Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta: Visual walkthrough

4.9.8 · D2 · Maths › Probability Theory & Statistics › Common continuous distributions — Uniform, Normal, Exponenti

Hum har symbol ko use karne se pehle build karte hain. Agar aapne kabhi $e$ , integral, ya letter $λ$ nahi dekha, toh line one se shuru karo aur skip mat karo.

Step 1 — "Random events in time" ki picture

KYA. Ek horizontal time-line $0$ se shuru karke draw karo. Uspe dots bikhero — har dot ek event hai (bus aana, atom decay karna, customer aana). Dots random girti hain, lekin ek steady long-run pace pe.

KYUN. Is page par har distribution secretly inhi dots ke baare mein ek sawaal hai. "Pehli dot tak kitna wait?" → Exponential. "Teesri dot tak kitna wait?" → Gamma. Jawaab dene se pehle, humein "steady pace" ko numerically pin down karna hoga.

PICTURE. Figure dekho. Dots line ke saath scattered hain; chhota brace wait $T$ mark karta hai shuru se pehli dot tak.

Figure — Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta

Dots ke baare mein hum jo ek rule assume karte hain (yahi hai Poisson Process): length $t$ ke ek stretch mein dots ki count ka mean $λ t$ ke saath Poisson distribution hoti hai, aur alag-alag stretches ek doosre ko influence nahi karte. Discrete Distributions se, mean $m$ wali Poisson count $N$ satisfy karti hai

$P (N = k) = shrink factor e^{- m} \cdot weight for exactly k \frac{m ^{k}}{k !} .$

$m = λ t$ length $t$ ke stretch mein expected count hai.
$e^{- m}$ matlab $e$ (number $2.718 \dots$ ) ko power $- m$ pe raise kiya; yeh $1$ se chhota positive number hai jo $m$ badhne pe $0$ ki taraf shrink karta hai.
$k!$ matlab $k \times (k - 1) \times \dots \times 1$ , $k$ cheezein arrange karne ke tarike ki count.

Humein iska sirf ek value chahiye: $k = 0$ .

Step 2 — "Abhi tak koi event nahi" se survival curve tak

KYA. Pucho: kya probability hai ki time ke pehle $t$ units mein zero dots aayein?

WHY. "Pehla event time $t$ tak nahi hua" bilkul wahi sentence hai jaise "wait $T$ , $t$ se zyada hai". Toh $P (N = 0)$ compute karna humein $P (T > t)$ free mein deta hai — koi nayi machinery nahi.

PICTURE. Figure mein empty stretch $[0, t]$ dikhti hai (koi dots nahi) aur curve $e^{- λ t}$ $1$ se $0$ ki taraf slide karta hai jab hum window badata hain.

Poisson formula mein $k = 0$ aur $m = λ t$ set karo. Kyunki $m^{0} = 1$ aur $0! = 1$ :

$P (N = 0) = e^{- λ t} \cdot \frac{( λ t ) ^{0}}{0 !} = e^{- λ t} .$

Curve ke dono edges padho (yeh degenerate checks hain):

$t = 0$ pe: $S (0) = e^{0} = 1$ . Tum pehle se negative time wait nahi kar sakte, toh obviously abhi bhi wait kar rahe ho. ✓
Jab $t \to \infty$ : $e^{- λ t} \to 0$ . Kaafi der wait karo aur event practically zaroor ho chuka hoga. ✓
Agar $λ$ bahut bada hai: curve turant gir jaati hai — dense dots, short waits. Agar $λ \to 0$ : curve $1$ ke paas flat rahti hai — tum shayad forever wait karo. Dono intuition se match karte hain.

Step 3 — Survival ko density mein convert karo (Exponential paida hota hai)

KYA. Survival curve ko wait $T$ ki probability density $f (t)$ mein convert karo.

KYUN. Ek density batati hai ki wait kitna concentrated hai — outcomes kahan ikatthi hoti hain. Wahan pahunchne ke liye humein ek tool chahiye: derivative, jo measure karta hai ki curve kitni tez change ho rahi hai.

Derivative hi kyun, kuch aur kyun nahi? Kyunki " $t$ ke paas ek tiny window mein kitni probability girti hai" exactly hai "accumulated probability ke change ki rate" — aur rate-of-change wohi hai jo derivative compute karta hai. Koi doosra tool woh sawaal answer nahi karta.

PICTURE. Top panel: $F (t) = 1 - e^{- λ t}$ $0$ se $1$ tak climb karta hai. Bottom panel: uski slope, decaying density $f (t) = λ e^{- λ t}$ . Climb ka sabse steep part density ke sabse tall part ke upar baitha hai.

Slope compute karo. $e^{- λ t}$ ki $t$ ke saath derivative $- λ e^{- λ t}$ hai (chain rule: inner slope $- λ$ neeche lao). Toh

$f (t) = \frac{d}{d t} (flat, slope 0 1 - slope - λ e^{- λ t} e^{- λ t}) = λ e^{- λ t}, t \geq 0.$

Step 4 — Mean wait, ek area ki tarah dekha

KYA. Average wait $μ = E [T]$ nikalo.

KYUN. Woh single number jo log actually chahte hain ("typically kitna lamba?"). Hum ise expectation integral se compute karte hain — weighted average ka continuous version.

PICTURE. Shaded region $t f (t)$ hai; uska balance point $1/ λ$ pe baitha hai, ek triangle fulcrum se mark kiya gaya.

Parts se integrate karne ki jagah, hum Gamma Function lete hain — woh tool jo exactly $t^{power} e^{- t}$ shape ke integrals ke liye bana hai:

$Γ (α) = \int_{0}^{\infty} t^{α - 1} e^{- t} d t, Γ (n) = (n - 1)! .$

Mean mein $u = λ t$ substitute karo (toh $t = u / λ$ , $d t = d u / λ$ ):

= \frac{1}{\lambda}\underbrace{\int_0^\infty u\,e^{-u}\,du}_{=\,\Gamma(2)\,=\,1!\,=\,1} = \frac{1}{\lambda}.$$ > [!formula] $\text{Exp}(\lambda)$ ka Mean aur Variance > $$\mu = \frac{1}{\lambda}, \qquad \operatorname{Var}(T) = \frac{1}{\lambda^2}.$$ > **Rate–mean flip:** rate $\lambda = 0.5$/min ⇒ mean wait $1/0.5 = 2$ min. Fast rate, short wait. > [!mistake] "$\lambda$ mean hai." > Yeh *lagta* sahi hai kyunki $\lambda$ ek hi knob hai. Lekin $\lambda$ ek pace hai; mean uska **reciprocal** $1/\lambda$ hai. Figure mein fulcrum dekho — woh $1/\lambda$ pe baitha hai, $\lambda$ pe nahi. --- ## Step 5 — Waits stack karo: Gamma, Exponentials ka sum hai **KYA.** Pehli dot ki jagah, **$\alpha$-vi** dot tak wait $Y$ poocho (abhi ke liye $\alpha$ ko pura number lo, jaise $\alpha = 3$). **KYUN.** Memorylessness se (parent §3), har dot ke baad clock reset ho jaata hai: process bhool jaata hai. Toh 3rd dot tak wait sirf **teen independent Exponential waits ek ke baad ek** hai. **PICTURE.** Teen colored bars — teen Exp$(\lambda)$ waits — head-to-tail rakhe; unki total length $Y$ hai. ![[deepdives/dd-maths-4.9.08-d2-s05.png]] $$Y \;=\; \underbrace{T_1}_{\text{Exp}(\lambda)} + \underbrace{T_2}_{\text{Exp}(\lambda)} + \underbrace{T_3}_{\text{Exp}(\lambda)}.$$ Independent parts ke means aur variances **add** hote hain: $$\mathbb{E}[Y] = 3\cdot\frac1\lambda = \frac{3}{\lambda}, \qquad \operatorname{Var}(Y) = 3\cdot\frac1{\lambda^2} = \frac{3}{\lambda^2}.$$ $3$ ki jagah general $\alpha$ rakho: $\ \mathbb{E}[Y]=\alpha/\lambda,\ \operatorname{Var}(Y)=\alpha/\lambda^2$. Yahi woh Gamma mean hai jo parent ne quote kiya — ab tum dekh sakte ho **kyun** yeh exponential answer ki $\alpha$ copies hai. --- ## Step 6 — Gamma density ki shape (aur kyun constant wohi hai) **KYA.** $Y \sim \text{Gamma}(\alpha,\lambda)$ ki density likhdo aur har piece samjho. **KYUN.** $\alpha$ exponentials ko jodne ke baad ab $0$ pe peak nahi hota — tumhe pehle kai events *collect* karne padte hain, isliye chhote totals unlikely ho jaate hain aur ek **hump** banta hai. Humein exact curve chahiye. **PICTURE.** Fixed $\lambda$ pe $\alpha = 1, 2, 3, 5$ ke liye densities. Jab $\alpha$ badhta hai, peak zero se utha jaata hai aur bump right ki taraf shift hoti hai aur badi hoti hai. ![[deepdives/dd-maths-4.9.08-d2-s06.png]] > [!formula] Gamma density, term by term > $$f(y) \;=\; \underbrace{\frac{\lambda^{\alpha}}{\Gamma(\alpha)}}_{\substack{\text{normalizer:}\\ \text{area}=1 \text{ karta hai}}}\;\underbrace{y^{\alpha-1}}_{\substack{\text{rising}\\ \text{start}}}\;\underbrace{e^{-\lambda y}}_{\substack{\text{falling}\\ \text{tail}}}, \qquad y > 0.$$ > - $y^{\alpha-1}$ density ko origin ke paas $0$ se **upar** kheenchta hai ($\alpha>1$ ke liye height $0$ se shuru hoti hai — tum $3$ events koi time mein finish nahi kar sakte). > - $e^{-\lambda y}$ eventually jeet jaata hai aur ise wapas neeche kheenchta hai. > - Unka product woh hump hai jo figure mein dikhta hai. **Constant $\lambda^\alpha/\Gamma(\alpha)$ kyun?** Yeh forced hai. $\int_0^\infty f = 1$ demand karo; $u=\lambda y$ substitute karne ke baad bacha hua integral exactly ek [[Gamma Function]] hai: $$\int_0^\infty y^{\alpha-1}e^{-\lambda y}\,dy \;=\; \frac{1}{\lambda^{\alpha}}\underbrace{\int_0^\infty u^{\alpha-1}e^{-u}\,du}_{=\;\Gamma(\alpha)} \;=\; \frac{\Gamma(\alpha)}{\lambda^{\alpha}},$$ isliye normalizer uska reciprocal hona chahiye, $\lambda^{\alpha}/\Gamma(\alpha)$. **Edge check karo:** $\alpha = 1$ pe, $\Gamma(1)=1$ aur $y^{0}=1$, jo $f(y)=\lambda e^{-\lambda y}$ deta hai — Step 3 se Exponential. Family apne aap mein close hoti hai. ✓ --- ## Step 7 — Baaki teen distributions kahan lagte hain **KYA.** Uniform, Normal aur Beta ko usi tree pe rakhdo taaki kuch bhi orphaned na lage. **KYUN.** Parent ne paanch stories present kiye; yahan bataya hai ki woh hum jo machinery abhi banaaye hain, usse kaise connect hote hain. **PICTURE.** Ek chhota tree: Poisson dots Exponential → Gamma feed karte hain; ek "flat ignorance" branch Uniform deta hai; "sum of many effects" branch CLT ke zariye Normal deta hai; aur Gamma pieces milke Beta bante hain. ![[deepdives/dd-maths-4.9.08-d2-s07.png]] - **Uniform** — rate story ko side rakhdo: $[a,b]$ ke andar *total ignorance* assume karo, jo flat $f = 1/(b-a)$ force karta hai. Uska variance $(b-a)^2/12$ pure algebra hai, koi dots nahi chahiye. - **Normal** — *bahut saare* independent waits (ya koi bhi chhote effects) jodo aur [[Central Limit Theorem]] total ko bell $N(\mu,\sigma^2)$ ki taraf moda deta hai. Waqai mein large $\alpha$ ke liye $\text{Gamma}(\alpha,\lambda)$ already Normal lagta hai — dekho Step 6 mein $\alpha=5$ curve symmetrize hona shuru karta hai. - **Beta** — agar $U\sim\text{Gamma}(\alpha,\lambda)$ aur $V\sim\text{Gamma}(\beta,\lambda)$ independently hain, toh $U/(U+V)\sim\text{Beta}(\alpha,\beta)$: $[0,1]$ mein ek random *proportion*, uska mean $\alpha/(\alpha+\beta)$ Gamma means ka echo karta hai. [[Beta Function]] $B(\alpha,\beta)=\Gamma(\alpha)\Gamma(\beta)/\Gamma(\alpha+\beta)$ uska normalizer hai, aur yeh [[Conjugate Priors in Bayesian Inference]] power karta hai. --- ## Ek-picture summary Ek diagram jo poori walk compress karta hai: dots → survival $e^{-\lambda t}$ → uski slope Exponential density hai → $\alpha$ stack karo → Gamma hump, means $1/\lambda$ aur $\alpha/\lambda$ mark kiye hue. ![[deepdives/dd-maths-4.9.08-d2-s08.png]] > [!recall]- Feynman retelling (simple words mein) > Socho beads ek string pe steady average pace $\lambda$ se girti hain. Chance ki string length $t$ tak *bead-free* hai woh hai $e^{-\lambda t}$ — $1$ se shuru hota hai (abhi tak koi wait nahi ki), $0$ ki taraf melt hota hai (kaafi wait karo aur ek bead zaroor girti hai). Woh kitna tez melt hota hai, instant by instant, woh pehli bead ke wait ki *density* hai: $\lambda e^{-\lambda t}$, shuru mein sabse tall. Uska balance point $1/\lambda$ pe baitha hai — typical wait rate ka *flip* hai, tez pace = chhota wait. Teesri bead chahiye? Teen waits ko naak-se-naak rakho; total tiny nahi ho sakta (pehle teen beads chahiye), isliye uski density ek hump hai: woh Gamma hai, mean $3/\lambda$ ke saath. "Flat karo" lagate ho toh Uniform milti hai; "chhotay bahut saare effects jodo" toh bell (Normal) aati hai; "do Gamma piles ko fraction mein compare karo" toh $0$ aur $1$ ke beech ek random probability milti hai, woh Beta. Ek bead ki string, paanch distributions. > [!recall]- Quick self-test > Pehle Poisson event tak wait ka survival ::: $S(t)=e^{-\lambda t}$ > Kyun Exponential density $\lambda e^{-\lambda t}$ hai ::: yeh CDF $F(t)=1-e^{-\lambda t}$ ka derivative (slope) hai > $\text{Exp}(\lambda)$ ka mean aur kyun flip hai ::: $1/\lambda$; $\lambda$ ek rate (pace) hai, mean uska reciprocal hai > $\text{Gamma}(\alpha,\lambda)$ ka mean aur uski picture ::: $\alpha/\lambda$; yeh $\alpha$ exponential waits end-to-end stacked hain, means add hote hain > $\Gamma(\alpha)$ density mein kya karta hai ::: area ko 1 pe normalize karta hai via $\int_0^\infty y^{\alpha-1}e^{-\lambda y}dy=\Gamma(\alpha)/\lambda^\alpha$