Worked examples — Uniform, Exponential, and Beta distributions

This page is a drill. The parent note Uniform, Exponential, and Beta distributions gave you the three PDFs and their formulas. Here we throw every kind of question at them — normal cases, edge cases, degenerate inputs, limits, a word problem, and an exam trap — and solve each one from the ground up.

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Two reminders before we start

Two objects will show up repeatedly, so let us pin them down now in plain words.

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The scenario matrix

Every question about these three distributions falls into one of these cells. Read the table, then we knock them out one by one.

#	Cell (scenario class)	Distribution	Where it bites
C1	Standard interval query — probability over a sub-range	Uniform	Just area of a rectangle
C2	Degenerate / zero-width input ($a=b$, or interval length $\to 0$)	Uniform	Density blows up, PDF undefined
C3	Negative & shifted support ($a<0<b$)	Uniform	Sign bookkeeping in mean/var
C4	Standard CDF query — "arrives within $t$"	Exponential	Plug into $1-e^{-\lambda t}$
C5	Memoryless twist — conditional "already waited"	Exponential	The past must cancel
C6	Limiting behaviour — $\lambda \to 0$ and $\lambda \to \infty$	Exponential	Mean $\to \infty$ or $\to 0$
C7	Prior = ignorance ($\alpha =\beta =1$) collapses to Uniform	Beta	Sanity bridge between families
C8	Bayesian update — observe data, get posterior mean	Beta	Add counts to $\alpha ,\beta$
C9	U-shaped / degenerate shape ($\alpha ,\beta <1$)	Beta	Density $\to \infty$ at ends
C10	Exam-style multi-step twist — mixes two distributions	Exp + Uniform	Chained reasoning

Let's fill them.

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Uniform: cells C1, C2, C3

1. Identify the distribution. "Equally likely at any point of an interval" is the exact phrase for $X\sim U(a,b)$ with $a=10$, $b=20$. Why this step? We must name the model before touching numbers — the whole method depends on the density being constant.
2. Write the density. $f(x)=\frac{1}{b-a}=\frac{1}{20-10}=\frac{1}{10}$ for $10\le x\le 20$. Why this step? The constant $\frac{1}{b-a}$ is the only value that makes the total area equal $1$ (see parent derivation).
3. Probability = area of a rectangle. For a flat PDF, $P(12\le X\le 17)$ is just height $\times$ width: $P=\frac{1}{10}\times (17-12)=\frac{5}{10}=0.5$ Why this step? Integrating a constant $\frac{1}{10}$ from $12$ to $17$ is base $\times$ height — no calculus needed.

Figure s01 below draws the flat blue PDF at height $\frac{1}{10}$ across $[10,20]$; the yellow shaded slab from $12$ to $17$ is exactly the area we computed — width $5$, height $\frac{1}{10}$, so area $0.5$.

Verify: The full window $[10,20]$ has area $\frac{1}{10}\times 10=1$ ✓. Our sub-window is exactly half the width, so probability $0.5$ matches the "half the box" picture. Units: minutes cancel, probability is dimensionless ✓.

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1. Look at the density formula. $f(x)=\frac{1}{b-a}$. As $b\to a$, the denominator $b-a\to 0^{+}$, so $f(x)\to +\infty$. Why this step? We test the formula at its boundary to find where it stops making sense.
2. Interpret. All the probability ($=1$) gets squeezed onto a shrinking interval, so the height must explode to keep area $=1$. Why this step? Total area is pinned at $1$; if width shrinks, height must grow to compensate.
3. Verdict for $U(3,3)$. The interval $[3,3]$ has width $0$, so $f=\frac{1}{0}$ is undefined. A single point has probability $0$ for any continuous variable, so a continuous uniform on one point does not exist — it degenerates to the infinitely-thin spike (Dirac delta) defined in the reminders above: all the certainty piled onto one value. Why this step? Naming the degenerate case protects you from ever writing $\frac{1}{b-a}$ with $a=b$.

Verify: Take $b=a+\epsilon$. Then $f=\frac{1}{\epsilon }$ and area $=\frac{1}{\epsilon }\cdot \epsilon =1$ for every $\epsilon >0$, but the limit $\epsilon \to 0$ gives $f\to \infty$ ✓. Degenerate case confirmed.

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1. Set $a=-0.1,b=0.1$. Width $b-a=0.1-(-0.1)=0.2$, so $f(x)=\frac{1}{0.2}=5$. Why this step? The subtraction $b-a$ handles the negative $a$ automatically; a common slip is writing $0.1-0.1=0$.
2. $P(W>0)$. The positive part is $[0,0.1]$, width $0.1$: $P=5\times 0.1=0.5$. Why this step? The window straddles $0$ symmetrically, so half the mass is positive — the rectangle area confirms it.
3. Mean. $\mu =\frac{a+b}{2}=\frac{-0.1+0.1}{2}=0$. Why this step? Symmetry around $0$ forces the midpoint to be $0$ — exactly the "no directional bias" that initialization wants.
4. Variance. ${\sigma }^2=\frac{(b-a{)}^2}{12}=\frac{(0.2{)}^2}{12}=\frac{0.04}{12}\approx 0.003333$. Why this step? Variance depends only on width, not position, so the negative sign never enters — the shift is irrelevant to spread.

Verify: $\frac{0.04}{12}=0.0033\overline{3}$ ✓. Mean $0$ and $P(W>0)=0.5$ are both consistent with a symmetric box centred at the origin ✓.

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Exponential: cells C4, C5, C6

1. Model. Time-until-next-event in a Poisson process is $T\sim \text{Exp}(\lambda )$, $\lambda =0.5$. Why this step? The waiting time of a Poisson process is exactly exponential — that is the parent's derivation.
2. (a) Use the CDF. The cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so $P(T\le 3)=1-e^{-0.5\cdot 3}=1-e^{-1.5}\approx 1-0.2231=\mathrm{0.7769.}$ Why this step? "Within $3$ s" = $P(T\le 3)$, which is precisely what the CDF returns.
3. (b) Invert the CDF. Set $F(t)=0.95$: $1-e^{-0.5t}=0.95\Rightarrow e^{-0.5t}=0.05\Rightarrow -0.5t=\ln (0.05)\Rightarrow t=\frac{-\ln 0.05}{0.5}\approx 5.991\text{s}.$ Why this step? A percentile is "the $t$ at which the CDF hits that fraction" — so we solve the CDF backwards.

Figure s02 below plots the blue CDF curve $1-e^{-0.5t}$; the yellow dashed line marks where it crosses $0.7769$ at $t=3$ (part a), and the red dashed line marks the $0.95$ level whose crossing gives $t\approx 5.99$ s (part b).

Verify: Plug $t=5.991$ back: $1-e^{-0.5\cdot 5.991}=1-e^{-2.996}\approx 1-0.05=0.95$ ✓. And $0.777>0.5$, meaning within one mean-wait ($1/\lambda =2$ s) plus more, most requests have arrived ✓. Units: $\lambda t$ is (per s)$\times$(s) = dimensionless ✓.

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1. Write the conditional. We want $P(T>4+3\mid T>4)$. Why this step? "Already waited $4$, then $3$ more" is literally a conditional survival probability.
2. Apply memorylessness. For the exponential (the only continuous distribution with this property): $P(T>s+t\mid T>s)=\frac{e^{-\lambda (s+t)}}{e^{-\lambda s}}=e^{-\lambda t}=P(T>t).$ Why this step? The $e^{-\lambda s}$ from the past cancels top and bottom — the process forgets the $4$ seconds you invested.
3. Compute. $P(T>3)=e^{-0.5\cdot 3}=e^{-1.5}\approx \mathrm{0.2231.}$ Why this step? After cancellation, only the future $3$ s survives, so it is the plain survival probability.

Verify: Direct route: $\frac{e^{-0.5\cdot 7}}{e^{-0.5\cdot 4}}=\frac{e^{-3.5}}{e^{-2}}=e^{-1.5}\approx 0.2231$ ✓. Identical to starting fresh — the sunk waiting time is worthless, which is the whole shock of memorylessness.

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1. Recall the mean. $E[T]=\frac{1}{\lambda }$. Why this step? All limiting behaviour of "typical wait" is read straight off $1/\lambda$.
2. (a) $\lambda \to 0^{+}$: events almost never happen, $E[T]=\frac{1}{\lambda }\to +\infty$. The PDF $\lambda e^{-\lambda t}$ flattens toward $0$ everywhere — an ever-lengthening, near-flat tail. Why this step? Rate $\to 0$ means "no events ever", so waiting time diverges — the opposite of a fast-data fit found by maximum likelihood estimation (MLE).
3. (b) $\lambda \to \infty$: events pour in, $E[T]=\frac{1}{\lambda }\to 0^{+}$. The PDF spikes hard near $t=0$ — almost all mass at instant arrival. Why this step? High rate crushes the wait toward zero, the exponential's version of a degenerate spike at $0$.

Verify: At $\lambda =0.001$, mean $=1000$ (huge) ✓; at $\lambda =1000$, mean $=0.001$ (tiny) ✓. Confirms "more frequent events → shorter waits", killing the common mistake of reading $\lambda$ as the wait.

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Beta: cells C7, C8, C9

1. Plug into the Beta PDF. $f(x)=\frac{1}{B(\alpha ,\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}$ with $\alpha =\beta =1$: $f(x)=\frac{1}{B(1,1)}{x}^0(1-x{)}^0=\frac{1}{B(1,1)}.$ Why this step? Setting the exponents to $\alpha -1=0$ and $\beta -1=0$ kills both $x$-terms, leaving a constant — the signature of a uniform.
2. Evaluate $B(1,1)$. Using the reminder $B(\alpha ,\beta )=\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}$: $B(1,1)=\frac{\Gamma (1)\Gamma (1)}{\Gamma (2)}=\frac{1\cdot 1}{1}=1$. Why this step? $\Gamma (1)=0!=1$ and $\Gamma (2)=1!=1$; this normalizer fixes the height.
3. Conclude. $f(x)=1$ on $[0,1]$ — exactly $U(0,1)$. Mean $=\frac{\alpha }{\alpha +\beta }=\frac{1}{2}$. Why this step? A flat height $1$ over $[0,1]$ is the uniform density; this is the bridge between the two families and why "flat prior = total ignorance" (see Bayesian Inference and Conjugate Priors).

Figure s03 below shows the green flat line at height $1$ over $[0,1]$ — the $\text{Beta}(1,1)$ curve — with the yellow dashed line marking its mean at $0.5$; it is visually identical to a uniform box.

Verify: Uniform mean $\frac{a+b}{2}=\frac{0+1}{2}=0.5$ equals Beta mean $\frac{1}{1+1}=0.5$ ✓, and both have constant density $1$ ✓.

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1. Prior. ${\alpha }_0=2,{\beta }_0=2$. Prior mean $=\frac{2}{2+2}=0.5$. Why this step? We record the starting belief before data — the anchor the update pulls away from.
2. Conjugacy update. Beta is conjugate to Bernoulli, so posterior is $\text{Beta}({\alpha }_0+\text{successes},{\beta }_0+\text{failures})$: $\text{Beta}(2+7,2+3)=\text{Beta}(9,5).$ Why this step? Because the Beta prior and Binomial likelihood multiply into another Beta, updating is just adding counts — the elegance the parent flagged (see Bayesian Inference and Conjugate Priors).
3. Posterior mean. $E[p]=\frac{\alpha }{\alpha +\beta }=\frac{9}{9+5}=\frac{9}{14}\approx \mathrm{0.6429.}$ Why this step? The posterior mean sits between the prior mean $0.5$ and the raw data proportion $\frac{7}{10}=0.7$ — data pulls the belief, prior tempers it.

Figure s04 below overlays the blue prior $\text{Beta}(2,2)$ (mean $0.5$) and the greener, taller posterior $\text{Beta}(9,5)$ (mean $0.643$); the red dotted line at $0.7$ is the raw data proportion, and you can see the posterior peak sitting between prior and data.

Verify: $\frac{9}{14}=0.6429$, and indeed $0.5<0.6429<0.7$ ✓. With more data the mean would slide closer to $0.7$; the prior's "pseudo-counts" of $2$ and $2$ act like $4$ extra observations.

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1. Substitute exponents. $\alpha -1=-0.5,\beta -1=-0.5$, so $f(x)\propto x^{-0.5}(1-x{)}^{-0.5}=\frac{1}{\sqrt{x}\sqrt{1-x}}.$ Why this step? Negative exponents put $x$ in the denominator, the opposite of the peaked cases — this is what creates the U.
2. Check the endpoints. As $x\to 0^{+}$, $\frac{1}{\sqrt{x}}\to \infty$; as $x\to 1^{-}$, $\frac{1}{\sqrt{1-x}}\to \infty$. Density diverges at both ends yet the total area stays finite ($=1$). Why this step? Diverging density is allowed as long as the integral converges — a subtle but legal edge case.
3. Find the normalizer. $B(0.5,0.5)=\frac{\Gamma (0.5)\Gamma (0.5)}{\Gamma (1)}=\frac{\sqrt{\pi }\cdot \sqrt{\pi }}{1}=\pi$, so the true PDF is $f(x)=\frac{1}{\pi \sqrt{x(1-x)}}$ (the arcsine distribution). Why this step? Dividing by $\pi$ is what makes the spiky curve enclose area exactly $1$, proving it is a genuine PDF.
4. Interpret. Mass piles up near $0$ and $1$, thin in the middle: this belief says "the coin is probably near-certainly biased one way or the other, just don't know which." Mean $=\frac{0.5}{0.5+0.5}=0.5$ by symmetry. Why this step? It is the "certainty near extremes" belief — the opposite of the $\text{Beta}(1,1)$ ignorance of C7.

Figure s05 below draws the red U-shaped $\text{Beta}(0.5,0.5)$ curve, with yellow arrows pointing at both endpoints where the density shoots to infinity while the shaded area under it stays finite.

Verify: ${\int }_0^1{x}^{-1/2}(1-x{)}^{-1/2}dx=\pi$, so $\frac{1}{\pi }{\int }_0^1{x}^{-1/2}(1-x{)}^{-1/2}dx=1$ ✓ — finite area despite infinite spikes at the two ends. Mean $0.5$ by symmetry ✓.

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The exam twist: cell C10

1. Freeze the conditioning. We are given $\lambda =2$, so the uniform step is done — we now work purely with $\text{Exp}(2)$. Why this step? Conditioning on the realized $\lambda$ turns a two-stage problem into a plain one-stage exponential; ignoring the "given" is the classic exam trap.
2. CDF query. $P(T\le 1)=1-e^{-\lambda t}=1-e^{-2\cdot 1}=1-e^{-2}\approx 1-0.1353=\mathrm{0.8647.}$ Why this step? Same cumulative-distribution-function machinery as C4, now with the specific $\lambda =2$.
3. Expected wait. $E[T]=\frac{1}{\lambda }=\frac{1}{2}=0.5$ s. Why this step? Given the rate, the mean is $1/\lambda$; no averaging over the prior is needed because we conditioned.
4. Sanity on the uniform stage. $\lambda =2$ is the midpoint of $U(1,3)$, whose mean is $\frac{1+3}{2}=2$ ✓ — so this draw is the most typical rate. Why this step? Confirms the given value is consistent with the first-stage distribution.

Verify: $1-e^{-2}=0.8647$ ✓; mean $0.5$ s means over half the mass lies below $t=1$, consistent with $P(T\le 1)=0.86$ ✓. Uniform mean $2$ matches the given $\lambda$ ✓.

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Recall Did every cell get filled?

Which cell did the memoryless example fill? ::: C5 — exponential conditional "already waited". Which example bridges Beta and Uniform? ::: C7 — $\text{Beta}(1,1)=U(0,1)$. In C9, where does the density diverge? ::: At both endpoints $x=0$ and $x=1$. In C10, why can we ignore the uniform stage? ::: Because we conditioned on the realized $\lambda =2$. What is the posterior in C8 after $7$ successes, $3$ failures on a $\text{Beta}(2,2)$ prior? ::: $\text{Beta}(9,5)$.