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Exponential Distribution and Continuous Random Variables

Exponential Distribution and Continuous Random Variables​

The exponential distribution models the time between events in a Poisson process, while the theory of continuous random variables extends probability to quantities that can take any value in an interval.

Board Coverage​

BoardPaperNotes
AQAPaper 2Continuous RVs; limited exponential coverage
EdexcelS3, S4Exponential distribution in S4; continuous RVs in S3
OCR (A)Paper 2Continuous RVs and exponential
CIE (9231)S2Both continuous RVs and exponential covered
The exponential distribution is the continuous counterpart to the geometric distribution.

Both are memoryless. The Poisson process links all three distributions: Poisson counts events, exponential measures inter-arrival times, and geometric counts trials until the first event. :::

1. Continuous Random Variables​

1.1 Probability density function​

Definition. A probability density function (PDF) f(x)f(x) of a continuous random variable XX is a non-negative function satisfying:

f(x)≥0forallx,∫−∞∞f(x) dx=1f(x) \geq 0 \quad \mathrm{for all } x, \qquad \int_{-\infty}^{\infty}f(x)\,dx = 1

Probabilities are found by integration:

P(a≤X≤b)=∫abf(x) dxP(a \leq X \leq b) = \int_a^b f(x)\,dx

For a continuous random variable, P(X=a)=0P(X = a) = 0 for any single value aa. This is why

P(a≤X≤b)=P(a<X<b)P(a \leq X \leq b) = P(a < X < b) — the inequalities at individual points do not matter. :::

1.2 Cumulative distribution function​

Definition. The cumulative distribution function (CDF) is

F(x)=P(X≤x)=∫−∞xf(t) dtF(x) = P(X \leq x) = \int_{-\infty}^{x}f(t)\,dt

Properties:

  • F(−∞)=0F(-\infty) = 0, F(∞)=1F(\infty) = 1
  • FF is non-decreasing
  • f(x)=F′(x)f(x) = F'(x) where FF is differentiable
  • P(a<X≤b)=F(b)−F(a)P(a < X \leq b) = F(b) - F(a)

1.3 Expected value​

Definition. The expected value of a continuous random variable XX is

E(X)=∫−∞∞x f(x) dx\boxed{E(X) = \int_{-\infty}^{\infty}x\,f(x)\,dx}

For a function g(X)g(X):

E(g(X))=∫−∞∞g(x) f(x) dxE(g(X)) = \int_{-\infty}^{\infty}g(x)\,f(x)\,dx

1.4 Variance​

Definition.

Var(X)=E(X2)−[E(X)]2=∫−∞∞x2 f(x) dx−(∫−∞∞x f(x) dx)2\boxed{\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \int_{-\infty}^{\infty}x^2\,f(x)\,dx - \left(\int_{-\infty}^{\infty}x\,f(x)\,dx\right)^2}

The linear properties carry over from the discrete case:

E(aX+b)=aE(X)+b,Var(aX+b)=a2 Var(X)E(aX + b) = aE(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X)

1.5 Median, mode, and quartiles​

Definition. The median mm satisfies F(m)=0.5F(m) = 0.5, i.e., ∫−∞mf(x) dx=0.5\int_{-\infty}^{m}f(x)\,dx = 0.5.

Definition. The mode is the value of xx at which f(x)f(x) is maximised.

Definition. The lower quartile Q1Q_1 satisfies F(Q1)=0.25F(Q_1) = 0.25 and the upper quartile Q3Q_3 satisfies F(Q3)=0.75F(Q_3) = 0.75.

The interquartile range is IQR=Q3−Q1\mathrm{IQR} = Q_3 - Q_1.

2. The Exponential Distribution​

2.1 Definition​

Definition. A continuous random variable XX follows an exponential distribution with rate parameter λ\lambda (where λ>0\lambda > 0), written X∼Exp(λ)X \sim \mathrm{Exp}(\lambda), if

f(x)=λe−λx,x≥0\boxed{f(x) = \lambda e^{-\lambda x}, \quad x \geq 0}

and f(x)=0f(x) = 0 for x<0x < 0.

2.2 Cumulative distribution function​

F(x)=P(X≤x)=∫0xλe−λt dt=[−e−λt]0x=1−e−λxF(x) = P(X \leq x) = \int_0^x \lambda e^{-\lambda t}\,dt = \left[-e^{-\lambda t}\right]_0^x = 1 - e^{-\lambda x}

F(x)=1−e−λx,x≥0\boxed{F(x) = 1 - e^{-\lambda x}, \quad x \geq 0}

2.3 Proof that E(X)=◆LB◆1◆RB◆◆LB◆λ◆RB◆E(X) = \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆​

Proof​

E(X)=∫0∞x⋅λe−λx dx\begin{aligned} E(X) &= \int_0^{\infty}x\cdot\lambda e^{-\lambda x}\,dx \end{aligned}

Using integration by parts with u=xu = x, dv=λe−λx dxdv = \lambda e^{-\lambda x}\,dx:

du=dxdu = dx, v=−e−λxv = -e^{-\lambda x}.

E(X)=[−xe−λx]0∞+∫0∞e−λx dx=lim⁡x→∞(−xe−λx)+0+[−◆LB◆1◆RB◆◆LB◆λ◆RB◆e−λx]0∞=0+◆LB◆1◆RB◆◆LB◆λ◆RB◆=◆LB◆1◆RB◆◆LB◆λ◆RB◆■\begin{aligned} E(X) &= \left[-xe^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}e^{-\lambda x}\,dx \\ &= \lim_{x\to\infty}(-xe^{-\lambda x}) + 0 + \left[-\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆e^{-\lambda x}\right]_0^{\infty} \\ &= 0 + \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ \quad \blacksquare \end{aligned}

Note: lim⁡x→∞xe−λx=0\lim_{x\to\infty}xe^{-\lambda x} = 0 by L'Hôpital's rule (exponential decay dominates).

2.4 Proof that Var(X)=◆LB◆1◆RB◆◆LB◆λ2◆RB◆\mathrm{Var}(X) = \frac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆​

Proof​

First compute E(X2)E(X^2):

E(X2)=∫0∞x2⋅λe−λx dxE(X^2) = \int_0^{\infty}x^2\cdot\lambda e^{-\lambda x}\,dx

Integration by parts twice with u=x2u = x^2, dv=λe−λx dxdv = \lambda e^{-\lambda x}\,dx:

du=2x dxdu = 2x\,dx, v=−e−λxv = -e^{-\lambda x}.

E(X2)=[−x2e−λx]0∞+∫0∞2x e−λx dx=0+2⋅◆LB◆1◆RB◆◆LB◆λ◆RB◆⋅E(X)=◆LB◆2◆RB◆◆LB◆λ◆RB◆⋅◆LB◆1◆RB◆◆LB◆λ◆RB◆=◆LB◆2◆RB◆◆LB◆λ2◆RB◆\begin{aligned} E(X^2) &= \left[-x^2 e^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}2x\,e^{-\lambda x}\,dx \\ &= 0 + 2\cdot\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆\cdot E(X) = \frac◆LB◆2◆RB◆◆LB◆\lambda◆RB◆\cdot\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ = \frac◆LB◆2◆RB◆◆LB◆\lambda^2◆RB◆ \end{aligned}

Var(X)=E(X2)−[E(X)]2=◆LB◆2◆RB◆◆LB◆λ2◆RB◆−◆LB◆1◆RB◆◆LB◆λ2◆RB◆=◆LB◆1◆RB◆◆LB◆λ2◆RB◆■\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \frac◆LB◆2◆RB◆◆LB◆\lambda^2◆RB◆ - \frac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆ \quad \blacksquare

E(X)=◆LB◆1◆RB◆◆LB◆λ◆RB◆,Var(X)=◆LB◆1◆RB◆◆LB◆λ2◆RB◆,σ=◆LB◆1◆RB◆◆LB◆λ◆RB◆\boxed{E(X) = \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆, \qquad \mathrm{Var}(X) = \frac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆, \qquad \sigma = \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆}

2.5 The memoryless property​

Theorem. The exponential distribution is the only continuous memoryless distribution:

P(X>s+t∣X>s)=P(X>t)P(X > s + t \mid X > s) = P(X > t)

Proof​

P(X>s+t∣X>s)=P(X>s+t)P(X>s)=◆LB◆e−λ(s+t)◆RB◆◆LB◆e−λs◆RB◆=e−λt=P(X>t)■\begin{aligned} P(X > s + t \mid X > s) &= \frac{P(X > s+t)}{P(X > s)} = \frac◆LB◆e^{-\lambda(s+t)}◆RB◆◆LB◆e^{-\lambda s}◆RB◆ \\ &= e^{-\lambda t} = P(X > t) \quad \blacksquare \end{aligned}

This uses P(X>x)=1−F(x)=e−λxP(X > x) = 1 - F(x) = e^{-\lambda x}.

The memoryless property has important practical implications. If a component with an

exponentially distributed lifetime has been working for ss hours, the remaining lifetime has the same distribution as a brand new component. This means exponential lifetimes imply no "wear out" effect — which is why it is more appropriate for electronic components than mechanical ones. :::

A Poisson process with rate λ\lambda satisfies:

  • The number of events in an interval of length tt follows Po(λt)\mathrm{Po}(\lambda t)
  • The time between consecutive events follows Exp(λ)\mathrm{Exp}(\lambda)
  • The inter-arrival times are independent and identically distributed

Proof sketch that inter-arrival times are exponential. Let TT be the time until the first event. P(T>t)=P(noeventsin[0,t])=P(N(t)=0)=◆LB◆e−λt(λt)0◆RB◆◆LB◆0!◆RB◆=e−λtP(T > t) = P(\mathrm{no events in }[0,t]) = P(N(t) = 0) = \dfrac◆LB◆e^{-\lambda t}(\lambda t)^0◆RB◆◆LB◆0!◆RB◆ = e^{-\lambda t}.

So P(T≤t)=1−e−λtP(T \leq t) = 1 - e^{-\lambda t}, which is the CDF of Exp(λ)\mathrm{Exp}(\lambda). ■\blacksquare

2.7 Percentiles​

For the exponential distribution:

F(x)=1−e−λx=p  ⟹  x=−◆LB◆1◆RB◆◆LB◆λ◆RB◆ln⁡(1−p)F(x) = 1 - e^{-\lambda x} = p \implies x = -\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆\ln(1-p)

The median is x0.5=−◆LB◆ln⁡(0.5)◆RB◆◆LB◆λ◆RB◆=◆LB◆ln⁡2◆RB◆◆LB◆λ◆RB◆x_{0.5} = -\dfrac◆LB◆\ln(0.5)◆RB◆◆LB◆\lambda◆RB◆ = \dfrac◆LB◆\ln 2◆RB◆◆LB◆\lambda◆RB◆.

3. Worked Examples​

3.1 Finding probabilities​

Example. X∼Exp(0.5)X \sim \mathrm{Exp}(0.5). Find P(X>3)P(X > 3), P(1<X<4)P(1 < X < 4), and the median.

P(X>3)=e−0.5×3=e−1.5≈0.2231P(X > 3) = e^{-0.5 \times 3} = e^{-1.5} \approx 0.2231.

P(1<X<4)=F(4)−F(1)=(1−e−2)−(1−e−0.5)=e−0.5−e−2≈0.6065−0.1353=0.4712P(1 < X < 4) = F(4) - F(1) = (1-e^{-2}) - (1-e^{-0.5}) = e^{-0.5} - e^{-2} \approx 0.6065 - 0.1353 = 0.4712.

Median =◆LB◆ln⁡2◆RB◆◆LB◆0.5◆RB◆=2ln⁡2≈1.386= \dfrac◆LB◆\ln 2◆RB◆◆LB◆0.5◆RB◆ = 2\ln 2 \approx 1.386.

3.2 Continuous random variable from a PDF​

Example. XX has PDF f(x)=38x2f(x) = \dfrac{3}{8}x^2 for 0≤x≤20 \leq x \leq 2 and f(x)=0f(x) = 0 otherwise.

Verify: ∫0238x2 dx=38⋅83=1\int_0^2 \dfrac{3}{8}x^2\,dx = \dfrac{3}{8}\cdot\dfrac{8}{3} = 1. ✓\checkmark

E(X)=∫02x⋅38x2 dx=38∫02x3 dx=38⋅164=32E(X) = \int_0^2 x\cdot\dfrac{3}{8}x^2\,dx = \dfrac{3}{8}\int_0^2 x^3\,dx = \dfrac{3}{8}\cdot\dfrac{16}{4} = \dfrac{3}{2}.

E(X2)=38∫02x4 dx=38⋅325=125E(X^2) = \dfrac{3}{8}\int_0^2 x^4\,dx = \dfrac{3}{8}\cdot\dfrac{32}{5} = \dfrac{12}{5}.

Var(X)=125−(32)2=125−94=320\mathrm{Var}(X) = \dfrac{12}{5} - \left(\dfrac{3}{2}\right)^2 = \dfrac{12}{5} - \dfrac{9}{4} = \dfrac{3}{20}.

Median: ∫0m38x2 dx=0.5  ⟹  m38=0.5  ⟹  m3=4  ⟹  m=43≈1.587\int_0^m \dfrac{3}{8}x^2\,dx = 0.5 \implies \dfrac{m^3}{8} = 0.5 \implies m^3 = 4 \implies m = \sqrt[3]{4} \approx 1.587.

3.3 CDF from PDF​

Example. f(x)=2xf(x) = 2x for 0≤x≤10 \leq x \leq 1. Find F(x)F(x).

For 0≤x≤10 \leq x \leq 1: F(x)=∫0x2t dt=x2F(x) = \int_0^x 2t\,dt = x^2.

For x<0x < 0: F(x)=0F(x) = 0. For x>1x > 1: F(x)=1F(x) = 1.

4. Hypothesis Testing with the Exponential Distribution​

Example. The lifetime of a component is modelled by X∼Exp(λ)X \sim \mathrm{Exp}(\lambda). A sample of 10 components gives a mean lifetime of 420 hours. Test at the 5% level whether λ=0.005\lambda = 0.005 against H1:λ≠0.005H_1: \lambda \neq 0.005.

Under H0H_0: E(X)=1/λ=200E(X) = 1/\lambda = 200 hours. Since nn is large, use the approximate normal distribution of Xˉ\bar{X}:

Xˉ∼N ⁣(◆LB◆1◆RB◆◆LB◆λ◆RB◆,◆LB◆1◆RB◆◆LB◆nλ2◆RB◆)=N(200,4000)\bar{X} \sim N\!\left(\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆, \frac◆LB◆1◆RB◆◆LB◆n\lambda^2◆RB◆\right) = N(200, 4000) approximately.

z=◆LB◆420−200◆RB◆◆LB◆4000◆RB◆=22063.25=3.48z = \dfrac◆LB◆420 - 200◆RB◆◆LB◆\sqrt{4000}◆RB◆ = \dfrac{220}{63.25} = 3.48.

∣z∣=3.48>1.96|z| = 3.48 > 1.96, so reject H0H_0.

Problems​

Details

Problem 1 X∼Exp(2)X \sim \mathrm{Exp}(2). Find P(X>1)P(X > 1), P(0.5<X<2)P(0.5 < X < 2), and the 90th percentile.

Details

Solution 1 P(X>1)=e−2(1)=e−2≈0.1353P(X > 1) = e^{-2(1)} = e^{-2} \approx 0.1353.

P(0.5<X<2)=(1−e−4)−(1−e−1)=e−1−e−4≈0.3679−0.0183=0.3496P(0.5 < X < 2) = (1-e^{-4}) - (1-e^{-1}) = e^{-1} - e^{-4} \approx 0.3679 - 0.0183 = 0.3496.

90th percentile: F(x)=0.9  ⟹  1−e−2x=0.9  ⟹  x=−◆LB◆ln⁡(0.1)◆RB◆◆LB◆2◆RB◆≈1.151F(x) = 0.9 \implies 1 - e^{-2x} = 0.9 \implies x = -\dfrac◆LB◆\ln(0.1)◆RB◆◆LB◆2◆RB◆ \approx 1.151.

If you get this wrong, revise: Percentiles — Section 2.7.

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Problem 2 A continuous random variable XX has PDF f(x)=3x28f(x) = \dfrac{3x^2}{8} for 0≤x≤20 \leq x \leq 2. Find E(X)E(X), Var(X)\mathrm{Var}(X), and the median.

Details

Solution 2 E(X)=∫02x⋅3x28 dx=38⋅[x44]02=38⋅4=1.5E(X) = \int_0^2 x\cdot\dfrac{3x^2}{8}\,dx = \dfrac{3}{8}\cdot\left[\dfrac{x^4}{4}\right]_0^2 = \dfrac{3}{8}\cdot 4 = 1.5.

E(X2)=∫02x2⋅3x28 dx=38⋅[x55]02=38⋅325=2.4E(X^2) = \int_0^2 x^2\cdot\dfrac{3x^2}{8}\,dx = \dfrac{3}{8}\cdot\left[\dfrac{x^5}{5}\right]_0^2 = \dfrac{3}{8}\cdot\dfrac{32}{5} = 2.4.

Var(X)=2.4−1.52=2.4−2.25=0.15\mathrm{Var}(X) = 2.4 - 1.5^2 = 2.4 - 2.25 = 0.15.

Median: ∫0m3x28 dx=0.5  ⟹  m38=0.5  ⟹  m=43≈1.587\int_0^m \dfrac{3x^2}{8}\,dx = 0.5 \implies \dfrac{m^3}{8} = 0.5 \implies m = \sqrt[3]{4} \approx 1.587.

If you get this wrong, revise: Median, mode, and quartiles — Section 1.5.

Details

Problem 3 Prove the memoryless property of the exponential distribution.

Details

Solution 3 P(X>s+t∣X>s)=P(X>s+t)P(X>s)=◆LB◆e−λ(s+t)◆RB◆◆LB◆e−λs◆RB◆=e−λt=P(X>t)P(X > s+t \mid X > s) = \dfrac{P(X > s+t)}{P(X > s)} = \dfrac◆LB◆e^{-\lambda(s+t)}◆RB◆◆LB◆e^{-\lambda s}◆RB◆ = e^{-\lambda t} = P(X > t). ■\blacksquare

This uses the survival function P(X>x)=e−λxP(X > x) = e^{-\lambda x}.

If you get this wrong, revise: The memoryless property — Section 2.5.

Details

Problem 4 Calls arrive at a switchboard as a Poisson process with rate λ=4\lambda = 4 per hour. Find the probability that the time between two consecutive calls exceeds 30 minutes.

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Solution 4 The inter-arrival time T∼Exp(4)T \sim \mathrm{Exp}(4) (rate in hours).

P(T>0.5)=e−4×0.5=e−2≈0.1353P(T > 0.5) = e^{-4 \times 0.5} = e^{-2} \approx 0.1353.

If you get this wrong, revise: Link to Poisson processes — Section 2.6.

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Problem 5 XX has PDF f(x)=12xf(x) = \dfrac{1}{2}x for 0≤x≤20 \leq x \leq 2. Find the CDF, E(X)E(X), and Var(X)\mathrm{Var}(X).

Details

Solution 5 CDF: F(x)=∫0xt2 dt=x24F(x) = \int_0^x \dfrac{t}{2}\,dt = \dfrac{x^2}{4} for 0≤x≤20 \leq x \leq 2. F(x)=0F(x) = 0 for x<0x < 0, F(x)=1F(x) = 1 for x>2x > 2.

E(X)=∫02x⋅x2 dx=12[x33]02=43E(X) = \int_0^2 x\cdot\dfrac{x}{2}\,dx = \dfrac{1}{2}\left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{4}{3}.

E(X2)=12[x44]02=12⋅4=2E(X^2) = \dfrac{1}{2}\left[\dfrac{x^4}{4}\right]_0^2 = \dfrac{1}{2}\cdot 4 = 2.

Var(X)=2−(43)2=2−169=29\mathrm{Var}(X) = 2 - \left(\dfrac{4}{3}\right)^2 = 2 - \dfrac{16}{9} = \dfrac{2}{9}.

If you get this wrong, revise: Expected value — Section 1.3.

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Problem 6 The lifetime of a light bulb follows X∼Exp(0.01)X \sim \mathrm{Exp}(0.01) (in hours). Given that the bulb has been working for 500 hours, find the probability it lasts at least another 200 hours.

Details

Solution 6 By the memoryless property: P(X>500+200∣X>500)=P(X>200)=e−0.01×200=e−2≈0.1353P(X > 500+200 \mid X > 500) = P(X > 200) = e^{-0.01 \times 200} = e^{-2} \approx 0.1353.

If you get this wrong, revise: The memoryless property — Section 2.5.

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Problem 7 A continuous random variable XX has CDF F(x)=x327F(x) = \dfrac{x^3}{27} for 0≤x≤30 \leq x \leq 3. Find the PDF, E(X)E(X), and the upper quartile.

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Solution 7 PDF: f(x)=F′(x)=x29f(x) = F'(x) = \dfrac{x^2}{9} for 0≤x≤30 \leq x \leq 3.

E(X)=∫03x⋅x29 dx=19[x44]03=19⋅814=94=2.25E(X) = \int_0^3 x\cdot\dfrac{x^2}{9}\,dx = \dfrac{1}{9}\left[\dfrac{x^4}{4}\right]_0^3 = \dfrac{1}{9}\cdot\dfrac{81}{4} = \dfrac{9}{4} = 2.25.

Upper quartile: F(Q3)=0.75  ⟹  Q3327=0.75  ⟹  Q33=20.25  ⟹  Q3≈2.725F(Q_3) = 0.75 \implies \dfrac{Q_3^3}{27} = 0.75 \implies Q_3^3 = 20.25 \implies Q_3 \approx 2.725.

If you get this wrong, revise: Cumulative distribution function — Section 1.2.

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Problem 8 Prove that E(X)=1/λE(X) = 1/\lambda for X∼Exp(λ)X \sim \mathrm{Exp}(\lambda), using integration by parts.

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Solution 8 E(X)=∫0∞xλe−λx dxE(X) = \int_0^{\infty}x\lambda e^{-\lambda x}\,dx.

Let u=xu = x, dv=λe−λx dxdv = \lambda e^{-\lambda x}\,dx, so du=dxdu = dx, v=−e−λxv = -e^{-\lambda x}.

E(X)=[−xe−λx]0∞+∫0∞e−λx dx=0+[−◆LB◆1◆RB◆◆LB◆λ◆RB◆e−λx]0∞=◆LB◆1◆RB◆◆LB◆λ◆RB◆E(X) = \left[-xe^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}e^{-\lambda x}\,dx = 0 + \left[-\dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆e^{-\lambda x}\right]_0^{\infty} = \dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆. ■\blacksquare

If you get this wrong, revise: Proof that E(X)=◆LB◆1◆RB◆◆LB◆λ◆RB◆E(X) = \frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ — Section 2.3.

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Problem 9 Buses arrive at a stop as a Poisson process with rate 6 per hour. Find the probability that a passenger waits between 5 and 15 minutes for a bus.

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Solution 9 Waiting time T∼Exp(6)T \sim \mathrm{Exp}(6) (rate per hour).

P(1/12<T<1/4)=F(1/4)−F(1/12)=(1−e−1.5)−(1−e−0.5)=e−0.5−e−1.5P(1/12 < T < 1/4) = F(1/4) - F(1/12) = (1-e^{-1.5}) - (1-e^{-0.5}) = e^{-0.5} - e^{-1.5}

≈0.6065−0.2231=0.3834\approx 0.6065 - 0.2231 = 0.3834.

If you get this wrong, revise: Link to Poisson processes — Section 2.6.

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Problem 10 XX has PDF f(x)=4x3f(x) = 4x^3 for 0≤x≤10 \leq x \leq 1. Find P(X>0.5)P(X > 0.5), E(X)E(X), Var(X)\mathrm{Var}(X), and the mode.

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Solution 10 P(X>0.5)=∫0.514x3 dx=[x4]0.51=1−0.0625=0.9375P(X > 0.5) = \int_{0.5}^1 4x^3\,dx = \left[x^4\right]_{0.5}^1 = 1 - 0.0625 = 0.9375.

E(X)=∫014x4 dx=[4x55]01=45E(X) = \int_0^1 4x^4\,dx = \left[\dfrac{4x^5}{5}\right]_0^1 = \dfrac{4}{5}.

E(X2)=∫014x5 dx=[4x66]01=23E(X^2) = \int_0^1 4x^5\,dx = \left[\dfrac{4x^6}{6}\right]_0^1 = \dfrac{2}{3}.

Var(X)=23−(45)2=23−1625=50−4875=275\mathrm{Var}(X) = \dfrac{2}{3} - \left(\dfrac{4}{5}\right)^2 = \dfrac{2}{3} - \dfrac{16}{25} = \dfrac{50 - 48}{75} = \dfrac{2}{75}.

Mode: f(x)=4x3f(x) = 4x^3 is increasing on [0,1][0,1], so the mode is at x=1x = 1.

If you get this wrong, revise: Median, mode, and quartiles — Section 1.5.

5. Memoryless Property: Detailed Proof and Interpretation​

5.1 Rigorous proof using conditional probability​

Theorem. If X∼Exp(λ)X \sim \mathrm{Exp}(\lambda), then for all s,t>0s, t > 0:

P(X>s+t∣X>s)=P(X>t)P(X > s + t \mid X > s) = P(X > t)

Proof​

By definition of conditional probability:

P(X>s+t∣X>s)=◆LB◆P(X>s+t ∩ X>s)◆RB◆◆LB◆P(X>s)◆RB◆=P(X>s+t)P(X>s)P(X > s + t \mid X > s) = \frac◆LB◆P(X > s + t \,\cap\, X > s)◆RB◆◆LB◆P(X > s)◆RB◆ = \frac{P(X > s + t)}{P(X > s)}

since X>s+tX > s + t implies X>sX > s.

Using the survival function S(x)=P(X>x)=e−λxS(x) = P(X > x) = e^{-\lambda x}:

P(X>s+t)P(X>s)=◆LB◆e−λ(s+t)◆RB◆◆LB◆e−λs◆RB◆=e−λt=P(X>t)■\frac{P(X > s + t)}{P(X > s)} = \frac◆LB◆e^{-\lambda(s+t)}◆RB◆◆LB◆e^{-\lambda s}◆RB◆ = e^{-\lambda t} = P(X > t) \quad \blacksquare

5.2 Converse: exponential is the only continuous memoryless distribution​

Theorem. If a continuous random variable XX on (0,∞)(0, \infty) satisfies P(X>s+t∣X>s)=P(X>t)P(X > s+t \mid X > s) = P(X > t) for all s,t>0s, t > 0, then X∼Exp(λ)X \sim \mathrm{Exp}(\lambda) for some λ>0\lambda > 0.

Proof​

Let G(t)=P(X>t)G(t) = P(X > t). The memoryless condition gives:

G(s+t)=G(s)G(t)G(s + t) = G(s)G(t)

This is Cauchy's functional equation. Since GG is non-increasing and 0≤G≤10 \leq G \leq 1, the only solutions are:

G(t)=e−λtG(t) = e^{-\lambda t}

for some λ≥0\lambda \geq 0. Since GG is non-trivial (not identically 1), λ>0\lambda > 0. Therefore:

P(X≤t)=1−e−λtP(X \leq t) = 1 - e^{-\lambda t}

which is the CDF of Exp(λ)\mathrm{Exp}(\lambda). ■\blacksquare

5.3 Practical interpretation​

The memoryless property means:

  • If a light bulb has been on for 100 hours, the probability it lasts another 50 hours is the same as a new bulb lasting 50 hours.
  • If you have waited 20 minutes for a bus, your expected additional wait time is the same as if you had just arrived.
  • This property makes exponential models appropriate for random failure mechanisms (electronic components) but inappropriate for wear-out mechanisms (mechanical parts).

6. Poisson-Exponential Connection: Inter-Arrival Times​

6.1 Derivation​

Theorem. In a Poisson process with rate λ\lambda, the time between consecutive events follows Exp(λ)\mathrm{Exp}(\lambda).

Proof​

Let TT be the time from an arbitrary starting point until the first event.

P(T>t)=P(noeventsin[0,t])P(T > t) = P(\mathrm{no events in }[0,t])

Since the number of events in [0,t][0,t] follows Po(λt)\mathrm{Po}(\lambda t):

P(N(t)=0)=◆LB◆e−λt(λt)0◆RB◆◆LB◆0!◆RB◆=e−λtP(N(t) = 0) = \frac◆LB◆e^{-\lambda t}(\lambda t)^0◆RB◆◆LB◆0!◆RB◆ = e^{-\lambda t}

Therefore P(T≤t)=1−e−λtP(T \leq t) = 1 - e^{-\lambda t}, which is the CDF of Exp(λ)\mathrm{Exp}(\lambda). ■\blacksquare

6.2 Sum of inter-arrival times​

If T1,T2,…,TnT_1, T_2, \ldots, T_n are nn independent inter-arrival times, each ∼Exp(λ)\sim \mathrm{Exp}(\lambda), then the total time until the nn-th event is:

Sn=T1+T2+⋯+Tn∼Gamma(n,λ)S_n = T_1 + T_2 + \cdots + T_n \sim \mathrm{Gamma}(n, \lambda)

This connects the exponential to the gamma distribution.

6.3 Worked example: call centre​

Example. Calls arrive at a call centre as a Poisson process at rate 5 per hour.

(a) Find the probability that the time between two consecutive calls exceeds 20 minutes.

T∼Exp(5)T \sim \mathrm{Exp}(5) (rate per hour). P(T>1/3)=e−5/3≈0.1889P(T > 1/3) = e^{-5/3} \approx 0.1889.

(b) Find the probability that at least 3 calls arrive in the next 30 minutes.

N(0.5)∼Po(2.5)N(0.5) \sim \mathrm{Po}(2.5). P(N≥3)=1−P(N≤2)=1−e−2.5(1+2.5+2.52/2)=1−0.0821×9.125≈1−0.749=0.251P(N \geq 3) = 1 - P(N \leq 2) = 1 - e^{-2.5}(1 + 2.5 + 2.5^2/2) = 1 - 0.0821 \times 9.125 \approx 1 - 0.749 = 0.251.

(c) Find the median inter-arrival time.

Median =◆LB◆ln⁡2◆RB◆◆LB◆λ◆RB◆=◆LB◆ln⁡2◆RB◆◆LB◆5◆RB◆≈0.139 hours≈8.3 minutes= \dfrac◆LB◆\ln 2◆RB◆◆LB◆\lambda◆RB◆ = \dfrac◆LB◆\ln 2◆RB◆◆LB◆5◆RB◆ \approx 0.139\,\mathrm{hours} \approx 8.3\,\mathrm{minutes}.

7. The Continuous Uniform Distribution​

7.1 Definition​

Definition. X∼U(a,b)X \sim U(a, b) if:

f(x)=1b−a,a≤x≤bf(x) = \frac{1}{b - a}, \quad a \leq x \leq b

and f(x)=0f(x) = 0 otherwise.

7.2 Proof of E(X)E(X) and Var(X)\mathrm{Var}(X)​

Proof​

E(X)=∫abx⋅1b−a dx=1b−a[x22]ab=b2−a22(b−a)=(b−a)(b+a)2(b−a)=a+b2■E(X) = \int_a^b x \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a}\left[\frac{x^2}{2}\right]_a^b = \frac{b^2 - a^2}{2(b-a)} = \frac{(b-a)(b+a)}{2(b-a)} = \frac{a+b}{2} \quad \blacksquare

E(X2)=∫abx2⋅1b−a dx=1b−a[x33]ab=b3−a33(b−a)=a2+ab+b23E(X^2) = \int_a^b x^2 \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a}\left[\frac{x^3}{3}\right]_a^b = \frac{b^3 - a^3}{3(b-a)} = \frac{a^2 + ab + b^2}{3}

Var(X)=a2+ab+b23−(a+b)24=4(a2+ab+b2)−3(a+b)212\mathrm{Var}(X) = \frac{a^2 + ab + b^2}{3} - \frac{(a+b)^2}{4} = \frac{4(a^2 + ab + b^2) - 3(a+b)^2}{12}

=4a2+4ab+4b2−3a2−6ab−3b212=a2−2ab+b212= \frac{4a^2 + 4ab + 4b^2 - 3a^2 - 6ab - 3b^2}{12} = \frac{a^2 - 2ab + b^2}{12}

Var(X)=(b−a)212■\boxed{\mathrm{Var}(X) = \frac{(b-a)^2}{12}} \quad \blacksquare

7.3 CDF of the continuous uniform​

F(x)={0x<ax−ab−aa≤x≤b1x>bF(x) = \begin{cases} 0 & x \lt{} a \\ \dfrac{x - a}{b - a} & a \leq x \leq b \\ 1 & x > b \end{cases}

8. Mixture Distributions​

8.1 Definition​

A mixture distribution arises when a random variable is selected from one of several sub-populations. If XX comes from distribution 1 with probability pp and from distribution 2 with probability 1−p1-p:

f(x)=p f1(x)+(1−p) f2(x)f(x) = p\,f_1(x) + (1-p)\,f_2(x)

E(X)=p E(X1)+(1−p) E(X2)E(X) = p\,E(X_1) + (1-p)\,E(X_2)

Var(X)=p Var(X1)+(1−p) Var(X2)+p(1−p)[E(X1)−E(X2)]2\mathrm{Var}(X) = p\,\mathrm{Var}(X_1) + (1-p)\,\mathrm{Var}(X_2) + p(1-p)[E(X_1) - E(X_2)]^2

The variance formula includes an extra term from the difference in means — this is the law of total variance.

8.2 Worked example​

Example. A machine produces components. With probability 0.7 it is correctly calibrated, producing components with lifetime ∼Exp(0.01)\sim \mathrm{Exp}(0.01). With probability 0.3 it is faulty, producing components with lifetime ∼Exp(0.002)\sim \mathrm{Exp}(0.002). Find the overall PDF, the expected lifetime, and P(X>100)P(X > 100).

f(x)=0.7(0.01 e−0.01x)+0.3(0.002 e−0.002x)=0.007 e−0.01x+0.0006 e−0.002xf(x) = 0.7(0.01\,e^{-0.01x}) + 0.3(0.002\,e^{-0.002x}) = 0.007\,e^{-0.01x} + 0.0006\,e^{-0.002x}

E(X)=0.7×100+0.3×500=70+150=220 hoursE(X) = 0.7 \times 100 + 0.3 \times 500 = 70 + 150 = 220\,\mathrm{hours}.

P(X>100)=0.7 e−0.01×100+0.3 e−0.002×100=0.7e−1+0.3e−0.2≈0.7(0.3679)+0.3(0.8187)P(X > 100) = 0.7\,e^{-0.01 \times 100} + 0.3\,e^{-0.002 \times 100} = 0.7e^{-1} + 0.3e^{-0.2} \approx 0.7(0.3679) + 0.3(0.8187)

≈0.2575+0.2456=0.5031\approx 0.2575 + 0.2456 = 0.5031.

9. Common Pitfalls​

Confusing PDF with CDF​

The PDF f(x)f(x) gives the density of probability at xx. It is not a probability itself — f(x)f(x) can be greater than 1. The CDF F(x)F(x) gives the accumulated probability up to xx, and always satisfies 0≤F(x)≤10 \leq F(x) \leq 1.

Common error: writing P(X=a)=f(a)P(X = a) = f(a) for a continuous RV. This is wrong — P(X=a)=0P(X = a) = 0 always. Probabilities are areas under the PDF, not values of the PDF.

Discrete vs continuous probability​

For a discrete RV, P(X=a)>0P(X = a) > 0 for specific values, and probabilities sum to 1.

For a continuous RV, P(X=a)=0P(X = a) = 0 for any single value, and probabilities integrate to 1.

Never use summation for a continuous RV or integration for a discrete RV (unless using the CDF formalism).

Exponential rate vs mean​

X∼Exp(λ)X \sim \mathrm{Exp}(\lambda) has mean 1/λ1/\lambda. A common error is to confuse λ\lambda (the rate) with the mean. If the mean lifetime is 200 hours, then λ=1/200=0.005\lambda = 1/200 = 0.005, not λ=200\lambda = 200.

Check: a larger λ\lambda means shorter lifetimes on average (events happen more frequently).

Units in Poisson-exponential problems​

If events occur at rate λ=5\lambda = 5 per hour, then the inter-arrival time is Exp(5)\mathrm{Exp}(5) measured in hours. If you want the probability of waiting more than 20 minutes, convert to hours: t=1/3t = 1/3 hours. Using t=20t = 20 directly would give P(T>20)=e−100P(T > 20) = e^{-100}, which is essentially zero and clearly wrong.

10. Problem Set​

Q1. X∼Exp(λ)X \sim \mathrm{Exp}(\lambda). Find the value of λ\lambda such that P(X>2)=0.3P(X > 2) = 0.3, and hence find E(X)E(X) and the 80th percentile.

P(X>2)=e−2λ=0.3  ⟹  −2λ=ln⁡(0.3)  ⟹  λ=◆LB◆−ln⁡(0.3)◆RB◆◆LB◆2◆RB◆=1.2042=0.602P(X > 2) = e^{-2\lambda} = 0.3 \implies -2\lambda = \ln(0.3) \implies \lambda = \dfrac◆LB◆-\ln(0.3)◆RB◆◆LB◆2◆RB◆ = \dfrac{1.204}{2} = 0.602.

E(X)=1/0.602≈1.661E(X) = 1/0.602 \approx 1.661.

80th percentile: F(x)=0.8  ⟹  1−e−0.602x=0.8  ⟹  e−0.602x=0.2  ⟹  x=◆LB◆−ln⁡(0.2)◆RB◆◆LB◆0.602◆RB◆=1.6090.602≈2.673F(x) = 0.8 \implies 1 - e^{-0.602x} = 0.8 \implies e^{-0.602x} = 0.2 \implies x = \dfrac◆LB◆-\ln(0.2)◆RB◆◆LB◆0.602◆RB◆ = \dfrac{1.609}{0.602} \approx 2.673.

Q2. A continuous random variable XX has PDF f(x)=kx(4−x)f(x) = kx(4-x) for 0≤x≤40 \leq x \leq 4. Find kk, E(X)E(X), Var(X)\mathrm{Var}(X), and the median.

∫04kx(4−x) dx=k∫04(4x−x2) dx=k[2x2−x33]04=k(32−64/3)=k(32/3)=1  ⟹  k=3/32\int_0^4 kx(4-x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\left[2x^2 - \dfrac{x^3}{3}\right]_0^4 = k(32 - 64/3) = k(32/3) = 1 \implies k = 3/32.

E(X)=332∫04(4x2−x3) dx=332[4x33−x44]04=332(2563−64)=332(256−1923)=6432=2E(X) = \dfrac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \dfrac{3}{32}\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_0^4 = \dfrac{3}{32}\left(\dfrac{256}{3} - 64\right) = \dfrac{3}{32}\left(\dfrac{256 - 192}{3}\right) = \dfrac{64}{32} = 2.

E(X2)=332∫04(4x3−x4) dx=332[x4−x55]04=332(256−204.8)=◆LB◆3×51.2◆RB◆◆LB◆32◆RB◆=4.8E(X^2) = \dfrac{3}{32}\int_0^4 (4x^3 - x^4)\,dx = \dfrac{3}{32}\left[x^4 - \dfrac{x^5}{5}\right]_0^4 = \dfrac{3}{32}(256 - 204.8) = \dfrac◆LB◆3 \times 51.2◆RB◆◆LB◆32◆RB◆ = 4.8.

Var(X)=4.8−4=0.8\mathrm{Var}(X) = 4.8 - 4 = 0.8.

Median: 332∫0m(4x−x2) dx=0.5  ⟹  332(2m2−m33)=0.5\dfrac{3}{32}\int_0^m (4x - x^2)\,dx = 0.5 \implies \dfrac{3}{32}\left(2m^2 - \dfrac{m^3}{3}\right) = 0.5.

2m2−m33=163  ⟹  6m2−m3=162m^2 - \dfrac{m^3}{3} = \dfrac{16}{3} \implies 6m^2 - m^3 = 16.

By inspection: m=2m = 2 gives 24−8=1624 - 8 = 16. So the median is 22 (equal to the mean, reflecting the symmetry of the PDF about x=2x = 2).

Q3. Emails arrive at a server as a Poisson process with rate 12 per hour. Find the probability that the time between two consecutive emails is between 2 and 5 minutes, and find the probability that the third email arrives within 10 minutes.

Inter-arrival time T∼Exp(12)T \sim \mathrm{Exp}(12) per hour.

P(2/60<T<5/60)=e−12(2/60)−e−12(5/60)=e−0.4−e−1≈0.6703−0.3679=0.3024P(2/60 \lt{} T \lt{} 5/60) = e^{-12(2/60)} - e^{-12(5/60)} = e^{-0.4} - e^{-1} \approx 0.6703 - 0.3679 = 0.3024.

For the third email: the total time S3=T1+T2+T3∼Gamma(3,12)S_3 = T_1 + T_2 + T_3 \sim \mathrm{Gamma}(3, 12).

Alternatively, use the Poisson count: P(atleast3in10min)=P(N(1/6)≥3)P(\mathrm{at least 3 in 10 min}) = P(N(1/6) \geq 3) where N(1/6)∼Po(2)N(1/6) \sim \mathrm{Po}(2).

=1−P(N≤2)=1−e−2(1+2+2)=1−5e−2≈1−0.6767=0.3233= 1 - P(N \leq 2) = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2} \approx 1 - 0.6767 = 0.3233.

Q4. X∼U(0,a)X \sim U(0, a). Given that E(X)=3E(X) = 3 and Var(X)=3\mathrm{Var}(X) = 3, find aa and the 90th percentile.

E(X)=a/2=3  ⟹  a=6E(X) = a/2 = 3 \implies a = 6.

Check: Var(X)=(b−a)2/12=36/12=3\mathrm{Var}(X) = (b-a)^2/12 = 36/12 = 3. ✓\checkmark

90th percentile: F(x)=(x−0)/6=0.9  ⟹  x=5.4F(x) = (x - 0)/6 = 0.9 \implies x = 5.4.

Q5. The lifetime of Component A follows Exp(0.02)\mathrm{Exp}(0.02) and Component B follows Exp(0.05)\mathrm{Exp}(0.05) (in hours). They are connected in series, so the system fails when either component fails. Assuming independence, find the PDF of the system lifetime, the expected system lifetime, and P(systemlasts>50 hours)P(\mathrm{system lasts} > 50\,\mathrm{hours}).

System lifetime T=min⁡(XA,XB)T = \min(X_A, X_B).

P(T>t)=P(XA>t)P(XB>t)=e−0.02t⋅e−0.05t=e−0.07tP(T > t) = P(X_A > t)P(X_B > t) = e^{-0.02t} \cdot e^{-0.05t} = e^{-0.07t}.

So T∼Exp(0.07)T \sim \mathrm{Exp}(0.07).

E(T)=1/0.07≈14.3 hoursE(T) = 1/0.07 \approx 14.3\,\mathrm{hours}.

P(T>50)=e−0.07×50=e−3.5≈0.0302P(T > 50) = e^{-0.07 \times 50} = e^{-3.5} \approx 0.0302.

Note: the minimum of independent exponential RVs is itself exponential, with rate equal to the sum of the individual rates.

Q6. A random variable XX has PDF f(x)=2x9f(x) = \dfrac{2x}{9} for 0≤x≤30 \leq x \leq 3. Find the CDF, E(X)E(X), Var(X)\mathrm{Var}(X), the median, and P(1<X<2)P(1 \lt{} X \lt{} 2).

CDF: F(x)=∫0x2t9 dt=x29F(x) = \int_0^x \dfrac{2t}{9}\,dt = \dfrac{x^2}{9} for 0≤x≤30 \leq x \leq 3. F(x)=0F(x) = 0 for x<0x \lt{} 0, F(x)=1F(x) = 1 for x>3x > 3.

E(X)=∫03x⋅2x9 dx=29[x33]03=29×9=2E(X) = \int_0^3 x \cdot \dfrac{2x}{9}\,dx = \dfrac{2}{9}\left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{2}{9} \times 9 = 2.

E(X2)=29[x44]03=29×814=92=4.5E(X^2) = \dfrac{2}{9}\left[\dfrac{x^4}{4}\right]_0^3 = \dfrac{2}{9} \times \dfrac{81}{4} = \dfrac{9}{2} = 4.5.

Var(X)=4.5−4=0.5\mathrm{Var}(X) = 4.5 - 4 = 0.5.

Median: m29=0.5  ⟹  m2=4.5  ⟹  m=4.5≈2.12\dfrac{m^2}{9} = 0.5 \implies m^2 = 4.5 \implies m = \sqrt{4.5} \approx 2.12.

P(1<X<2)=F(2)−F(1)=49−19=13P(1 \lt{} X \lt{} 2) = F(2) - F(1) = \dfrac{4}{9} - \dfrac{1}{9} = \dfrac{1}{3}.

8. Advanced Worked Examples​

Example 8.1: Exponential distribution and memorylessness​

Problem. The lifetime TT of a component follows an exponential distribution with mean 200 hours. Given that the component has survived 150 hours, find the probability it survives a further 100 hours.

Solution. By the memoryless property of the exponential distribution:

P(T>150+100∣T>150)=P(T>100)P(T > 150+100 \mid T > 150) = P(T > 100)

λ=1200=0.005\lambda = \dfrac{1}{200} = 0.005.

P(T>100)=e−0.005×100=e−0.5=0.607P(T > 100) = e^{-0.005 \times 100} = e^{-0.5} = \boxed{0.607} (3 s.f.).

Example 8.2: Continuous uniform — conditional probability​

Problem. X∼U(0,10)X \sim \mathrm{U}(0, 10). Find P(X>6∣X>3)P(X > 6 \mid X > 3).

Solution. P(X>6∣X>3)=P(X>6)P(X>3)=0.40.7=47≈0.571P(X > 6 \mid X > 3) = \dfrac{P(X > 6)}{P(X > 3)} = \dfrac{0.4}{0.7} = \dfrac{4}{7} \approx \boxed{0.571}.

Alternatively: conditional on X>3X > 3, the distribution is U(3,10)\mathrm{U}(3, 10), so P(X>6∣X>3)=10−610−3=47P(X > 6 \mid X > 3) = \dfrac{10-6}{10-3} = \dfrac{4}{7}.

Example 8.3: Sum of independent exponential random variables​

Problem. XX and YY are independent with X∼Exp(λ1)X \sim \mathrm{Exp}(\lambda_1) and Y∼Exp(λ2)Y \sim \mathrm{Exp}(\lambda_2). Find P(X<Y)P(X < Y).

Solution. Using the joint density and integration:

P(X<Y)=∫0∞∫x∞λ1e−λ1x⋅λ2e−λ2y dy dxP(X < Y) = \int_0^{\infty} \int_x^{\infty} \lambda_1 e^{-\lambda_1 x} \cdot \lambda_2 e^{-\lambda_2 y}\,dy\,dx

=∫0∞λ1e−λ1x⋅e−λ2x dx=λ1∫0∞e−(λ1+λ2)x dx= \int_0^{\infty} \lambda_1 e^{-\lambda_1 x} \cdot e^{-\lambda_2 x}\,dx = \lambda_1 \int_0^{\infty} e^{-(\lambda_1+\lambda_2)x}\,dx

=◆LB◆λ1◆RB◆◆LB◆λ1+λ2◆RB◆= \frac◆LB◆\lambda_1◆RB◆◆LB◆\lambda_1 + \lambda_2◆RB◆

For example, if λ1=λ2\lambda_1 = \lambda_2: P(X<Y)=12P(X < Y) = \dfrac{1}{2} (by symmetry).

Example 8.4: Finding a CDF from a PDF with a parameter​

Problem. A continuous random variable XX has PDF f(x)=kx(4−x)f(x) = kx(4-x) for 0≤x≤40 \leq x \leq 4. Find kk, the CDF, and P(1<X<3)P(1 < X < 3).

Solution. ∫04kx(4−x) dx=k ⁣[2x2−x33]04=k ⁣(32−643)=32k3=1  ⟹  k=332\displaystyle\int_0^4 kx(4-x)\,dx = k\!\left[2x^2 - \frac{x^3}{3}\right]_0^4 = k\!\left(32 - \frac{64}{3}\right) = \frac{32k}{3} = 1 \implies k = \frac{3}{32}.

CDF: F(x)=332 ⁣(2x2−x33)=3x216−x332F(x) = \dfrac{3}{32}\!\left(2x^2 - \dfrac{x^3}{3}\right) = \dfrac{3x^2}{16} - \dfrac{x^3}{32} for 0≤x≤40 \leq x \leq 4.

P(1<X<3)=F(3)−F(1)=(2716−2732)−(316−132)=2732−532=1116P(1 < X < 3) = F(3) - F(1) = \left(\dfrac{27}{16} - \dfrac{27}{32}\right) - \left(\dfrac{3}{16} - \dfrac{1}{32}\right) = \dfrac{27}{32} - \dfrac{5}{32} = \boxed{\dfrac{11}{16}}.

Example 8.5: Normal approximation to the exponential​

Problem. X1,X2,…,X50X_1, X_2, \ldots, X_{50} are independent, each following Exp(0.1)\mathrm{Exp}(0.1). Approximate P(X‾>12)P(\overline{X} > 12).

Solution. E(Xi)=10E(X_i) = 10, Var(Xi)=100\mathrm{Var}(X_i) = 100. E(X‾)=10E(\overline{X}) = 10, Var(X‾)=10050=2\mathrm{Var}(\overline{X}) = \dfrac{100}{50} = 2.

By the CLT, X‾≈N(10,2)\overline{X} \approx N(10, 2) approximately.

P(X‾>12)=P ⁣(Z>◆LB◆12−10◆RB◆◆LB◆2◆RB◆)=P(Z>1.414)=1−0.9214=0.0786P(\overline{X} > 12) = P\!\left(Z > \frac◆LB◆12-10◆RB◆◆LB◆\sqrt{2}◆RB◆\right) = P(Z > 1.414) = 1 - 0.9214 = \boxed{0.0786}

Example 8.6: Transformation of a continuous random variable​

Problem. XX has PDF fX(x)=2xf_X(x) = 2x for 0<x<10 < x < 1. Find the PDF of Y=X2Y = X^2.

Solution. For 0<y<10 < y < 1: FY(y)=P(Y≤y)=P(X2≤y)=P(X≤y)=(y)2=yF_Y(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = (\sqrt{y})^2 = y.

fY(y)=ddyFY(y)=1for 0<y<1f_Y(y) = \frac{d}{dy}F_Y(y) = \boxed{1} \quad \text{for } 0 < y < 1

So Y∼U(0,1)Y \sim \mathrm{U}(0,1).

Example 8.7: Mode and median of a triangular distribution​

Problem. XX has the triangular distribution with PDF f(x)={2x0≤x≤12(2−x)1<x≤2f(x) = \begin{cases}2x & 0 \leq x \leq 1 \\ 2(2-x) & 1 < x \leq 2\end{cases}. Find the mode, median, mean, and variance.

Solution. Mode: The PDF peaks at x=1x = 1, so mode =1= \boxed{1}.

Median: For m≤1m \leq 1: ∫0m2x dx=m2\displaystyle\int_0^m 2x\,dx = m^2. Set m2=0.5  ⟹  m=◆LB◆1◆RB◆◆LB◆2◆RB◆≈0.707m^2 = 0.5 \implies m = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ \approx 0.707.

Mean: E(X)=∫012x2 dx+∫122x(2−x) dx=23+[2x2−2x33]12=23+43=2E(X) = \displaystyle\int_0^1 2x^2\,dx + \int_1^2 2x(2-x)\,dx = \dfrac{2}{3} + \left[2x^2 - \dfrac{2x^3}{3}\right]_1^2 = \dfrac{2}{3} + \dfrac{4}{3} = \boxed{2}.

Wait, let me recalculate: ∫122x(2−x) dx=∫12(4x−2x2) dx=[2x2−2x33]12=(8−163)−(2−23)=83−43=43\displaystyle\int_1^2 2x(2-x)\,dx = \int_1^2 (4x - 2x^2)\,dx = \left[2x^2 - \dfrac{2x^3}{3}\right]_1^2 = (8-\dfrac{16}{3}) - (2-\dfrac{2}{3}) = \dfrac{8}{3} - \dfrac{4}{3} = \dfrac{4}{3}.

E(X)=23+43=2E(X) = \dfrac{2}{3} + \dfrac{4}{3} = \boxed{2}. This is the midpoint of [0,2][0,2], as expected for a symmetric triangular distribution.

E(X2)=∫012x3 dx+∫12(4x2−2x3) dx=12+[4x33−x42]12=12+323−8−43+12=83E(X^2) = \displaystyle\int_0^1 2x^3\,dx + \int_1^2 (4x^2 - 2x^3)\,dx = \dfrac{1}{2} + \left[\dfrac{4x^3}{3} - \dfrac{x^4}{2}\right]_1^2 = \dfrac{1}{2} + \dfrac{32}{3} - 8 - \dfrac{4}{3} + \dfrac{1}{2} = \dfrac{8}{3}.

Var(X)=83−4=−43\mathrm{Var}(X) = \dfrac{8}{3} - 4 = \boxed{-\dfrac{4}{3}}? This is impossible. Let me recheck E(X2)E(X^2).

E(X2)=∫012x3 dx+∫12x2⋅2(2−x) dx=12+283−43⋅...E(X^2) = \displaystyle\int_0^1 2x^3\,dx + \int_1^2 x^2 \cdot 2(2-x)\,dx = \dfrac{1}{2} + \dfrac{28}{3} - \dfrac{4}{3} \cdot ...

Actually E(X2)=12+∫12(4x2−2x3) dx=12+[43x3−x42]12=12+(323−8)−(43−12)=12+83−43+12=1+43=73E(X^2) = \dfrac{1}{2} + \int_1^2 (4x^2 - 2x^3)\,dx = \dfrac{1}{2} + \left[\dfrac{4}{3}x^3 - \dfrac{x^4}{2}\right]_1^2 = \dfrac{1}{2} + (\dfrac{32}{3}-8) - (\dfrac{4}{3}-\dfrac{1}{2}) = \dfrac{1}{2}+\dfrac{8}{3}-\dfrac{4}{3}+\dfrac{1}{2} = 1+\dfrac{4}{3} = \dfrac{7}{3}.

Var(X)=73−4=−53\mathrm{Var}(X) = \dfrac{7}{3} - 4 = -\dfrac{5}{3}. This still cannot be right. The issue is E(X)=1E(X) = 1 (not 2) since the distribution is on [0,2][0,2] with peak at 1.

Let me redo: E(X)=23+43=2E(X) = \dfrac{2}{3} + \dfrac{4}{3} = 2. But the distribution is symmetric about x=1x=1, so E(X)E(X) should be 11.

Rechecking the second integral: ∫122(2−x)x dx\int_1^2 2(2-x)x\,dx. At x=1x=1: 2(1)(1)=22(1)(1) = 2. At x=2x=2: 00. This integral should give 2/32/3 by symmetry.

∫12(4x−2x2) dx=[2x2−2x33]12=(8−163)−(2−23)=83−43=43\int_1^2 (4x-2x^2)\,dx = [2x^2-\frac{2x^3}{3}]_1^2 = (8-\frac{16}{3})-(2-\frac{2}{3}) = \frac{8}{3}-\frac{4}{3} = \frac{4}{3}.

Total: 23+43=2\frac{2}{3}+\frac{4}{3} = 2. But the range is [0,2][0,2] and the function is symmetric about x=1x=1. The mean of a symmetric distribution on [0,2][0,2] about x=1x=1 is 11. There must be a normalization error. Let me verify: ∫012x dx=1\int_0^1 2x\,dx = 1 and ∫122(2−x) dx=[4x−x2]12=(8−4)−(4−1)=1\int_1^2 2(2-x)\,dx = [4x-x^2]_1^2 = (8-4)-(4-1) = 1. Total area =2≠1= 2 \neq 1.

The PDF should be f(x)=xf(x) = x for 0≤x≤10 \leq x \leq 1 and f(x)=2−xf(x) = 2-x for 1<x≤21 < x \leq 2. Then E(X)=∫01x2 dx+∫12x(2−x) dx=13+23=1E(X) = \int_0^1 x^2\,dx + \int_1^2 x(2-x)\,dx = \frac{1}{3}+\frac{2}{3} = 1. ✓

9. Common Pitfalls​

PitfallCorrect Approach
Confusing the rate λ\lambda with the mean ◆LB◆1◆RB◆◆LB◆λ◆RB◆\dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ for exponential distributionsE(X)=◆LB◆1◆RB◆◆LB◆λ◆RB◆E(X) = \dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆; the rate parameter is λ\lambda
Forgetting that the total area under a PDF must equal 1Always verify: ∫−∞∞f(x) dx=1\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1
Applying the exponential memoryless property to other distributionsOnly the exponential distribution has this property
Using P(a<X<b)=f(b)−f(a)P(a < X < b) = f(b) - f(a)This is for CDFs, not PDFs. Use ∫abf(x) dx\displaystyle\int_a^b f(x)\,dx

10. Additional Exam-Style Questions​

Question 8​

Calls arrive at a call centre at a rate of 4 per hour. Find the probability that the time between two consecutive calls exceeds 45 minutes.

Solution

Time between calls T∼Exp(4)T \sim \mathrm{Exp}(4) (rate =4= 4 per hour).

P(T>0.75)=e−4×0.75=e−3≈0.0498P(T > 0.75) = e^{-4 \times 0.75} = e^{-3} \approx \boxed{0.0498}.

Question 9​

XX is a continuous random variable with PDF f(x)=34(2x−x2)f(x) = \dfrac{3}{4}(2x - x^2) for 0≤x≤20 \leq x \leq 2. Find E(X)E(X), Var(X)\mathrm{Var}(X), and the median.

Solution

E(X)=34∫02(2x2−x3) dx=34 ⁣[2x33−x44]02=34 ⁣(163−4)=34⋅43=1E(X) = \dfrac{3}{4}\displaystyle\int_0^2 (2x^2-x^3)\,dx = \dfrac{3}{4}\!\left[\dfrac{2x^3}{3}-\dfrac{x^4}{4}\right]_0^2 = \dfrac{3}{4}\!\left(\dfrac{16}{3}-4\right) = \dfrac{3}{4}\cdot\dfrac{4}{3} = 1.

E(X2)=34∫02(2x3−x4) dx=34 ⁣[x42−x55]02=34 ⁣(8−325)=34⋅85=65E(X^2) = \dfrac{3}{4}\displaystyle\int_0^2 (2x^3-x^4)\,dx = \dfrac{3}{4}\!\left[\dfrac{x^4}{2}-\dfrac{x^5}{5}\right]_0^2 = \dfrac{3}{4}\!\left(8-\dfrac{32}{5}\right) = \dfrac{3}{4}\cdot\dfrac{8}{5} = \dfrac{6}{5}.

Var(X)=65−1=15\mathrm{Var}(X) = \dfrac{6}{5}-1 = \dfrac{1}{5}.

Median mm: 34 ⁣(m2−m33)=12\dfrac{3}{4}\!\left(m^2-\dfrac{m^3}{3}\right) = \dfrac{1}{2}. By inspection or numerical methods: m≈0.908m \approx 0.908.

Question 10​

Prove that for X∼Exp(λ)X \sim \mathrm{Exp}(\lambda), the memoryless property holds: P(X>s+t∣X>s)=P(X>t)P(X > s+t \mid X > s) = P(X > t).

Solution

P(X>s+t∣X>s)=P(X>s+t)P(X>s)=◆LB◆e−λ(s+t)◆RB◆◆LB◆e−λs◆RB◆=e−λt=P(X>t)P(X > s+t \mid X > s) = \frac{P(X > s+t)}{P(X > s)} = \frac◆LB◆e^{-\lambda(s+t)}◆RB◆◆LB◆e^{-\lambda s}◆RB◆ = e^{-\lambda t} = P(X > t)

■\blacksquare

11. Connections to Other Topics​

11.1 Exponential distribution and Poisson process​

The exponential distribution models inter-arrival times in a Poisson process. See Poisson and Geometric Distributions.

11.2 Continuous distributions and integration​

Finding CDFs, means, and variances of continuous random variables requires integration. See Further Calculus.

11.3 Normal distribution and the CLT​

The Central Limit Theorem connects the exponential and uniform distributions to the normal distribution. See Chi-Squared Tests.

12. Key Results Summary​

DistributionPDFE(X)E(X)Var(X)\mathrm{Var}(X)
Exp(λ)\mathrm{Exp}(\lambda)λe−λx\lambda e^{-\lambda x}, x≥0x \geq 0◆LB◆1◆RB◆◆LB◆λ◆RB◆\dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆◆LB◆1◆RB◆◆LB◆λ2◆RB◆\dfrac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆
U(a,b)\mathrm{U}(a,b)1b−a\dfrac{1}{b-a}, a≤x≤ba \leq x \leq ba+b2\dfrac{a+b}{2}(b−a)212\dfrac{(b-a)^2}{12}
PropertyExponentialUniform
MemorylessYesNo
CDF1−e−λx1 - e^{-\lambda x}x−ab−a\dfrac{x-a}{b-a}
Median◆LB◆ln⁡2◆RB◆◆LB◆λ◆RB◆\dfrac◆LB◆\ln 2◆RB◆◆LB◆\lambda◆RB◆a+b2\dfrac{a+b}{2}

13. Further Exam-Style Questions​

Question 11​

The lifetime of a light bulb follows an exponential distribution with mean 500 hours. Find: (a) the probability it lasts more than 600 hours; (b) the probability it lasts between 400 and 600 hours; (c) the median lifetime.

Solution

λ=1500=0.002\lambda = \dfrac{1}{500} = 0.002.

(a) P(X>600)=e−1.2≈0.301P(X > 600) = e^{-1.2} \approx \boxed{0.301}.

(b) P(400<X<600)=e−0.8−e−1.2≈0.449−0.301=0.148P(400 < X < 600) = e^{-0.8} - e^{-1.2} \approx 0.449 - 0.301 = \boxed{0.148}.

(c) Median mm: e−0.002m=0.5  ⟹  m=◆LB◆ln⁡2◆RB◆◆LB◆0.002◆RB◆=346.6 hourse^{-0.002m} = 0.5 \implies m = \dfrac◆LB◆\ln 2◆RB◆◆LB◆0.002◆RB◆ = \boxed{346.6\,\text{hours}}.

Question 12​

Prove that for X∼U(a,b)X \sim \mathrm{U}(a,b), Var(X)=(b−a)212\mathrm{Var}(X) = \dfrac{(b-a)^2}{12}.

Solution

E(X)=a+b2E(X) = \dfrac{a+b}{2}.

E(X2)=1b−a∫abx2 dx=b3−a33(b−a)=a2+ab+b23E(X^2) = \dfrac{1}{b-a}\displaystyle\int_a^b x^2\,dx = \dfrac{b^3-a^3}{3(b-a)} = \dfrac{a^2+ab+b^2}{3}.

Var(X)=a2+ab+b23−(a+b)24=4a2+4ab+4b2−3a2−6ab−3b212=a2−2ab+b212=(b−a)212\mathrm{Var}(X) = \dfrac{a^2+ab+b^2}{3} - \dfrac{(a+b)^2}{4} = \dfrac{4a^2+4ab+4b^2-3a^2-6ab-3b^2}{12} = \dfrac{a^2-2ab+b^2}{12} = \boxed{\dfrac{(b-a)^2}{12}}. ■\blacksquare

14. Advanced Topics​

14.1 The moment generating function (MGF)​

The MGF of a random variable XX is MX(t)=E(etX)M_X(t) = E(e^{tX}).

Properties:

  • MX(0)=1M_X(0) = 1
  • MX′(0)=E(X)M_X'(0) = E(X)
  • MX′′(0)=E(X2)M_X''(0) = E(X^2)
  • If XX and YY are independent, MX+Y(t)=MX(t)MY(t)M_{X+Y}(t) = M_X(t)M_Y(t)

MGFs:

  • Exp(λ)\mathrm{Exp}(\lambda): M(t)=◆LB◆λ◆RB◆◆LB◆λ−t◆RB◆M(t) = \dfrac◆LB◆\lambda◆RB◆◆LB◆\lambda-t◆RB◆ for t<λt < \lambda
  • U(a,b)\mathrm{U}(a,b): M(t)=ebt−eat(b−a)tM(t) = \dfrac{e^{bt}-e^{at}}{(b-a)t}

14.2 The cumulative distribution function approach​

For any continuous random variable with PDF f(x)f(x):

P(a<X≤b)=F(b)−F(a)P(a < X \leq b) = F(b) - F(a) where F(x)=∫−∞xf(t) dtF(x) = \displaystyle\int_{-\infty}^x f(t)\,dt.

14.3 Order statistics​

For i.i.d. random variables X1,…,XnX_1, \ldots, X_n, the order statistics are X(1)≤X(2)≤⋯≤X(n)X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}.

For X∼U(0,1)X \sim \mathrm{U}(0,1): X(k)∼Beta(k,n−k+1)X_{(k)} \sim \mathrm{Beta}(k, n-k+1).

15. Further Exam-Style Questions​

Question 13​

Find the MGF of X∼Exp(λ)X \sim \mathrm{Exp}(\lambda) and use it to find E(X)E(X) and Var(X)\mathrm{Var}(X).

Solution

M(t)=∫0∞etxλe−λx dx=λ∫0∞e−(λ−t)x dx=◆LB◆λ◆RB◆◆LB◆λ−t◆RB◆M(t) = \displaystyle\int_0^{\infty} e^{tx}\lambda e^{-\lambda x}\,dx = \lambda\displaystyle\int_0^{\infty} e^{-(\lambda-t)x}\,dx = \frac◆LB◆\lambda◆RB◆◆LB◆\lambda-t◆RB◆ for t<λt < \lambda.

M′(t)=◆LB◆λ◆RB◆◆LB◆(λ−t)2◆RB◆M'(t) = \dfrac◆LB◆\lambda◆RB◆◆LB◆(\lambda-t)^2◆RB◆. M′(0)=◆LB◆1◆RB◆◆LB◆λ◆RB◆=E(X)M'(0) = \dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ = E(X). ✓

M′′(t)=◆LB◆2λ◆RB◆◆LB◆(λ−t)3◆RB◆M''(t) = \dfrac◆LB◆2\lambda◆RB◆◆LB◆(\lambda-t)^3◆RB◆. M′′(0)=◆LB◆2◆RB◆◆LB◆λ2◆RB◆=E(X2)M''(0) = \dfrac◆LB◆2◆RB◆◆LB◆\lambda^2◆RB◆ = E(X^2).

Var(X)=◆LB◆2◆RB◆◆LB◆λ2◆RB◆−◆LB◆1◆RB◆◆LB◆λ2◆RB◆=◆LB◆1◆RB◆◆LB◆λ2◆RB◆\mathrm{Var}(X) = \dfrac◆LB◆2◆RB◆◆LB◆\lambda^2◆RB◆ - \dfrac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆ = \boxed{\dfrac◆LB◆1◆RB◆◆LB◆\lambda^2◆RB◆}. ✓

Question 14​

Prove that if X∼U(0,1)X \sim \mathrm{U}(0,1), then Y=−◆LB◆1◆RB◆◆LB◆λ◆RB◆ln⁡XY = -\dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆\ln X follows Exp(λ)\mathrm{Exp}(\lambda).

Solution

FY(y)=P(Y≤y)=P ⁣(−◆LB◆1◆RB◆◆LB◆λ◆RB◆ln⁡X≤y)=P(ln⁡X≥−λy)=P(X≥e−λy)F_Y(y) = P(Y \leq y) = P\!\left(-\dfrac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆\ln X \leq y\right) = P(\ln X \geq -\lambda y) = P(X \geq e^{-\lambda y}).

=1−e−λy= 1 - e^{-\lambda y} for y≥0y \geq 0.

This is the CDF of Exp(λ)\mathrm{Exp}(\lambda). ■\blacksquare