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

Reminders you will lean on (all proven in the parent). Note the support (where the density is nonzero) each time — outside it, $f = 0$ , which prevents edge-case slips:

Here $Φ$ is the "cumulative area under the standard bell curve up to a point" — the function that answers "what fraction of a $N (0, 1)$ variable sits at or below this $z$ ?" We use these standard values throughout: $Φ (0) = 0.5, Φ (1) = 0.8413, Φ (1.5) = 0.9332, Φ (2) = 0.9772, Φ (- 1) = 0.1587$ .

Level 1 — Recognition

L1.1

State which of the five distributions best models each and give its one-line "story": (a) the exact time you wait for the first raindrop to hit a sensor firing at a constant average rate; (b) an unknown coin's bias $p$ (a probability, so it lives in $[0, 1]$ ); (c) a number you know only to lie between $3$ and $9$ with no further preference.

Recall Solution

(a) Exponential — waiting time to the first event of a Poisson Process with constant rate. (b) Beta — a random probability/proportion on $[0, 1]$ . (c) Uniform $U (3, 9)$ — "maximum ignorance inside an interval," flat density.

L1.2

For $X \sim U (3, 9)$ , write down $f (x)$ , then compute $μ$ and $Var (X)$ using only the boxed formulas (no integrals).

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

A help desk answers calls as a Poisson process at rate $λ = 0.4$ per minute. Let $X$ be the wait for the first answered call. (a) Mean wait? (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 are $N (μ = 170, σ^{2} = 100)$ (so $σ = 10$ ). Find $P (X \leq 185)$ and $P (X > 150)$ .

Recall Solution

Standardize (tables only work for $N (0, 1)$ ). $z = \frac{185 - 170}{10} = 1.5 \Rightarrow P (X \leq 185) = Φ (1.5) = 0.9332$ . $z = \frac{150 - 170}{10} = - 2 \Rightarrow P (X \leq 150) = Φ (- 2) = 1 - Φ (2) = 0.0228$ , so $P (X > 150) = 1 - 0.0228 = 0.9772$ .

The figure below shows this exact problem: the blue curve is $N (170, 1 0^{2})$ , the yellow shaded region is the answer to (a), $P (X \leq 185) = 0.9332$ , and the red line at $x = 150$ marks where the second question cuts off — everything to its right, $P (X > 150) = 0.9772$ , is the answer to (b). Notice how far into the left tail $x = 150$ sits: two full standard deviations below the mean, so only a sliver ( $0.0228$ ) lies below it.

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

L2.3

$Y \sim Gamma (α = 3, λ = 2)$ (wait for the 3rd event, rate 2). Give $μ$ and $Var (Y)$ two ways: the formulas, and the "sum of exponentials" argument.

Recall Solution

Formula: $μ = \frac{α}{λ} = \frac{3}{2} = 1.5$ , $Var = \frac{α}{λ ^{2}} = \frac{3}{4} = 0.75$ . Sum argument: $Y = E_{1} + E_{2} + E_{3}$ with each $E_{i} \sim Exp (2)$ , mean $\frac{1}{2}$ , var $\frac{1}{4}$ . Means add: $3 \cdot \frac{1}{2} = 1.5$ . For independent variables variances add: $3 \cdot \frac{1}{4} = 0.75$ . Same answer.

Level 3 — Analysis

L3.1 (memorylessness)

A machine's lifetime is $Exp (λ)$ . It has already run $s = 4$ years. Show $P (lasts another t ∣ alive at 4)$ does not depend on the $4$ , and evaluate for $λ = 0.1, t = 3$ .

Recall Solution

$P (X > s + t ∣ X > s) = \frac{P ( X > s + t )}{P ( X > s )} = \frac{e ^{- λ (s + t)}}{e ^{- λ s}} = e^{- λ t} .$ The $e^{- λ s}$ cancels — the elapsed $s$ vanishes. That is memorylessness: the machine "forgets" it is 4 years old. Value: $e^{- 0.1 \cdot 3} = e^{- 0.3} \approx 0.7408$ .

L3.2 (68–95–99.7 rule)

Heights are $N (170, 1 0^{2})$ cm. Without a full table, estimate $P (160 \leq X \leq 180)$ and $P (150 \leq X \leq 190)$ .

Recall Solution

$160$ and $180$ are $μ \pm 1 σ$ ⇒ by the empirical rule $P \approx 0.68$ . $150$ and $190$ are $μ \pm 2 σ$ ⇒ $P \approx 0.95$ . Check with $Φ$ : $Φ (1) - Φ (- 1) = 0.8413 - 0.1587 = 0.6826 \approx 0.68$ . ✓

L3.3 (Beta shape reasoning)

For $Beta (2, 5)$ , compute $μ$ and $Var$ , and explain from the story why the mean sits left of $0.5$ .

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: think of $α - 1 = 1$ "successes" and $β - 1 = 4$ "failures" of weight. More failure mass drags the balance point toward $0$ , so $μ < 0.5$ .

The figure makes the "pulled left" story visible: the green curve is the $Beta (2, 5)$ density on its support $0 < x < 1$ , the white dashed line marks the midpoint $0.5$ , and the yellow line marks the mean $\frac{2}{7} \approx 0.286$ . Because $β = 5$ dumps far more probability mass near $0$ than $α = 2$ puts near $1$ , the balance point (mean) sits noticeably to the left of the midpoint — exactly what the algebra predicts.

Level 4 — Synthesis

L4.1 (Gamma ↔ Exponential bridge)

Show algebraically that $Gamma (α = 1, λ)$ is $Exp (λ)$ , using the density and $Γ (1) = 1$ .

Recall Solution

Gamma density (support $x > 0$ ): $f (x) = \frac{λ ^{α}}{Γ ( α )} x^{α - 1} e^{- λ x}$ . Put $α = 1$ : $f (x) = \frac{λ ^{1}}{Γ ( 1 )} x^{0} e^{- λ x} = \frac{λ}{1} \cdot 1 \cdot e^{- λ x} = λ e^{- λ x}, x > 0,$ using the Gamma Function fact $Γ (1) = \int_{0}^{\infty} t^{1 - 1} e^{- t} d t = \int_{0}^{\infty} t^{0} e^{- t} d t = \int_{0}^{\infty} e^{- t} d t = 1$ (the general definition is $Γ (α) = \int_{0}^{\infty} t^{α - 1} e^{- t} d t$ ; here $α = 1$ makes the power $t^{0} = 1$ ) and $x^{0} = 1$ . This is exactly the Exponential density on $t \geq 0$ . So Exponential is the $α = 1$ member of the Gamma family. ✓

L4.2 (Uniform → Exponential via inverse CDF)

Let $U \sim U (0, 1)$ . Define $X = - \frac{1}{λ} ln (1 - U)$ . Show $X \sim Exp (λ)$ by computing its CDF. Then, with $λ = 2$ and a drawn value $U = 0.75$ , compute $X$ .

Recall Solution

Find $P (X \leq x)$ for $x \geq 0$ (the support of an Exponential): $P (X \leq x) = P (- \frac{1}{λ} ln (1 - U) \leq x) .$ Multiply both sides by $- λ$ . Since $λ > 0$ , the factor $- λ$ is negative, and multiplying an inequality by a negative number flips its direction ( $\leq$ becomes $\geq$ ): $ln (1 - U) \geq - λ x .$ Now exponentiate. The function $e^{(\cdot)}$ is increasing, so it preserves the direction (no flip): $1 - U \geq e^{- λ x} ⟺ U \leq 1 - e^{- λ x} .$ (The last rearrangement subtracts $U$ and $e^{- λ x}$ across, which again flips once — check: $1 - U \geq e^{- λ x} \Rightarrow 1 - e^{- λ x} \geq U$ .) Since $U$ is uniform on $[0, 1]$ , $P (U \leq u) = u$ for $0 \leq u \leq 1$ , so $P (X \leq x) = 1 - e^{- λ x} = F_{Exp} (x) . ✓$ 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)

Show $Beta (1, 1) = U (0, 1)$ from the density, using $B (1, 1)$ and the Beta Beta Function identity $B (α, β) = \frac{Γ ( α ) Γ ( β )}{Γ ( α + β )}$ .

Recall Solution

$B (1, 1) = \frac{Γ ( 1 ) Γ ( 1 )}{Γ ( 2 )} = \frac{1 \cdot 1}{1} = 1$ (using $Γ (2) = 1! = 1$ ). Beta density with $α = β = 1$ , on its support $0 < x < 1$ : $f (x) = \frac{x ^{0} ( 1 - x ) ^{0}}{B ( 1 , 1 )} = \frac{1}{1} = 1, 0 < x < 1,$ which is the constant density of $U (0, 1)$ . ✓ So Uniform is the "flat" Beta.

Level 5 — Mastery

L5.1 (derive the Exponential mean from scratch)

Prove $E [X] = \frac{1}{λ}$ for $X \sim Exp (λ)$ using integration by parts, showing every step. Then confirm via the Gamma Function identity.

Recall Solution

$μ = \int_{0}^{\infty} t λ e^{- λ t} d t (support t \geq 0) .$ Parts: let $u = t$ (so $d u = d t$ ) and $d v = λ e^{- λ t} d t$ (so $v = - e^{- λ t}$ ). $μ = [- t e^{- λ t}]_{0}^{\infty} + \int_{0}^{\infty} e^{- λ t} d t .$ Boundary term: at $t \to \infty$ , $t e^{- λ t} \to 0$ (exponential beats linear); at $t = 0$ it is $0$ . So the bracket is $0$ . 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: $μ = λ \int_{0}^{\infty} t^{1} e^{- λ t} d t = λ \cdot \frac{Γ ( 2 )}{λ ^{2}} = λ \cdot \frac{1}{λ ^{2}} = \frac{1}{λ}$ , using $Γ (2) = 1$ .

L5.2 (derive the Gamma mean)

Starting from $f (x) = \frac{λ ^{α}}{Γ ( α )} x^{α - 1} e^{- λ x}$ (support $x > 0$ ), prove $E [X] = \frac{α}{λ}$ .

Recall Solution

$E [X] = \frac{λ ^{α}}{Γ ( α )} \int_{0}^{\infty} x^{α} e^{- λ x} d x .$ Evaluate the integral with the substitution $u = λ x$ . Compute the differential: differentiating gives $d u = λ d x$ , so $d x = \frac{d u}{λ}$ , and also $x = \frac{u}{λ}$ , hence $x^{α} = \frac{u ^{α}}{λ ^{α}}$ . The limits are unchanged ( $x : 0 \to \infty$ gives $u : 0 \to \infty$ since $λ > 0$ ). Substituting: $\int_{0}^{\infty} x^{α} e^{- λ x} d x = \int_{0}^{\infty} \frac{u ^{α}}{λ ^{α}} e^{- u} \frac{d u}{λ} = \frac{1}{λ ^{α + 1}} \int_{0}^{\infty} u^{α} e^{- u} d u = \frac{Γ ( α + 1 )}{λ ^{α + 1}},$ where the factor $\frac{1}{λ ^{α}}$ (from $x^{α}$ ) times $\frac{1}{λ}$ (from $d x$ ) produces the $\frac{1}{λ ^{α + 1}}$ , and the remaining integral is the Gamma Function with exponent $α + 1$ . So $E [X] = \frac{λ ^{α}}{Γ ( α )} \cdot \frac{Γ ( α + 1 )}{λ ^{α + 1}} = \frac{Γ ( α + 1 )}{λ Γ ( α )}$ . Using the recursion $Γ (α + 1) = α Γ (α)$ (proved by parts in the parent), the $Γ (α)$ cancels: $E [X] = \frac{α Γ ( α )}{λ Γ ( α )} = \frac{α}{λ} . ✓$ Sanity: $α = 1$ gives $\frac{1}{λ}$ , matching L5.1.

L5.3 (Beta variance, full derivation)

Prove $Var (X) = \frac{α β}{( α + β ) ^{2} ( α + β + 1 )}$ for $Beta (α, β)$ , given $E [X] = \frac{α}{α + β}$ and $E [X^{2}] = \frac{α ( α + 1 )}{( α + β ) ( α + β + 1 )}$ . Evaluate for $Beta (2, 5)$ .

Recall Solution

$Var = E [X^{2}] - μ^{2}$ . Let $s = α + β$ to keep it clean. $Var = \frac{α ( α + 1 )}{s ( s + 1 )} - \frac{α ^{2}}{s ^{2}} .$ Common denominator $s^{2} (s + 1)$ : $= \frac{α ( α + 1 ) s - α ^{2} ( s + 1 )}{s ^{2} ( s + 1 )} = \frac{α [ ( α + 1 ) s - α ( s + 1 ) ]}{s ^{2} ( s + 1 )} .$ Inside: $(α + 1) s - α s - α = α s + s - α s - α = s - α = β$ . $Var = \frac{α β}{s ^{2} ( s + 1 )} = \frac{α β}{( α + β ) ^{2} ( α + β + 1 )} . ✓$ $Beta (2, 5)$ : $\frac{2 \cdot 5}{7 ^{2} \cdot 8} = \frac{10}{392} = \frac{5}{196} \approx 0.02551$ .

Recall Where these connect

The Gamma-as-sum-of-Exponentials picture (L2.3, L4.1) is one step from the Central Limit Theorem intuition, and the Beta-as-random-probability (L1.1b, L3.3, L4.3) is the engine behind Conjugate Priors in Bayesian Inference. The rate story (L2.1, L3.1) comes straight out of the Poisson Process.