Exponential Distribution and Continuous Random Variables

This topic explores fundamental concepts that shape our understanding of the world.

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	Paper	Notes
AQA	Paper 2	Continuous RVs; limited exponential coverage
Edexcel	S3, S4	Exponential distribution in S4; continuous RVs in S3
OCR (A)	Paper 2	Continuous RVs and exponential
CIE (9231)	S2	Both continuous RVs and exponential covered

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

$f(x) \geq 0 \quad \mathrm{for all } x, \qquad \int_{-\infty}^{\infty}f(x)\,dx = 1$

Probabilities are found by integration:

$P(a \leq X \leq b) = \int_a^b f(x)\,dx$

### 1.2 Cumulative distribution function

Definition. The cumulative distribution function (CDF) is

$F(x) = P(X \leq x) = \int_{-\infty}^{x}f(t)\,dt$

Properties:

• $F(-\infty) = 0$, $F(\infty) = 1$
• $F$ is non-decreasing
• $f(x) = F"(x)$ where $F$ is differentiable
• $P(a < X \leq b) = F(b) - F(a)$

Definition. The expected value of a continuous random variable $X$ is

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

For a function $g(X)$:

$E(g(X)) = \int_{-\infty}^{\infty}g(x)\,f(x)\,dx$

Definition.

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

The linear properties carry over from the discrete case:

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

Definition. The median $m$ satisfies $F(m) = 0.5$I.e., $\int_{-\infty}^{m}f(x)\,dx = 0.5$.

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

Definition. The lower quartile $Q_1$ satisfies $F(Q_1) = 0.25$ and the upper quartile $Q_3$ satisfies $F(Q_3) = 0.75$.

The interquartile range is $\mathrm{IQR} = Q_3 - Q_1$.

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Definition. A continuous random variable $X$ follows an exponential distribution with rate Parameter $\lambda$ (where $\lambda > 0$), written $X \sim \mathrm{Exp}(\lambda)$If

$\boxed{f(x) = \lambda e^{-\lambda x}, \quad x \geq 0}$

And $f(x) = 0$ for $x < 0$.

$F(x) = P(X \leq x) = \int_0^x \lambda e^{-\lambda t}\,dt = \left[-e^{-\lambda t}\right]_0^x = 1 - e^{-\lambda x}$

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

$$\begin{aligned} E(X) &= \int_0^{\infty}x\cdot\lambda e^{-\lambda x}\,dx \end{aligned}$$

Using integration by parts with $u = x$, $dv = \lambda e^{-\lambda x}\,dx$:

$du = dx$, $v = -e^{-\lambda x}$.

$$\begin{aligned} E(X) &= \left[-xe^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}e^{-\lambda x}\,dx \\ &= \lim_{x\to\infty}(-xe^{-\lambda x}) + 0 + \left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^{\infty} \\ &= 0 + \frac{1}{\lambda} = \frac{1}{\lambda} \quad \blacksquare \end{aligned}$$

Note: $\lim_{x\to\infty}xe^{-\lambda x} = 0$ by L’Hôpital’s rule (exponential decay dominates).

First compute $E(X^2)$:

$E(X^2) = \int_0^{\infty}x^2\cdot\lambda e^{-\lambda x}\,dx$

Integration by parts twice with $u = x^2$, $dv = \lambda e^{-\lambda x}\,dx$:

$du = 2x\,dx$, $v = -e^{-\lambda x}$.

$$\begin{aligned} E(X^2) &= \left[-x^2 e^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}2x\,e^{-\lambda x}\,dx \\ &= 0 + 2\cdot\frac{1}{\lambda}\cdot E(X) = \frac{2}{\lambda}\cdot\frac{1}{\lambda} = \frac{2}{\lambda^2} \end{aligned}$$

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2} \quad \blacksquare$

$\boxed{E(X) = \frac{1}{\lambda}, \qquad \mathrm{Var}(X) = \frac{1}{\lambda^2}, \qquad \sigma = \frac{1}{\lambda}}$

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

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

$$\begin{aligned} P(X > s + t \mid X > s) &= \frac{P(X > s+t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} \\ &= e^{-\lambda t} = P(X > t) \quad \blacksquare \end{aligned}$$

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

### 2.6 Link to Poisson processes

A Poisson process with rate $\lambda$ satisfies:

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

Proof sketch that inter-arrival times are exponential. Let $T$ be the time until the first Event. $P(T > t) = P(\mathrm{no events in }[0,t]) = P(N(t) = 0) = \dfrac{e^{-\lambda t}(\lambda t)^0}{0!} = e^{-\lambda t}$.

So $P(T \leq t) = 1 - e^{-\lambda t}$Which is the CDF of $\mathrm{Exp}(\lambda)$. $\blacksquare$

For the exponential distribution:

$F(x) = 1 - e^{-\lambda x} = p \implies x = -\frac{1}{\lambda}\ln(1-p)$

The median is $x_{0.5} = -\dfrac{\ln(0.5)}{\lambda} = \dfrac{\ln 2}{\lambda}$.

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

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

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

Median $= \dfrac{\ln 2}{0.5} = 2\ln 2 \approx 1.386$.

Example. $X$ has PDF $f(x) = \dfrac{3}{8}x^2$ for $0 \leq x \leq 2$ and $f(x) = 0$ otherwise.

Verify: $\int_0^2 \dfrac{3}{8}x^2\,dx = \dfrac{3}{8}\cdot\dfrac{8}{3} = 1$. $\checkmark$

$E(X) = \int_0^2 x\cdot\dfrac{3}{8}x^2\,dx = \dfrac{3}{8}\int_0^2 x^3\,dx = \dfrac{3}{8}\cdot\dfrac{16}{4} = \dfrac{3}{2}$.

$E(X^2) = \dfrac{3}{8}\int_0^2 x^4\,dx = \dfrac{3}{8}\cdot\dfrac{32}{5} = \dfrac{12}{5}$.

$\mathrm{Var}(X) = \dfrac{12}{5} - \left(\dfrac{3}{2}\right)^2 = \dfrac{12}{5} - \dfrac{9}{4} = \dfrac{3}{20}$.

Median: $\int_0^m \dfrac{3}{8}x^2\,dx = 0.5 \implies \dfrac{m^3}{8} = 0.5 \implies m^3 = 4 \implies m = \sqrt[3]{4} \approx 1.587$.

Example. $f(x) = 2x$ for $0 \leq x \leq 1$. Find $F(x)$.

For $0 \leq x \leq 1$: $F(x) = \int_0^x 2t\,dt = x^2$.

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

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

Under $H_0$: $E(X) = 1/\lambda = 200$ hours. Since $n$ is large, use the approximate normal Distribution of $\bar{X}$:

$\bar{X} \sim N\!\left(\frac{1}{\lambda}, \frac{1}{n\lambda^2}\right) = N(200, 4000)$ Approximately.

$z = \dfrac{420 - 200}{\sqrt{4000}} = \dfrac{220}{63.25} = 3.48$.

$|z| = 3.48 > 1.96$So reject $H_0$.

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Problem 1

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

Solution 1

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

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

90th percentile: $F(x) = 0.9 \implies 1 - e^{-2x} = 0.9 \implies x = -\dfrac{\ln(0.1)}{2} \approx 1.151$.

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

Problem 2

A continuous random variable $X$ has PDF $f(x) = \dfrac{3x^2}{8}$ for $0 \leq x \leq 2$. Find $E(X)$, $\mathrm{Var}(X)$And the median.

Solution 2

$E(X) = \int_0^2 x\cdot\dfrac{3x^2}{8}\,dx = \dfrac{3}{8}\cdot\left[\dfrac{x^4}{4}\right]_0^2 = \dfrac{3}{8}\cdot 4 = 1.5$.

$E(X^2) = \int_0^2 x^2\cdot\dfrac{3x^2}{8}\,dx = \dfrac{3}{8}\cdot\left[\dfrac{x^5}{5}\right]_0^2 = \dfrac{3}{8}\cdot\dfrac{32}{5} = 2.4$.

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

Median: $\int_0^m \dfrac{3x^2}{8}\,dx = 0.5 \implies \dfrac{m^3}{8} = 0.5 \implies m = \sqrt[3]{4} \approx 1.587$.

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

Problem 3

Prove the memoryless property of the exponential distribution.

Solution 3

$P(X > s+t \mid X > s) = \dfrac{P(X > s+t)}{P(X > s)} = \dfrac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t)$. $\blacksquare$

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

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

Problem 4

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

Solution 4

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

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

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

Problem 5

$X$ has PDF $f(x) = \dfrac{1}{2}x$ for $0 \leq x \leq 2$. Find the CDF, $E(X)$And $\mathrm{Var}(X)$.

Solution 5

CDF: $F(x) = \int_0^x \dfrac{t}{2}\,dt = \dfrac{x^2}{4}$ for $0 \leq x \leq 2$. $F(x) = 0$ for $x < 0$, $F(x) = 1$ for $x > 2$.

$E(X) = \int_0^2 x\cdot\dfrac{x}{2}\,dx = \dfrac{1}{2}\left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{4}{3}$.

$E(X^2) = \dfrac{1}{2}\left[\dfrac{x^4}{4}\right]_0^2 = \dfrac{1}{2}\cdot 4 = 2$.

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

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

Problem 6

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

Solution 6

By the memoryless property: $P(X > 500+200 \mid X > 500) = P(X > 200) = e^{-0.01 \times 200} = e^{-2} \approx 0.1353$.

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

Problem 7

A continuous random variable $X$ has CDF $F(x) = \dfrac{x^3}{27}$ for $0 \leq x \leq 3$. Find the PDF, $E(X)$And the upper quartile.

Solution 7

PDF: $f(x) = F'(x) = \dfrac{x^2}{9}$ for $0 \leq x \leq 3$.

$E(X) = \int_0^3 x\cdot\dfrac{x^2}{9}\,dx = \dfrac{1}{9}\left[\dfrac{x^4}{4}\right]_0^3 = \dfrac{1}{9}\cdot\dfrac{81}{4} = \dfrac{9}{4} = 2.25$.

Upper quartile: $F(Q_3) = 0.75 \implies \dfrac{Q_3^3}{27} = 0.75 \implies Q_3^3 = 20.25 \implies Q_3 \approx 2.725$.

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

Problem 8

Prove that $E(X) = 1/\lambda$ for $X \sim \mathrm{Exp}(\lambda)$Using integration by parts.

Solution 8

$E(X) = \int_0^{\infty}x\lambda e^{-\lambda x}\,dx$.

Let u = x$$dv = \lambda e^{-\lambda x}\,dxSo du = dx$$v = -e^{-\lambda x}.

$E(X) = \left[-xe^{-\lambda x}\right]_0^{\infty} + \int_0^{\infty}e^{-\lambda x}\,dx = 0 + \left[-\dfrac{1}{\lambda}e^{-\lambda x}\right]_0^{\infty} = \dfrac{1}{\lambda}$. $\blacksquare$

If you get this wrong, revise: Proof that $E(X) = \frac{1}{\lambda}$ — Section 2.3.

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.

Solution 9

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

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

$\approx 0.6065 - 0.2231 = 0.3834$.

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

Problem 10

$X$ has PDF $f(x) = 4x^3$ for $0 \leq x \leq 1$. Find $P(X > 0.5)$$E(X)$$\mathrm{Var}(X)$And the mode.

Solution 10

$P(X > 0.5) = \int_{0.5}^1 4x^3\,dx = \left[x^4\right]_{0.5}^1 = 1 - 0.0625 = 0.9375$.

$E(X) = \int_0^1 4x^4\,dx = \left[\dfrac{4x^5}{5}\right]_0^1 = \dfrac{4}{5}$.

$E(X^2) = \int_0^1 4x^5\,dx = \left[\dfrac{4x^6}{6}\right]_0^1 = \dfrac{2}{3}$.

$\mathrm{Var}(X) = \dfrac{2}{3} - \left(\dfrac{4}{5}\right)^2 = \dfrac{2}{3} - \dfrac{16}{25} = \dfrac{50 - 48}{75} = \dfrac{2}{75}$.

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

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

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

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

By definition of conditional probability:

$P(X > s + t \mid X > s) = \frac{P(X > s + t \,\cap\, X > s)}{P(X > s)} = \frac{P(X > s + t)}{P(X > s)}$

Since $X > s + t$ implies $X > s$.

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

$\frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t) \quad \blacksquare$

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

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

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

This is Cauchy’s functional equation. Since $G$ is non-increasing and $0 \leq G \leq 1$The only Solutions are:

$G(t) = e^{-\lambda t}$

For some $\lambda \geq 0$. Since $G$ is non-trivial (not identically 1), $\lambda > 0$. Therefore:

$P(X \leq t) = 1 - e^{-\lambda t}$

Which is the CDF of $\mathrm{Exp}(\lambda)$. $\blacksquare$

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).

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

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

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

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

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

Therefore $P(T \leq t) = 1 - e^{-\lambda t}$Which is the CDF of $\mathrm{Exp}(\lambda)$. $\blacksquare$

If $T_1, T_2, \ldots, T_n$ are $n$ independent inter-arrival times, each $\sim \mathrm{Exp}(\lambda)$Then the total time until the $n$-th event is:

$S_n = T_1 + T_2 + \cdots + T_n \sim \mathrm{Gamma}(n, \lambda)$

This connects the exponential to the gamma distribution.

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.

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

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

$N(0.5) \sim \mathrm{Po}(2.5)$. $P(N \geq 3) = 1 - P(N \leq 2) = 1 - e^{-2.5}(1 + 2.5 + 2.5^2/2) = 1 - 0.0821 \times 9.125 \approx 1 - 0.749 = 0.251$.

(c) Find the median inter-arrival time.

Median $= \dfrac{\ln 2}{\lambda} = \dfrac{\ln 2}{5} \approx 0.139\,\mathrm{hours} \approx 8.3\,\mathrm{minutes}$.

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

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

And $f(x) = 0$ otherwise.

$E(X) = \int_a^b x \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a}\left[\frac{x^2}{2}\right]_a^b = \frac{b^2 - a^2}{2(b-a)} = \frac{(b-a)(b+a)}{2(b-a)} = \frac{a+b}{2} \quad \blacksquare$

$E(X^2) = \int_a^b x^2 \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a}\left[\frac{x^3}{3}\right]_a^b = \frac{b^3 - a^3}{3(b-a)} = \frac{a^2 + ab + b^2}{3}$

$\mathrm{Var}(X) = \frac{a^2 + ab + b^2}{3} - \frac{(a+b)^2}{4} = \frac{4(a^2 + ab + b^2) - 3(a+b)^2}{12}$

$= \frac{4a^2 + 4ab + 4b^2 - 3a^2 - 6ab - 3b^2}{12} = \frac{a^2 - 2ab + b^2}{12}$

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

$F(x) = \begin{cases} 0 & x \lt a \\ \dfrac{x - a}{b - a} & a \leq x \leq b \\ 1 & x > b \end{cases}$

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A mixture distribution arises when a random variable is selected from one of several Sub-populations. If $X$ comes from distribution 1 with probability $p$ and from distribution 2 with Probability $1-p$:

$f(x) = p\,f_1(x) + (1-p)\,f_2(x)$

$E(X) = p\,E(X_1) + (1-p)\,E(X_2)$

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

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

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

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

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

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

$\approx 0.2575 + 0.2456 = 0.5031$.

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The PDF $f(x)$ gives the density of probability at $x$. It is not a probability itself — $f(x)$ Can be greater than 1. The CDF $F(x)$ gives the accumulated probability up to $x$And always Satisfies $0 \leq F(x) \leq 1$.

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

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

For a continuous RV, $P(X = a) = 0$ 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).

$X \sim \mathrm{Exp}(\lambda)$ has mean $1/\lambda$. A common error is to confuse $\lambda$ (the Rate) with the mean. If the mean lifetime is 200 hours, then $\lambda = 1/200 = 0.005$Not $\lambda = 200$.

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

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

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

$P(X > 2) = e^{-2\lambda} = 0.3 \implies -2\lambda = \ln(0.3) \implies \lambda = \dfrac{-\ln(0.3)}{2} = \dfrac{1.204}{2} = 0.602$.

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

80th percentile: $F(x) = 0.8 \implies 1 - e^{-0.602x} = 0.8 \implies e^{-0.602x} = 0.2 \implies x = \dfrac{-\ln(0.2)}{0.602} = \dfrac{1.609}{0.602} \approx 2.673$.

Q2. A continuous random variable $X$ has PDF $f(x) = kx(4-x)$ for $0 \leq x \leq 4$. Find $k$, $E(X)$, $\mathrm{Var}(X)$And the median.

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

$E(X) = \dfrac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \dfrac{3}{32}\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_0^4 = \dfrac{3}{32}\left(\dfrac{256}{3} - 64\right) = \dfrac{3}{32}\left(\dfrac{256 - 192}{3}\right) = \dfrac{64}{32} = 2$.

$E(X^2) = \dfrac{3}{32}\int_0^4 (4x^3 - x^4)\,dx = \dfrac{3}{32}\left[x^4 - \dfrac{x^5}{5}\right]_0^4 = \dfrac{3}{32}(256 - 204.8) = \dfrac{3 \times 51.2}{32} = 4.8$.

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

Median: $\dfrac{3}{32}\int_0^m (4x - x^2)\,dx = 0.5 \implies \dfrac{3}{32}\left(2m^2 - \dfrac{m^3}{3}\right) = 0.5$.

$2m^2 - \dfrac{m^3}{3} = \dfrac{16}{3} \implies 6m^2 - m^3 = 16$.

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

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 $T \sim \mathrm{Exp}(12)$ per hour.

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

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

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

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

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

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

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

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

Q5. The lifetime of Component A follows $\mathrm{Exp}(0.02)$ and Component B follows $\mathrm{Exp}(0.05)$ (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 $P(\mathrm{system lasts} > 50\,\mathrm{hours})$.

System lifetime $T = \min(X_A, X_B)$.

$P(T > t) = P(X_A > t)P(X_B > t) = e^{-0.02t} \cdot e^{-0.05t} = e^{-0.07t}$.

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

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

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

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

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

CDF: $F(x) = \int_0^x \dfrac{2t}{9}\,dt = \dfrac{x^2}{9}$ for $0 \leq x \leq 3$. $F(x) = 0$ for $x \lt 0$, $F(x) = 1$ for $x > 3$.

$E(X) = \int_0^3 x \cdot \dfrac{2x}{9}\,dx = \dfrac{2}{9}\left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{2}{9} \times 9 = 2$.

$E(X^2) = \dfrac{2}{9}\left[\dfrac{x^4}{4}\right]_0^3 = \dfrac{2}{9} \times \dfrac{81}{4} = \dfrac{9}{2} = 4.5$.

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

Median: $\dfrac{m^2}{9} = 0.5 \implies m^2 = 4.5 \implies m = \sqrt{4.5} \approx 2.12$.

$P(1 \lt X \lt 2) = F(2) - F(1) = \dfrac{4}{9} - \dfrac{1}{9} = \dfrac{1}{3}$.

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Problem. The lifetime $T$ 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:

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

$\lambda = \dfrac{1}{200} = 0.005$.

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

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

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

Alternatively: conditional on $X > 3$The distribution is $\mathrm{U}(3, 10)$So $P(X > 6 \mid X > 3) = \dfrac{10-6}{10-3} = \dfrac{4}{7}$.

Problem. $X$ and $Y$ are independent with $X \sim \mathrm{Exp}(\lambda_1)$ and $Y \sim \mathrm{Exp}(\lambda_2)$. Find $P(X < Y)$.

Solution. Using the joint density and integration:

$P(X < Y) = \int_0^{\infty} \int_x^{\infty} \lambda_1 e^{-\lambda_1 x} \cdot \lambda_2 e^{-\lambda_2 y}\,dy\,dx$

$= \int_0^{\infty} \lambda_1 e^{-\lambda_1 x} \cdot e^{-\lambda_2 x}\,dx = \lambda_1 \int_0^{\infty} e^{-(\lambda_1+\lambda_2)x}\,dx$

$= \frac{\lambda_1}{\lambda_1 + \lambda_2}$

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

Problem. A continuous random variable $X$ has PDF $f(x) = kx(4-x)$ for $0 \leq x \leq 4$. Find $k$The CDF, and $P(1 < X < 3)$.

Solution. $\displaystyle\int_0^4 kx(4-x)\,dx = k\!\left[2x^2 - \frac{x^3}{3}\right]_0^4 = k\!\left(32 - \frac{64}{3}\right) = \frac{32k}{3} = 1 \implies k = \frac{3}{32}$.

CDF: $F(x) = \dfrac{3}{32}\!\left(2x^2 - \dfrac{x^3}{3}\right) = \dfrac{3x^2}{16} - \dfrac{x^3}{32}$ For $0 \leq x \leq 4$.

$P(1 < X < 3) = F(3) - F(1) = \left(\dfrac{27}{16} - \dfrac{27}{32}\right) - \left(\dfrac{3}{16} - \dfrac{1}{32}\right) = \dfrac{27}{32} - \dfrac{5}{32} = \boxed{\dfrac{11}{16}}$.

Problem. $X_1, X_2, \ldots, X_{50}$ are independent, each following $\mathrm{Exp}(0.1)$. Approximate $P(\overline{X} > 12)$.

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

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

$P(\overline{X} > 12) = P\!\left(Z > \frac{12-10}{\sqrt{2}}\right) = P(Z > 1.414) = 1 - 0.9214 = \boxed{0.0786}$

Problem. $X$ has PDF $f_X(x) = 2x$ for $0 < x < 1$. Find the PDF of $Y = X^2$.

Solution. For $0 < y < 1$: $F_Y(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = (\sqrt{y})^2 = y$.

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

So $Y \sim \mathrm{U}(0,1)$.

Problem. $X$ has the triangular distribution with PDF $f(x) = \begin{cases}2x & 0 \leq x \leq 1 \\ 2(2-x) & 1 < x \leq 2\end{cases}$. Find the mode, Median, mean, and variance.

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

Median: For $m \leq 1$: $\displaystyle\int_0^m 2x\,dx = m^2$. Set $m^2 = 0.5 \implies m = \dfrac{1}{\sqrt{2}} \approx 0.707$.

Mean: $E(X) = \displaystyle\int_0^1 2x^2\,dx + \int_1^2 2x(2-x)\,dx = \dfrac{2}{3} + \left[2x^2 - \dfrac{2x^3}{3}\right]_1^2 = \dfrac{2}{3} + \dfrac{4}{3} = \boxed{2}$.

Wait, let me recalculate: $\displaystyle\int_1^2 2x(2-x)\,dx = \int_1^2 (4x - 2x^2)\,dx = \left[2x^2 - \dfrac{2x^3}{3}\right]_1^2 = (8-\dfrac{16}{3}) - (2-\dfrac{2}{3}) = \dfrac{8}{3} - \dfrac{4}{3} = \dfrac{4}{3}$.

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

$E(X^2) = \displaystyle\int_0^1 2x^3\,dx + \int_1^2 (4x^2 - 2x^3)\,dx = \dfrac{1}{2} + \left[\dfrac{4x^3}{3} - \dfrac{x^4}{2}\right]_1^2 = \dfrac{1}{2} + \dfrac{32}{3} - 8 - \dfrac{4}{3} + \dfrac{1}{2} = \dfrac{8}{3}$.

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

$E(X^2) = \displaystyle\int_0^1 2x^3\,dx + \int_1^2 x^2 \cdot 2(2-x)\,dx = \dfrac{1}{2} + \dfrac{28}{3} - \dfrac{4}{3} \cdot ...$

Actually $E(X^2) = \dfrac{1}{2} + \int_1^2 (4x^2 - 2x^3)\,dx = \dfrac{1}{2} + \left[\dfrac{4}{3}x^3 - \dfrac{x^4}{2}\right]_1^2 = \dfrac{1}{2} + (\dfrac{32}{3}-8) - (\dfrac{4}{3}-\dfrac{1}{2}) = \dfrac{1}{2}+\dfrac{8}{3}-\dfrac{4}{3}+\dfrac{1}{2} = 1+\dfrac{4}{3} = \dfrac{7}{3}$.

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

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

Rechecking the second integral: $\int_1^2 2(2-x)x\,dx$. At $x=1$: $2(1)(1) = 2$. At $x=2$: $0$. This Integral should give $2/3$ by symmetry.

$\int_1^2 (4x-2x^2)\,dx = [2x^2-\frac{2x^3}{3}]_1^2 = (8-\frac{16}{3})-(2-\frac{2}{3}) = \frac{8}{3}-\frac{4}{3} = \frac{4}{3}$.

Total: $\frac{2}{3}+\frac{4}{3} = 2$. But the range is $[0,2]$ and the function is symmetric about $x=1$. The mean of a symmetric distribution on $[0,2]$ about $x=1$ is $1$. There must be a Normalization error. Let me verify: $\int_0^1 2x\,dx = 1$ and $\int_1^2 2(2-x)\,dx = [4x-x^2]_1^2 = (8-4)-(4-1) = 1$. Total area $= 2 \neq 1$.

The PDF should be $f(x) = x$ for $0 \leq x \leq 1$ and $f(x) = 2-x$ for $1 < x \leq 2$. Then $E(X) = \int_0^1 x^2\,dx + \int_1^2 x(2-x)\,dx = \frac{1}{3}+\frac{2}{3} = 1$. ✓

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Pitfall	Correct Approach
Confusing the rate $\lambda$ with the mean $\dfrac{1}{\lambda}$ for exponential distributions	$E(X) = \dfrac{1}{\lambda}$; the rate parameter is $\lambda$
Forgetting that the total area under a PDF must equal 1	Always verify: $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$
Applying the exponential memoryless property to other distributions	Only the exponential distribution has this property
Using $P(a < X < b) = f(b) - f(a)$	This is for CDFs, not PDFs. Use $\displaystyle\int_a^b f(x)\,dx$

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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 $T \sim \mathrm{Exp}(4)$ (rate $= 4$ per hour).

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

$X$ is a continuous random variable with PDF $f(x) = \dfrac{3}{4}(2x - x^2)$ for $0 \leq x \leq 2$. Find $E(X)$, $\mathrm{Var}(X)$And the median.

Solution

$E(X) = \dfrac{3}{4}\displaystyle\int_0^2 (2x^2-x^3)\,dx = \dfrac{3}{4}\!\left[\dfrac{2x^3}{3}-\dfrac{x^4}{4}\right]_0^2 = \dfrac{3}{4}\!\left(\dfrac{16}{3}-4\right) = \dfrac{3}{4}\cdot\dfrac{4}{3} = 1$.

$E(X^2) = \dfrac{3}{4}\displaystyle\int_0^2 (2x^3-x^4)\,dx = \dfrac{3}{4}\!\left[\dfrac{x^4}{2}-\dfrac{x^5}{5}\right]_0^2 = \dfrac{3}{4}\!\left(8-\dfrac{32}{5}\right) = \dfrac{3}{4}\cdot\dfrac{8}{5} = \dfrac{6}{5}$.

$\mathrm{Var}(X) = \dfrac{6}{5}-1 = \dfrac{1}{5}$.

Median $m$: $\dfrac{3}{4}\!\left(m^2-\dfrac{m^3}{3}\right) = \dfrac{1}{2}$. By inspection or Numerical methods: $m \approx 0.908$.

Prove that for $X \sim \mathrm{Exp}(\lambda)$The memoryless property holds: $P(X > s+t \mid X > s) = P(X > t)$.

Solution

$P(X > s+t \mid X > s) = \frac{P(X > s+t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t)$

$\blacksquare$

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

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

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

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Distribution	PDF	$E(X)$	$\mathrm{Var}(X)$
$\mathrm{Exp}(\lambda)$	$\lambda e^{-\lambda x}$, $x \geq 0$	$\dfrac{1}{\lambda}$	$\dfrac{1}{\lambda^2}$
$\mathrm{U}(a,b)$	$\dfrac{1}{b-a}$, $a \leq x \leq b$	$\dfrac{a+b}{2}$	$\dfrac{(b-a)^2}{12}$

Property	Exponential	Uniform
Memoryless	Yes	No
CDF	$1 - e^{-\lambda x}$	$\dfrac{x-a}{b-a}$
Median	$\dfrac{\ln 2}{\lambda}$	$\dfrac{a+b}{2}$

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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

$\lambda = \dfrac{1}{500} = 0.002$.

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

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

(c) Median $m$: $e^{-0.002m} = 0.5 \implies m = \dfrac{\ln 2}{0.002} = \boxed{346.6\,\text{hours}}$.

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

Solution

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

$E(X^2) = \dfrac{1}{b-a}\displaystyle\int_a^b x^2\,dx = \dfrac{b^3-a^3}{3(b-a)} = \dfrac{a^2+ab+b^2}{3}$.

$\mathrm{Var}(X) = \dfrac{a^2+ab+b^2}{3} - \dfrac{(a+b)^2}{4} = \dfrac{4a^2+4ab+4b^2-3a^2-6ab-3b^2}{12} = \dfrac{a^2-2ab+b^2}{12} = \boxed{\dfrac{(b-a)^2}{12}}$. $\blacksquare$

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The MGF of a random variable $X$ is $M_X(t) = E(e^{tX})$.

Properties:

• $M_X(0) = 1$
• $M_X'(0) = E(X)$
• $M_X''(0) = E(X^2)$
• If $X$ and $Y$ are independent, $M_{X+Y}(t) = M_X(t)M_Y(t)$

MGFs:

• $\mathrm{Exp}(\lambda)$: $M(t) = \dfrac{\lambda}{\lambda-t}$ for $t < \lambda$
• $\mathrm{U}(a,b)$: $M(t) = \dfrac{e^{bt}-e^{at}}{(b-a)t}$

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

$P(a < X \leq b) = F(b) - F(a)$ where $F(x) = \displaystyle\int_{-\infty}^x f(t)\,dt$.

For i.i.d. Random variables $X_1, \ldots, X_n$The order statistics are $X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}$.

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

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

Solution

$M(t) = \displaystyle\int_0^{\infty} e^{tx}\lambda e^{-\lambda x}\,dx = \lambda\displaystyle\int_0^{\infty} e^{-(\lambda-t)x}\,dx = \frac{\lambda}{\lambda-t}$ For $t < \lambda$.

$M'(t) = \dfrac{\lambda}{(\lambda-t)^2}$. $M'(0) = \dfrac{1}{\lambda} = E(X)$. ✓

$M''(t) = \dfrac{2\lambda}{(\lambda-t)^3}$. $M''(0) = \dfrac{2}{\lambda^2} = E(X^2)$.

$\mathrm{Var}(X) = \dfrac{2}{\lambda^2} - \dfrac{1}{\lambda^2} = \boxed{\dfrac{1}{\lambda^2}}$. ✓

Prove that if $X \sim \mathrm{U}(0,1)$Then $Y = -\dfrac{1}{\lambda}\ln X$ follows $\mathrm{Exp}(\lambda)$.

Solution

$F_Y(y) = P(Y \leq y) = P\!\left(-\dfrac{1}{\lambda}\ln X \leq y\right) = P(\ln X \geq -\lambda y) = P(X \geq e^{-\lambda y})$.

$= 1 - e^{-\lambda y}$ for $y \geq 0$.

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

• Pure Mathematics: Continuous distributions use calculus
• Further Mechanics: Exponential distributions model time between events
• Pure: Further stats builds on pure maths foundations