Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta: Exercises

4.9.8 · D4 · Maths › Probability Theory & Statistics › Common continuous distributions — Uniform, Normal, Exponenti

Kuch reminders jinhe tum baar baar use karoge (sab parent mein prove hue hain). Har baar support (jahaan density nonzero hai) dhyan rakho — uske bahar $f = 0$ hota hai, jo edge-case mistakes se bachata hai:

Yahan $Φ$ hai "standard bell curve ke neeche ka cumulative area ek point tak" — woh function jo jawab deta hai " $N (0, 1)$ variable ka kitna fraction is $z$ ke equal ya neeche baithta hai?" Hum poore note mein yeh standard values use karenge: $Φ (0) = 0.5, Φ (1) = 0.8413, Φ (1.5) = 0.9332, Φ (2) = 0.9772, Φ (- 1) = 0.1587$ .

Level 1 — Recognition

L1.1

Batao inme se kaun sa paanch distributions best model karta hai aur uski ek-line "story" do: (a) woh exact time jab tum pehli raindrop ka wait karte ho jo constant average rate pe fire hone wale sensor ko hit kare; (b) ek unknown coin ka bias $p$ (ek probability, toh yeh $[0, 1]$ mein rehta hai); (c) ek number jiske baare mein tum sirf itna jaante ho ki woh $3$ aur $9$ ke beech hai, koi aur preference nahi.

Recall Solution

(a) Exponential — constant rate wale Poisson Process ke pehle event tak ka waiting time. (b) Beta — $[0, 1]$ pe ek random probability/proportion. (c) Uniform $U (3, 9)$ — "ek interval ke andar maximum ignorance," flat density.

L1.2

$X \sim U (3, 9)$ ke liye, $f (x)$ likho, phir sirf boxed formulas use karke $μ$ aur $Var (X)$ compute karo (koi integrals nahi).

Recall Solution

$f (x) = \frac{1}{9 - 3} = \frac{1}{6}$ for $3 \leq x \leq 9$ , else $0$ . $μ = \frac{3 + 9}{2} = 6$ . $Var = \frac{( 9 - 3 ) ^{2}}{12} = \frac{36}{12} = 3$ .

Level 2 — Application

L2.1

Ek help desk calls ko Poisson process ke roop mein $λ = 0.4$ per minute rate pe answer karta hai. Maan lo $X$ pehli answered call ka wait hai. (a) Mean wait kitna hai? (b) $P (X > 5)$ ? (c) $P (X \leq 2)$ ?

Recall Solution

$X \sim Exp (0.4)$ , support $t \geq 0$ . (a) $μ = \frac{1}{0.4} = 2.5$ min. (b) survival: $P (X > 5) = e^{- 0.4 \cdot 5} = e^{- 2} \approx 0.1353$ . (c) CDF: $P (X \leq 2) = 1 - e^{- 0.4 \cdot 2} = 1 - e^{- 0.8} \approx 0.5507$ .

L2.2

Exam scores $N (μ = 170, σ^{2} = 100)$ hain (toh $σ = 10$ ). $P (X \leq 185)$ aur $P (X > 150)$ nikalo.

Recall Solution

Standardize karo (tables sirf $N (0, 1)$ ke liye kaam karti hain). $z = \frac{185 - 170}{10} = 1.5 \Rightarrow P (X \leq 185) = Φ (1.5) = 0.9332$ . $z = \frac{150 - 170}{10} = - 2 \Rightarrow P (X \leq 150) = Φ (- 2) = 1 - Φ (2) = 0.0228$ , toh $P (X > 150) = 1 - 0.0228 = 0.9772$ .

Neeche wali figure exactly yahi problem dikhati hai: blue curve $N (170, 1 0^{2})$ hai, yellow shaded region (a) ka answer hai, $P (X \leq 185) = 0.9332$ , aur $x = 150$ pe red line mark karti hai jahaan doosra question cut off hota hai — uske right mein sab kuch, $P (X > 150) = 0.9772$ , (b) ka answer hai. Dhyan do ki $x = 150$ left tail mein kitna andar hai: mean se neeche poore do standard deviations, toh sirf ek sliver ( $0.0228$ ) uske neeche hai.

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

L2.3

$Y \sim Gamma (α = 3, λ = 2)$ (3rd event ka wait, rate 2). $μ$ aur $Var (Y)$ do tarike se do: formulas se, aur "sum of exponentials" argument se.

Recall Solution

Formula: $μ = \frac{α}{λ} = \frac{3}{2} = 1.5$ , $Var = \frac{α}{λ ^{2}} = \frac{3}{4} = 0.75$ . Sum argument: $Y = E_{1} + E_{2} + E_{3}$ jahan har $E_{i} \sim Exp (2)$ hai, mean $\frac{1}{2}$ , var $\frac{1}{4}$ . Means add hote hain: $3 \cdot \frac{1}{2} = 1.5$ . Independent variables ke liye variances add hote hain: $3 \cdot \frac{1}{4} = 0.75$ . Same answer.

Level 3 — Analysis

L3.1 (memorylessness)

Ek machine ki lifetime $Exp (λ)$ hai. Woh already $s = 4$ saal chal chuki hai. Dikhao ki $P (aur t saal chalega ∣ 4 pe zinda)$ us $4$ pe depend nahi karta, aur $λ = 0.1, t = 3$ ke liye evaluate karo.

Recall Solution

$P (X > s + t ∣ X > s) = \frac{P ( X > s + t )}{P ( X > s )} = \frac{e ^{- λ (s + t)}}{e ^{- λ s}} = e^{- λ t} .$ $e^{- λ s}$ cancel ho jaata hai — elapsed $s$ gayab ho jaata hai. Yahi memorylessness hai: machine "bhool jaati hai" ki woh 4 saal purani hai. Value: $e^{- 0.1 \cdot 3} = e^{- 0.3} \approx 0.7408$ .

L3.2 (68–95–99.7 rule)

Heights $N (170, 1 0^{2})$ cm hain. Bina poori table ke, $P (160 \leq X \leq 180)$ aur $P (150 \leq X \leq 190)$ estimate karo.

Recall Solution

$160$ aur $180$ , $μ \pm 1 σ$ hain ⇒ empirical rule se $P \approx 0.68$ . $150$ aur $190$ , $μ \pm 2 σ$ hain ⇒ $P \approx 0.95$ . $Φ$ se check karo: $Φ (1) - Φ (- 1) = 0.8413 - 0.1587 = 0.6826 \approx 0.68$ . ✓

L3.3 (Beta shape reasoning)

$Beta (2, 5)$ ke liye, $μ$ aur $Var$ compute karo, aur story se explain karo ki mean $0.5$ ke left kyun baithta hai.

Recall Solution

$μ = \frac{α}{α + β} = \frac{2}{2 + 5} = \frac{2}{7} \approx 0.2857$ . $Var = \frac{α β}{( α + β ) ^{2} ( α + β + 1 )} = \frac{2 \cdot 5}{7 ^{2} \cdot 8} = \frac{10}{392} = \frac{5}{196} \approx 0.0255$ . Story: socho $α - 1 = 1$ "successes" aur $β - 1 = 4$ "failures" ke weight ke roop mein. Zyada failure mass balance point ko $0$ ki taraf kheenchta hai, isliye $μ < 0.5$ .

Figure "pulled left" story ko visible banati hai: green curve $Beta (2, 5)$ density hai apne support $0 < x < 1$ pe, white dashed line midpoint $0.5$ mark karti hai, aur yellow line mean $\frac{2}{7} \approx 0.286$ mark karti hai. Kyunki $β = 5$ near $0$ pe zyada probability mass daalta hai compared to $α = 2$ jo near $1$ pe daalta hai, balance point (mean) midpoint ke noticeably left baithta hai — exactly wahi jo algebra predict karta hai.

Level 4 — Synthesis

L4.1 (Gamma ↔ Exponential bridge)

Algebraically dikhao ki $Gamma (α = 1, λ)$ hai hi $Exp (λ)$ , density aur $Γ (1) = 1$ use karke.

Recall Solution

Gamma density (support $x > 0$ ): $f (x) = \frac{λ ^{α}}{Γ ( α )} x^{α - 1} e^{- λ x}$ . $α = 1$ rakho: $f (x) = \frac{λ ^{1}}{Γ ( 1 )} x^{0} e^{- λ x} = \frac{λ}{1} \cdot 1 \cdot e^{- λ x} = λ e^{- λ x}, x > 0,$ Gamma Function fact $Γ (1) = \int_{0}^{\infty} t^{1 - 1} e^{- t} d t = \int_{0}^{\infty} t^{0} e^{- t} d t = \int_{0}^{\infty} e^{- t} d t = 1$ use kiya (general definition hai $Γ (α) = \int_{0}^{\infty} t^{α - 1} e^{- t} d t$ ; yahan $α = 1$ power ko $t^{0} = 1$ bana deta hai) aur $x^{0} = 1$ . Yeh exactly $t \geq 0$ pe Exponential density hai. Toh Exponential, Gamma family ka $α = 1$ member hai. ✓

L4.2 (Uniform → Exponential via inverse CDF)

Maan lo $U \sim U (0, 1)$ . Define karo $X = - \frac{1}{λ} ln (1 - U)$ . CDF compute karke dikhao ki $X \sim Exp (λ)$ hai. Phir, $λ = 2$ aur drawn value $U = 0.75$ ke saath, $X$ compute karo.

Recall Solution

$x \geq 0$ ke liye $P (X \leq x)$ nikalo (Exponential ka support): $P (X \leq x) = P (- \frac{1}{λ} ln (1 - U) \leq x) .$ Dono sides ko $- λ$ se multiply karo. Kyunki $λ > 0$ hai, factor $- λ$ negative hai, aur inequality ko negative number se multiply karne se direction flip hoti hai ( $\leq$ ban jaata hai $\geq$ ): $ln (1 - U) \geq - λ x .$ Ab exponentiate karo. Function $e^{(\cdot)}$ increasing hai, toh yeh direction preserve karta hai (koi flip nahi): $1 - U \geq e^{- λ x} ⟺ U \leq 1 - e^{- λ x} .$ (Last rearrangement mein $U$ aur $e^{- λ x}$ across subtract kiye, jo ek baar phir flip karta hai — check karo: $1 - U \geq e^{- λ x} \Rightarrow 1 - e^{- λ x} \geq U$ .) Kyunki $U$ , $[0, 1]$ pe uniform hai, $P (U \leq u) = u$ for $0 \leq u \leq 1$ , toh $P (X \leq x) = 1 - e^{- λ x} = F_{Exp} (x) . ✓$ Numeric: $λ = 2, U = 0.75$ : $X = - \frac{1}{2} ln (1 - 0.75) = - \frac{1}{2} ln (0.25) = - \frac{1}{2} (- 1.386294) \approx 0.6931$ .

L4.3 (Beta = Uniform special case)

Density se dikhao ki $Beta (1, 1) = U (0, 1)$ hai, $B (1, 1)$ aur Beta Beta Function identity $B (α, β) = \frac{Γ ( α ) Γ ( β )}{Γ ( α + β )}$ use karke.

Recall Solution

$B (1, 1) = \frac{Γ ( 1 ) Γ ( 1 )}{Γ ( 2 )} = \frac{1 \cdot 1}{1} = 1$ ( $Γ (2) = 1! = 1$ use kiya). Beta density with $α = β = 1$ , apne support $0 < x < 1$ pe: $f (x) = \frac{x ^{0} ( 1 - x ) ^{0}}{B ( 1 , 1 )} = \frac{1}{1} = 1, 0 < x < 1,$ jo $U (0, 1)$ ki constant density hai. ✓ Toh Uniform "flat" Beta hai.

Level 5 — Mastery

L5.1 (Exponential mean scratch se derive karo)

Integration by parts use karke prove karo ki $E [X] = \frac{1}{λ}$ for $X \sim Exp (λ)$ , har step dikhao. Phir Gamma Function identity se confirm karo.

Recall Solution

$μ = \int_{0}^{\infty} t λ e^{- λ t} d t (support t \geq 0) .$ Parts: maan lo $u = t$ (toh $d u = d t$ ) aur $d v = λ e^{- λ t} d t$ (toh $v = - e^{- λ t}$ ). $μ = [- t e^{- λ t}]_{0}^{\infty} + \int_{0}^{\infty} e^{- λ t} d t .$ Boundary term: $t \to \infty$ pe, $t e^{- λ t} \to 0$ (exponential linear ko beat karta hai); $t = 0$ pe yeh $0$ hai. Toh bracket $0$ hai. Remaining integral: $\int_{0}^{\infty} e^{- λ t} d t = [- \frac{1}{λ} e^{- λ t}]_{0}^{\infty} = \frac{1}{λ}$ . Hence $μ = \frac{1}{λ}$ . ✓ Gamma check: $μ = λ \int_{0}^{\infty} t^{1} e^{- λ t} d t = λ \cdot \frac{Γ ( 2 )}{λ ^{2}} = λ \cdot \frac{1}{λ ^{2}} = \frac{1}{λ}$ , $Γ (2) = 1$ use kiya.

L5.2 (Gamma mean derive karo)

$f (x) = \frac{λ ^{α}}{Γ ( α )} x^{α - 1} e^{- λ x}$ (support $x > 0$ ) se shuru karke, prove karo ki $E [X] = \frac{α}{λ}$ hai.

Recall Solution

$E [X] = \frac{λ ^{α}}{Γ ( α )} \int_{0}^{\infty} x^{α} e^{- λ x} d x .$ Integral ko substitution $u = λ x$ se evaluate karo. Differential compute karo: differentiate karne se $d u = λ d x$ milta hai, toh $d x = \frac{d u}{λ}$ , aur $x = \frac{u}{λ}$ bhi, hence $x^{α} = \frac{u ^{α}}{λ ^{α}}$ . Limits unchanged rehte hain ( $x : 0 \to \infty$ deta hai $u : 0 \to \infty$ kyunki $λ > 0$ hai). Substitute karo: $\int_{0}^{\infty} x^{α} e^{- λ x} d x = \int_{0}^{\infty} \frac{u ^{α}}{λ ^{α}} e^{- u} \frac{d u}{λ} = \frac{1}{λ ^{α + 1}} \int_{0}^{\infty} u^{α} e^{- u} d u = \frac{Γ ( α + 1 )}{λ ^{α + 1}},$ jahaan factor $\frac{1}{λ ^{α}}$ ( $x^{α}$ se) times $\frac{1}{λ}$ ( $d x$ se) produce karta hai $\frac{1}{λ ^{α + 1}}$ , aur remaining integral Gamma Function hai exponent $α + 1$ ke saath. Toh $E [X] = \frac{λ ^{α}}{Γ ( α )} \cdot \frac{Γ ( α + 1 )}{λ ^{α + 1}} = \frac{Γ ( α + 1 )}{λ Γ ( α )}$ . Recursion $Γ (α + 1) = α Γ (α)$ use karke (parent mein parts se prove hua), $Γ (α)$ cancel ho jaata hai: $E [X] = \frac{α Γ ( α )}{λ Γ ( α )} = \frac{α}{λ} . ✓$ Sanity: $α = 1$ deta hai $\frac{1}{λ}$ , L5.1 se match karta hai.

L5.3 (Beta variance, full derivation)

Prove karo ki $Var (X) = \frac{α β}{( α + β ) ^{2} ( α + β + 1 )}$ for $Beta (α, β)$ , given $E [X] = \frac{α}{α + β}$ aur $E [X^{2}] = \frac{α ( α + 1 )}{( α + β ) ( α + β + 1 )}$ . $Beta (2, 5)$ ke liye evaluate karo.

Recall Solution

$Var = E [X^{2}] - μ^{2}$ . Clean rakhne ke liye $s = α + β$ lo. $Var = \frac{α ( α + 1 )}{s ( s + 1 )} - \frac{α ^{2}}{s ^{2}} .$ Common denominator $s^{2} (s + 1)$ : $= \frac{α ( α + 1 ) s - α ^{2} ( s + 1 )}{s ^{2} ( s + 1 )} = \frac{α [ ( α + 1 ) s - α ( s + 1 ) ]}{s ^{2} ( s + 1 )} .$ Andar: $(α + 1) s - α s - α = α s + s - α s - α = s - α = β$ . $Var = \frac{α β}{s ^{2} ( s + 1 )} = \frac{α β}{( α + β ) ^{2} ( α + β + 1 )} . ✓$ $Beta (2, 5)$ : $\frac{2 \cdot 5}{7 ^{2} \cdot 8} = \frac{10}{392} = \frac{5}{196} \approx 0.02551$ .

Recall Yeh sab kahan connect hote hain

Gamma-as-sum-of-Exponentials picture (L2.3, L4.1) Central Limit Theorem intuition se ek step door hai, aur Beta-as-random-probability (L1.1b, L3.3, L4.3) Conjugate Priors in Bayesian Inference ka engine hai. Rate story (L2.1, L3.1) seedha Poisson Process se aata hai.