Binomial distribution — PMF, mean, variance

2.7.13 · Maths › Statistics & Probability — Intermediate

The four conditions (BINS)

PMF ko scratch se derive karna

HOW — pehle ek specific outcome banao. Maano pehle $k$ trials succeed karte hain aur baaki fail:

k S S \dots S n - k F F \dots F

Kyunki trials independent hain, probabilities ko multiply karo:

P (this exact sequence) = k p \cdot p \dots p \cdot n - k (1 - p) \dots (1 - p) = p^{k} (1 - p)^{n - k} .

Yeh step kyun? Independence ki wajah se ek joint event ki probability ek product mein factor ho jaati hai.

Ab arrangements count karo. Koi bhi sequence jisme $k$ successes hain (kisi bhi position mein) usi probability $p^{k} (1 - p)^{n - k}$ ki hogi. Aise sequences ki count = yeh choose karne ke tarike ki $n$ slots mein se kaun se $k$ slots successes hain:

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

Yeh step kyun? Humein sirf count $k$ ki parwah hai, order ki nahi, isliye hum sab equally-likely orderings pe sum karte hain.

Figure — Binomial distribution — PMF, mean, variance

Mean: slick tarika (indicator trick)

HOW. $X = X_{1} + X_{2} + \dots + X_{n}$ likho, jahan $X_{i} = 1$ agar trial $i$ succeed kare, warna $0$ . Yeh indicator (Bernoulli) variables hain.

Ek indicator ke liye:

E [X_{i}] = 1 \cdot p + 0 \cdot (1 - p) = p .

Yeh step kyun? Expectation $\sum (value) \times (prob)$ hai; sirf "1·p" wala term bachta hai.

Linearity of expectation se (bina independence ke bhi kaam karta hai):

E [X] = i = 1 \sum n E [X_{i}] = n p .

Variance: independence chahiye

HOW. Ek indicator ke liye, $X_{i}^{2} = X_{i}$ (kyunki $0^{2} = 0, 1^{2} = 1$ ), isliye $E [X_{i}^{2}] = p$ . Tab:

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

Yeh step kyun? Definition $Var = E [X^{2}] - (E [X])^{2}$ ; identity $X_{i}^{2} = X_{i}$ key shortcut hai.

Independence ⇒ variances add hoti hain:

Var (X) = i = 1 \sum n Var (X_{i}) = n p (1 - p) .

Worked examples

Common mistakes

Active recall

Recall Reveal se pehle answer karo

Q: Mean mein $(1 - p)$ kyun nahi hai lekin variance mein hai? Mean linearity se aata hai: har trial average mein $p$ add karta hai → $n p$ . Ek Bernoulli ka variance $p - p^{2} = p (1 - p)$ hota hai; independence se yeh add ho jaate hain → $n p (1 - p)$ . Extra $q$ measure karta hai ki outcomes kitne spread out hain, $p = 0.5$ par maximum hota hai.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum 10 thumbtacks uchalte ho. Har ek same chance se point-up land karta hai. "Kitne point-up?" ek binomial question hai. Average mein tumhe (chance × tacks ki number) point-ups milte hain — woh $n p$ hai. Ek single tack ka up hona doosron ko change nahi karta, isliye hum har tack ki chhoti si "wiggle" of uncertainty ko simply add kar sakte hain total wiggle paane ke liye — woh variance $n p (1 - p)$ hai. $(k n)$ bas yeh hai ki "kitne alag-alag tacks up-wale ho sakte hain."

Flashcards

Kaun si char conditions ek variable ko binomial banati hain?

BINS — Binary trials, Independent, fixed Number n, Same probability p.

Binomial PMF batao.

P (X = k) = (k n) p^{k} (1 - p)^{n - k}

for

k = 0, \dots, n

PMF mein

(k n)

kahan se aata hai?

Yeh

n

trials mein

k

successes ke orderings ki number count karta hai, jo sab equally likely hote hain.

Bin (n, p)

ke liye

E [X]

kya hai aur isse kaise derive karte hain?

n p

;

X = \sum X_{i}

Bernoulli indicators ke through, jahan

E [X_{i}] = p

aur linearity of expectation use hoti hai.

Bin (n, p)

ke liye

Var (X)

kya hai?

n p (1 - p)

; har Bernoulli ka variance

p (1 - p)

hota hai, independence se add hote hain.

Variance ko independence kyun chahiye lekin mean ko nahi?

Mean linearity use karta hai (hamesha hold karta hai); sum ka variance tabhi add hota hai jab covariances zero hon, yaani independent trials.

Binomial variance kis

p

par maximise hoti hai?

p = 0.5

par,

n p \cdot 0.25

deta hai;

p = 0

1

par zero.

Prove karo ki PMF 1 tak sum karta hai.

\sum_{k} (k n) p^{k} (1 - p)^{n - k} = (p + (1 - p))^{n} = 1

Binomial Theorem se.

Ek success/fail experiment binomial kab NAHI hota?

Jab p change ho ya trials dependent hon (jaise without replacement sampling → hypergeometric).

"At least one" success compute karne ka best tarika?

Complement use karo:

1 - P (zero successes) = 1 - (1 - p)^{n}

Connections

Bernoulli distribution — binomial ka $n = 1$ building block.
Binomial Theorem — $(k n)$ provide karta hai aur prove karta hai ki total probability $= 1$ hai.
Linearity of expectation — bina messy sums ke mean deta hai.
Variance and covariance — kyun independence se variances add hoti hain.
Poisson distribution — binomial ka limit jab $n \to \infty, p \to 0, n p = λ$ fixed ho.
Normal approximation to binomial — large $n$ ke liye, $Bin (n, p) \approx N (n p, n p (1 - p))$ .
Hypergeometric distribution — "without replacement" wala cousin.