo kehta hai $(a + b)^{n} = \sum_{k} (k n) a^{k} b^{n - k}$ . Yahan $a = p$ , $b = q$ , aur kyunki $p + q = 1$ (Step 1!), hume milta hai

(p + q)^{n} = 1^{n} = 1. ✓

Do building blocks — width picture $p + q = 1$ aur counting picture $(k n)$ — perfectly snap together hote hain.

PICTURE. $n = 5, p = 0.5$ ke liye full bar chart: beech ke kareeb peak karta ek symmetric hump, saare bars $1$ sum karte hue.

Step 5 — Mean, ek balance point ki tarah drawn

KYA. Mean $E [X]$ us bar chart ka balance point hai — woh jagah jahan agar bars ek plank par weights hote, toh woh level tip hota. Hum claim karte hain ki yeh $n p$ par baitha hai.

KYUN slick proof brute force se behtar hai. Ugly sum $\sum_{k} k (k n) p^{k} q^{n - k}$ ki jagah, total count ko chhote pieces mein split karo. Maano $X_{i} = 1$ agar trial $i$ succeed kare, warna $0$ — ek indicator. Tab

X = X_{1} + X_{2} + \dots + X_{n}

Har indicator ka average easy hai:

E [X_{i}] = 1 \cdot p + 0 \cdot q = p

$1 \cdot p$ ::: value $1$ (success) times uski chance $p$ .
$0 \cdot q$ ::: value $0$ kuch contribute nahi karta.

Ab Linearity of expectation — rule "sum ka average, averages ka sum hota hai," jo tab bhi hold karta hai jab pieces interact karein — deta hai

E [X] = n terms p + p + \dots + p = n p

PICTURE. Bar chart jisme $x = n p$ par ek triangular fulcrum rakha hua hai; plank balance karta hai.

Step 6 — Variance, aur kyun ise independence chahiye

KYA. Variance spread measure karta hai — bars typically balance point se kitni door baithe hain. Hum claim karte hain ki yeh $n p (1 - p) = n pq$ ke equal hai.

KYUN. Pehle ek indicator ka spread nikalo. Kyunki $X_{i}$ sirf $0$ ya $1$ hota hai, squaring kuch nahi badlata: $0^{2} = 0$ , $1^{2} = 1$ , toh $X_{i}^{2} = X_{i}$ aur $E [X_{i}^{2}] = p$ . Tab

Var (X_{i}) = E [X_{i}^{2}] - (E [X_{i}])^{2} = p - p^{2} = p (1 - p) = pq

$E [X_{i}^{2}] = p$ ::: shortcut $X_{i}^{2} = X_{i}$ se.
$(E [X_{i}])^{2} = p^{2}$ ::: mean, squared.

Ab pieces ke variances add karna sirf tab legal hai jab trials independent hon — woh BINS mein I phir se hai, aur yahi hai jo covariance terms ko vanish karta hai. Independence ke saath:

Var (X) = n terms pq + pq + \dots + pq = n p (1 - p)

PICTURE. Do bar charts side by side: $p = 0.5$ (sabse wide hump — maximum spread) versus $p = 0.9$ (skewed, narrow — chhota spread). Hump ki width hi variance hai.

Step 7 — Degenerate cases: pictures tab bhi kaam karti hain

KYA. $p$ aur $k$ ko edges tak push karo aur confirm karo ki kuch toot nahi raha.

$p = 0$ (kabhi succeed nahi karta). $P (X = 0) = (0 n) 0^{0} 1^{n} = 1$ ; baaki har bar $0$ hai. Mean $= n \cdot 0 = 0$ , variance $= n \cdot 0 \cdot 1 = 0$ . $k = 0$ par height $1$ ka ek single bar — ek certainty, koi spread nahi.
$p = 1$ (hamesha succeed karta hai). Mirror image: $k = n$ par height $1$ ka ek bar. Mean $= n$ , variance $= 0$ .
General mein $k = 0$ . $(0 n) = 1$ toh $P (X = 0) = q^{n}$ — saare trials fail hone chahiye, sirf ek story. Yeh exactly "at least one" complement engine hai: $P (X \geq 1) = 1 - q^{n}$ .
$n = 1$ . $(k 1)$ sirf do bars deta hai: $k = 0$ par $q$ , $k = 1$ par $p$ . Mean $p$ , variance $pq$ — hum Step 1 ke single Bernoulli distribution par wapas aa gaye.

KYUN yeh dikhao. Ek formula jis par tumhara trust sirf "nice middle" mein hai, woh formula nahi hai jis par tum trust karte ho. Edges wahan hain jahan galtiyan chhupti hain — yahan sab ek single spike (ek sure thing) mein ya ek Bernoulli trial par wapas reduce ho jaate hain.

PICTURE. Teen chhote charts: $p = 0$ spike, $p = 1$ spike, aur $n = 1$ do-bar Bernoulli.

Ek-picture summary

Is page ki poori cheez ek single frame mein: ek trial $p$ aur $q$ mein split hoti hai (Step 1) → ek story ke saath multiply karke $p^{k} q^{n - k}$ milta hai (Step 2) → $(k n)$ se stories count karo (Step 3) → bars $(p + q)^{n}$ ke zariye $1$ sum hote hain (Step 4) → woh $n p$ par balance karte hain (Step 5) → woh $n pq$ se spread hote hain (Step 6).

Recall Feynman retelling — ek dost ko batao

Socho ek thoda tedha coin $n$ baar flip kar rahe ho. Ek flip ke do outcomes hain jinki chances $p$ aur $q$ ek strip ko end to end bharte hain — woh ek mein add hote hain. Flips ka ek pura run ek tree mein ek path hai; kyunki flips ek doosre se baat nahi karte, tum path ke saath chhoti chances multiply karte ho, ek run ke liye $p^{k} q^{n - k}$ milta hai jisme $k$ heads hain. Lekin bahut saare alag runs mein same number of heads hote hain — unhe tum " $n$ choose $k$ " se count karte ho, jo bas yeh hai ki " $n$ slots mein $k$ heads kitne tareekon se rakh sakta hoon." Count ko per-run weight se multiply karo aur tumhare paas poora bar chart hai. Woh bars ek mein sum hone chahiye, aur hote hain, kyunki unhe sum karna secretly $(p + q)^{n} = 1^{n} = 1$ expand karna hai. Chart kahan balance karta hai? Har flip apna average $p$ se khud akele upar khichta hai, toh $n$ flips $n p$ dete hain — yahi mean hai, aur isse koi farak nahi padta ki flips interact karte hain ya nahi. Hump kitna wide hai? Har flip $p (1 - p)$ se wobble karta hai, aur kyunki woh independent hain tumhe un wobbles ko simply add karne ki permission hai: $n p (1 - p)$ . $p$ ko $0$ ya $1$ set karo aur wobble mar jaata hai — ek sure thing mein ek tall bar hota hai aur koi spread nahi.

Connections

Bernoulli distribution — Step 1 aur $n = 1$ degenerate case hi Bernoulli hain.
Binomial Theorem — Step 4 mein "bars 1 mein sum hote hain" ke peeche ka engine.
Linearity of expectation — woh shortcut jo Step 5 mein $n p$ deta hai.
Variance and covariance — kyun independence Step 6 mein wobbles ko add hone deta hai.
Poisson distribution — bar chart kya ban jaata hai jab $n \to \infty, p \to 0, n p$ fixed ho.
Normal approximation to binomial — woh smooth curve jiske paas hump large $n$ ke liye approach karta hai.
Hypergeometric distribution — without-replacement cousin jahan Step 2 ka multiplying toot jaata hai.