Continuous Probability Distributions and Applied Statistics

Review of Continuous Uniform Distribution

• Definition & Parameters:

  • A continuous random variable $X$ is uniformly distributed on the interval $[a,b]$, denoted as $X \sim U(a,b)$, if its probability density function is constant over that interval.

• Probability Density Function (PDF):     $f(x) = \begin{cases} \frac{1}{b-a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$

• Cumulative Distribution Function (CDF):     $F(x) = P(X \le x) = \begin{cases} 0 & x < a \\ \frac{x-a}{b-a} & a \le x \le b \\ 1 & x > b \end{cases}$

• Expectation & Variance:     $E(X) = \frac{a+b}{2}$$V(X) = \frac{(b-a)^2}{12}$

• Moment-Generating Function (MGF):     $M(t) = \begin{cases} \frac{e^{tb} - e^{ta}}{t(b-a)} & t \neq 0 \\ 1 & t = 0 \end{cases}$

• Worked Problems & Applications:

  • Point Selection on Line Segment: A point is chosen at random on $[0,5]$. The probability that the chosen point lies between $1$ and $7/2$ ($3.5$) is:         $P(1 \le X \le 3.5) = \frac{3.5 - 1}{5 - 0} = \frac{2.5}{5} = 0.5$
  • Rectangular Glass Panels: Panel length $X \sim U(40, 60)\,\text{cm}$ and width $Y \sim U(20, 30)\,\text{cm}$. Cuts are made by two separate robotic arms on different assembly lines, making length and width completely independent. The expected area of a panel is:         $E(XY) = E(X) \times E(Y) = \left(\frac{40+60}{2}\right) \times \left(\frac{20+30}{2}\right) = 50 \times 25 = 1250\,\text{cm}^2$
  • MGF Parameter Identification: Continuous random variable $X$ with MGF $M(t) = \frac{e^{5t}-e^{4t}}{t}$ for $t \neq 0$ and $M(0)=1$. Matching this to the general uniform MGF yields parameters $a = 4$ and $b = 5$ (so $X \sim U(4,5)$):
    1. $E(X) = \frac{4+5}{2} = 4.5$
    2. $P(4.2 < X \le 4.7) = \frac{4.7 - 4.2}{5 - 4} = 0.5$
  • Transformations of Uniform Variables ($X \sim U(0,1)$):
    1. $Y = a + (b-a)X$: CDF $P(Y \le y) = P\left(a + (b-a)X \le y\right) = P\left(X \le \frac{y-a}{b-a}\right) = \frac{y-a}{b-a}$ for $a \le y \le b$. This is a continuous uniform distribution $U(a,b)$.
    2. $2X + 1$: Evaluates to $U(1,3)$, which is a continuous uniform distribution.
    3. $X^2$: CDF $P(X^2 \le y) = P(X \le \sqrt{y}) = \sqrt{y}$ for $0 \le y \le 1$. The PDF is $f(y) = \frac{1}{2\sqrt{y}}$, which is NOT a continuous uniform distribution.

The Normal Distribution

• Definition & Parameters:

  • A continuous random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ (or variance $\sigma^2$), denoted as $X \sim N(\mu, \sigma^2)$, where $-\infty < \mu < \infty$ and $\sigma > 0$.
  • Mathematical constants involved: $e \approx 2.71828$ and $\pi \approx 3.14159$.

• Probability Density Function (PDF):     $f(x; \mu, \sigma) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, \quad -\infty < x < \infty$

• Density Curve Characteristics:

  • Commonly known as the bell curve or Gaussian distribution, characterized by a symmetric, single-peaked graph.
  • The curve is symmetric about $x = \mu$, where its height is maximized.
  • The mean, median, and mode are all equal to $\mu$.
  • $\mu$ is the location parameter; $\sigma$ is the scale parameter.
  • Asymptotic limits: $\lim_{x \to \infty} f(x) = 0$ and $\lim_{x \to -\infty} f(x) = 0$.
  • The name "Normal" stems from the pattern naturally surfacing across diverse fields (human heights, measurement errors, yields); it does not imply that other distributions are abnormal.

• Proof of PDF Validity:

  • Non-Negativity: Because $\sigma > 0$ and $e^u > 0$ for all real $u$, $f(x; \mu, \sigma) \ge 0$ for all $x \in (-\infty, \infty)$.
  • Total Integral Equals 1: Let $I = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} dx$. Substituting $z = \frac{x-\mu}{\sigma}$ yields:         $I = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz$         Square the integral $I$:         $I^2 = \left(\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2} du\right)\left(\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-v^2/2} dv\right) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(u^2+v^2)/2} du\,dv$         Convert to polar coordinates with $u = r \cos(\theta)$, $v = r \sin(\theta)$, $u^2+v^2 = r^2$, and $du\,dv = r\,dr\,d\theta$:         $I^2 = \frac{1}{2\pi} \int_0^{2\pi} \int_0^\infty e^{-r^2/2} r\,dr\,d\theta = \frac{1}{2\pi} \int_0^{2\pi} \left[-e^{-r^2/2}\right]_0^\infty d\theta = \frac{1}{2\pi} \int_0^{2\pi} 1\,d\theta = 1$         Since $I > 0$, $I = 1$, confirming $f(x)$ is a valid probability density function.

• Moment-Generating Function (MGF) Derivation:

  • By definition:         $M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{tx - \frac{(x-\mu)^2}{2\sigma^2}} dx$
  • Completing the square in the exponent:         $tx - \frac{(x-\mu)^2}{2\sigma^2} = \frac{2\sigma^2 tx - (x^2 - 2\mu x + \mu^2)}{2\sigma^2} = -\frac{x^2 - 2(\mu + \sigma^2 t)x + \mu^2}{2\sigma^2}$$= -\frac{[x - (\mu + \sigma^2 t)]^2 - (\mu + \sigma^2 t)^2 + \mu^2}{2\sigma^2} = -\frac{[x - (\mu + \sigma^2 t)]^2}{2\sigma^2} + \mu t + \frac{1}{2}\sigma^2 t^2$
  • Substituting back into the MGF integral:         $M_X(t) = e^{\mu t + \frac{1}{2}\sigma^2 t^2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{[x - (\mu + \sigma^2 t)]^2}{2\sigma^2}} dx$
  • The integral equals $1$ because it represents the total area under a normal PDF with mean $(\mu + \sigma^2 t)$ and variance $\sigma^2$.
  • Final MGF expression:         $M_X(t) = e^{\mu t + \frac{1}{2}\sigma^2 t^2}$

• Derivation of Mean and Variance from MGF:

  • First derivative:         $M_X'(t) = e^{\mu t + \frac{1}{2}\sigma^2 t^2} \times (\mu + \sigma^2 t) = M_X(t)(\mu + \sigma^2 t)$
  • Expected Value:         $E(X) = M_X'(0) = e^0 \times (\mu + 0) = \mu$
  • Second derivative:         $M_X''(t) = M_X'(t)(\mu + \sigma^2 t) + M_X(t)(\sigma^2) = M_X(t)[(\mu + \sigma^2 t)^2 + \sigma^2]$
  • Second Moment:         $E(X^2) = M_X''(0) = e^0 \times [(\mu + 0)^2 + \sigma^2] = \mu^2 + \sigma^2$
  • Variance:         $V(X) = E(X^2) - [E(X)]^2 = (\mu^2 + \sigma^2) - \mu^2 = \sigma^2$

• Linear Combination Properties:

  • Linear Transformation: If $X \sim N(\mu, \sigma^2)$, then $Y = aX + b \sim N(a\mu + b, a^2\sigma^2)$ for $a \neq 0$.
  • Linear Combination of Independent Normals: If $X_i \sim N(\mu_i, \sigma_i^2)$ ($i = 1, 2, \dots, n$) are independent continuous random variables, then:         $\sum_{i=1}^n a_i X_i \sim N\left(\sum_{i=1}^n a_i \mu_i, \sum_{i=1}^n a_i^2 \sigma_i^2\right)$

The Standard Normal Distribution and Probability Calculations

• Definition & Notation:

  • The standard normal distribution is the special case with $\mu = 0$ and $\sigma = 1$, denoted by random variable $Z \sim N(0,1)$.

• Probability Density Function (PDF):     $f(z; 0, 1) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty$

  • The standard normal curve ($z$ curve) has inflection points at $z = 1$ and $z = -1$.

• Cumulative Distribution Function (CDF):     $\Phi(z) = P(Z \le z) = \int_{-\infty}^z f(y; 0, 1)\,dy$

• Standardization Proposition:
  • If $X \sim N(\mu, \sigma^2)$, then $Z = \frac{X - \mu}{\sigma} \sim N(0,1)$.
  • Proof via MGF:         $M_Z(t) = E\left[e^{t\left(\frac{X-\mu}{\sigma}\right)}\right] = e^{-\frac{\mu t}{\sigma}} M_X\left(\frac{t}{\sigma}\right) = e^{-\frac{\mu t}{\sigma}} e^{\mu\left(\frac{t}{\sigma}\right) + \frac{1}{2}\sigma^2\left(\frac{t}{\sigma}\right)^2} = e^{\frac{1}{2}t^2}$         This matches the MGF of $N(0,1)$.
  • Probability computations for arbitrary $X \sim N(\mu, \sigma^2)$:         $P(a \le X \le b) = P\left(\frac{a-\mu}{\sigma} \le Z \le \frac{b-\mu}{\sigma}\right) = \Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)$$P(X \le a) = \Phi\left(\frac{a-\mu}{\sigma}\right)$$P(X \ge b) = 1 - \Phi\left(\frac{b-\mu}{\sigma}\right)$

• Symmetry and Interval Calculations:
  • Due to symmetry: $P(Z \le -z) = P(Z \ge z) = 1 - \Phi(z)$.
  • $P(Z \le 1.25) = \Phi(1.25) = 0.8944$
  • $P(Z \le -1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056$
  • $P(-0.38 \le Z \le 1.25) = \Phi(1.25) - \Phi(-0.38) = \Phi(1.25) - [1 - \Phi(0.38)] = 0.8944 - [1 - 0.6480] = 0.5424$

Critical Values, Percentiles, and Empirical Rule for Normal Distributions

• Definition of $z$ Critical Values:
  • The critical value $z_\alpha$ denotes the point on the $z$-axis such that an area of $\alpha$ lies strictly to its right under the standard normal curve:         $P(Z \ge z_\alpha) = \alpha$
  • $z_\alpha$ is the $100(1-\alpha)\text{th}$ percentile of the standard normal distribution.
  • Lower-tail critical value notation: $\eta(\alpha) = -z_\alpha$ where $\Phi(\eta(\alpha)) = \alpha$

• Common Standard Normal Critical Values ($z_\alpha$):

  • 90th Percentile (Upper-tail $\alpha = 0.10$): $z_{0.10} = 1.28$
  • 95th Percentile (Upper-tail $\alpha = 0.05$): $z_{0.05} = 1.645$
  • 97.5th Percentile (Upper-tail $\alpha = 0.025$): $z_{0.025} = 1.96$
  • 99th Percentile (Upper-tail $\alpha = 0.01$): $z_{0.01} = 2.33$
  • 99.5th Percentile (Upper-tail $\alpha = 0.005$): $z_{0.005} = 2.58$
  • 99.9th Percentile (Upper-tail $\alpha = 0.001$): $z_{0.001} = 3.08$
  • 99.95th Percentile (Upper-tail $\alpha = 0.0005$): $z_{0.0005} = 3.27$

• Percentiles of Arbitrary Normal Distributions:

  • To find the $(100p)\text{th}$ percentile $\eta_X(p)$ of $X \sim N(\mu, \sigma^2)$, scale and shift the $(100p)\text{th}$ percentile $\eta_Z(p)$ of $N(0,1)$:         $\eta_X(p) = \mu + \eta_Z(p) \times \sigma$

• Empirical Rule ($68-95-99.7$ Rule):

  • For any normally distributed variable $X \sim N(\mu, \sigma^2)$:         $P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.68$$P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.95$$P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.997$

• Detailed Worked Problems:
  • Bolt Thread Length:
    • Within $1.5$ standard deviations of mean:             $P(|X-\mu| \le 1.5\sigma) = P(|Z| \le 1.5) = \Phi(1.5) - \Phi(-1.5) = 0.9332 - (1 - 0.9332) = 0.8664$
    • Between $1$ and $2$ SDs from mean:             $P(1\sigma < |X-\mu| < 2\sigma) = P(1 < |Z| < 2) = 2[\Phi(2) - \Phi(1)] = 2[0.9772 - 0.8413] = 0.2718$
  • Yield Strength of A36 Steel ($X \sim N(43, 4.5^2)$) :
    • Probability yield strength is at least $40$ ksi:             $P(X \ge 40) = P\left(Z \ge \frac{40-43}{4.5}\right) = P(Z \ge -0.67) = \Phi(0.67) = 0.7486$
  • Serum Zinc Level ($X \sim N(90, 15^2)$) :
    • Conditional probability $P(X > 120 \mid X > 105)$:             $P(X > 120 \mid X > 105) = \frac{P(X > 120)}{P(X > 105)} = \frac{P\left(Z > \frac{120-90}{15}\right)}{P\left(Z > \frac{105-90}{15}\right)} = \frac{P(Z > 2)}{P(Z > 1)} = \frac{1 - 0.9772}{1 - 0.8413} = \frac{0.0228}{0.1587} \approx 0.1437$
  • Substrate Concentration ($X \sim N(0.3, 0.06^2)$) :
    1. Exceeds $0.25$: $P(X > 0.25) = P\left(Z > \frac{0.25-0.3}{0.06}\right) = P(Z > -0.83) = \Phi(0.83) = 0.7967$
    2. At most $0.10$: $P(X \le 0.10) = P\left(Z \le \frac{0.10-0.3}{0.06}\right) = P(Z \le -3.33) = \Phi(-3.33) = 0.0004$
    3. Largest $5\%$ characterization: $\eta_X(0.95) = \mu + z_{0.05} \times \sigma = 0.3 + 1.645 \times 0.06 = 0.3987$
  • Oil Percolation Time (Mean $\mu = 20\,\text{s}$, $P(X > 31.5) = 0.25$):         $P\left(Z > \frac{31.5-20}{\sigma}\right) = 0.25 \implies \frac{11.5}{\sigma} = z_{0.25} = 0.67 \implies \sigma = \frac{11.5}{0.67} \approx 17.16\,\text{seconds}$
  • Lithium-Ion Battery Operating Lifetime ($X \sim N(1200, 150^2)$):
    • Qualifying lifetime threshold for extended warranty (top $10\%$ longest-lasting):             $\eta_X(0.90) = \mu + z_{0.10} \times \sigma = 1200 + 1.28 \times 150 = 1392\,\text{hours}$
  • Roasted Coffee Beans Weight ($X \sim N(250, 4^2)$) :
    • Probability a single bag weighs less than $245\,\text{g}$:             $p = P(X < 245) = P\left(Z < \frac{245-250}{4}\right) = P(Z < -1.25) = 1 - 0.8944 = 0.1056$
    • Probability in a sample of $n = 15$ bags that exactly $3$ bags weigh less than $245\,\text{g}$ ($Y \sim \text{Bin}(15, 0.1056)$) :             $P(Y = 3) = \binom{15}{3} (0.1056)^3 (1 - 0.1056)^{12} = 455 \times 0.001177 \times 0.2598 \approx 0.1392$

The Gamma Function and Gamma Distribution

• Motivation:

  • Normal distributions assume symmetry and allow negative values. Real-world continuous data (e.g., waiting times, insurance claims) is strictly positive ($x \ge 0$) and right-skewed. The gamma family provides a flexible model for positive, skewed data.

• The Gamma Function:

  • For $\alpha > 0$, the gamma function $\Gamma(\alpha)$ is defined as:         $\Gamma(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x}\,dx$

• Properties of the Gamma Function:

  1. $\Gamma(1) = 1$
  2. $\Gamma(\alpha) = (\alpha - 1)\Gamma(\alpha - 1)$ for any $\alpha > 1$
  3. $\Gamma(n) = (n - 1)!$ for any positive integer $n$
  4. $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$

• Gamma Distribution Definition & Parameters:

  • A continuous random variable $X$ has a gamma distribution, denoted as $X \sim \Gamma(\alpha, \beta)$, if its PDF is:         $f(x; \alpha, \beta) = \begin{cases} \frac{1}{\beta^\alpha \Gamma(\alpha)} x^{\alpha-1} e^{-x/\beta} & x \ge 0 \\ 0 & \text{otherwise} \end{cases}$         where $\alpha > 0$ is the shape parameter and $\beta > 0$ is the scale parameter.

• Moment-Generating Function (MGF) Derivation:     $M_X(t) = E[e^{tX}] = \int_0^\infty e^{tx} \frac{1}{\beta^\alpha \Gamma(\alpha)} x^{\alpha-1} e^{-x/\beta}\,dx = \frac{1}{\beta^\alpha \Gamma(\alpha)} \int_0^\infty x^{\alpha-1} e^{-x\left(\frac{1}{\beta} - t\right)}\,dx$

  • For the integral to converge, the exponent multiplier must be positive: $\frac{1}{\beta} - t > 0 \implies t < \frac{1}{\beta}$.
  • Substitute $u = x \left(\frac{1-\beta t}{\beta}\right)$, so $x = \frac{u\beta}{1-\beta t}$ and $dx = \frac{\beta}{1-\beta t}\,du$:         $M_X(t) = \frac{1}{\beta^\alpha \Gamma(\alpha)} \int_0^\infty \left(\frac{u\beta}{1-\beta t}\right)^{\alpha-1} e^{-u} \left(\frac{\beta}{1-\beta t}\right) du = \frac{1}{\beta^\alpha \Gamma(\alpha)} \frac{\beta^\alpha}{(1-\beta t)^\alpha} \int_0^\infty u^{\alpha-1} e^{-u}\,du$
  • Since $\int_0^\infty u^{\alpha-1} e^{-u}\,du = \Gamma(\alpha)$, terms cancel to yield:         $M_X(t) = (1 - \beta t)^{-\alpha}, \quad \text{for } t < \frac{1}{\beta}$

• Mean and Variance Derivation:

  • First derivative: $M_X'(t) = -\alpha(1 - \beta t)^{-\alpha-1}(-\beta) = \alpha\beta(1 - \beta t)^{-\alpha-1}$
  • Expected Value: $E(X) = M_X'(0) = \alpha\beta$
  • Second derivative: $M_X''(t) = \alpha\beta(-\alpha-1)(1 - \beta t)^{-\alpha-2}(-\beta) = \alpha(\alpha+1)\beta^2(1 - \beta t)^{-\alpha-2}$
  • Second Moment: $E(X^2) = M_X''(0) = \alpha(\alpha+1)\beta^2$
  • Variance: $V(X) = E(X^2) - [E(X)]^2 = \alpha(\alpha+1)\beta^2 - \alpha^2\beta^2 = \alpha\beta^2$

• Core Properties of Gamma Variables:

  • Sum of Independent Gammas: If $X_i \sim \Gamma(\alpha_i, \beta)$ ($i = 1, 2, \dots, n$) are independent, then:         $\sum_{i=1}^n X_i \sim \Gamma\left(\sum_{i=1}^n \alpha_i, \beta\right)$
  • Scaling Transformation: If $X \sim \Gamma(\alpha, \beta)$ and $k > 0$, then:         $kX \sim \Gamma(\alpha, k\beta)$

• Poisson Process Connection:

  • Gamma models the total waiting time until the $\alpha$\text{-th} event in a Poisson process.
  • $\beta$ represents the average time between occurrences ($1/\beta$ is the rate parameter in average events per unit time).
  • CDF Connection: The probability that the $\alpha$\text{-th} event occurs by time $x$ equals $1$ minus the probability that fewer than $\alpha$ events occur in time $x$:         $F(x) = P(X \le x) = 1 - P(\text{fewer than } \alpha \text{ events in time } x) = 1 - \sum_{k=0}^{\alpha-1} P(Y = k), \quad Y \sim \text{Pois}\left(\frac{x}{\beta}\right)$

Special Cases of the Gamma Distribution: Standard Gamma

• Taxonomy of Gamma Special Cases:

  • Standard Gamma: $\beta = 1 \implies X \sim \Gamma(\alpha, 1)$
  • Exponential: $\alpha = 1 \implies X \sim \Gamma(1, \beta)$
  • Chi-Squared: $\alpha = \frac{\nu}{2}, \beta = 2 \implies X \sim \Gamma\left(\frac{\nu}{2}, 2\right)$

• Standard Gamma Distribution ($X \sim \Gamma(\alpha, 1)$):

  • Probability Density Function:         $f(x; \alpha) = \begin{cases} \frac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)} & x \ge 0 \\ 0 & \text{otherwise} \end{cases}$
  • Density Curve Shape as $\alpha$ varies:
    • $\alpha < 1$: Unbounded near $x = 0$, no finite peak.
    • $\alpha = 1$: Maximum at $x = 0$, decays exponentially.
    • $\alpha > 1$: Mound-shaped, starts at $0$, peaks at mode $x = \alpha - 1$, and becomes increasingly symmetric/bell-shaped as $\alpha$ grows.

• Incomplete Gamma Function & Probability Calculations:

  • The CDF of a standard gamma variable $X \sim \Gamma(\alpha, 1)$ is the incomplete gamma function:         $F(x; \alpha) = \int_0^x \frac{y^{\alpha-1} e^{-y}}{\Gamma(\alpha)}\,dy, \quad x > 0$
  • For arbitrary $X \sim \Gamma(\alpha, \beta)$, standardizing yields:         $P(X \le x) = F\left(\frac{x}{\beta}; \alpha\right)$$P(a \le X \le b) = F\left(\frac{b}{\beta}; \alpha\right) - F\left(\frac{a}{\beta}; \alpha\right)$

• Worked Problems:

  • Customer Arrivals at Shop (Poisson rate $30/\text{hr} \implies \beta = 2\,\text{minutes}$ per customer):
    1. Wait more than $5\,\text{minutes}$ before both of the first 2 customers arrive ($\alpha = 2$, $X \sim \Gamma(2, 2)$):             $P(X > 5) = 1 - P(X \le 5) = 1 - F\left(\frac{5}{2}; 2\right) = 1 - F(2.5; 2)$             From Table A.4 (Incomplete Gamma Function Table), $F(2.5; 2) = 0.713$.             $P(X > 5) = 1 - 0.713 = 0.287$
    2. Average time before $3$ customers arrive:             $E(X) = \alpha\beta = 3 \times 2 = 6\,\text{minutes}$
  • Electronic Component Lifetime ($E(X) = 30\,\text{hours}$, $\sigma = \sqrt{300}\,\text{hours} \implies V(X) = 300$):
    • Parameter solving:             $\alpha\beta = 30, \quad \alpha\beta^2 = 300 \implies \beta = \frac{300}{30} = 10, \quad \alpha = \frac{30}{10} = 3 \implies X \sim \Gamma(3, 10)$
    1. Probability component lasts at least $30\,\text{hours}$:             $P(X \ge 30) = 1 - F\left(\frac{30}{10}; 3\right) = 1 - F(3; 3) = 1 - 0.577 = 0.423$
    2. Probability component lasts between $20$ and $30\,\text{hours}$:             $P(20 \le X \le 30) = F\left(\frac{30}{10}; 3\right) - F\left(\frac{20}{10}; 3\right) = F(3; 3) - F(2; 3) = 0.577 - 0.323 = 0.254$

The Exponential Distribution

• Definition & Parameters:

  • The exponential distribution is a special case of gamma with $\alpha = 1$ and $\beta = \frac{1}{\lambda}$, denoted as $X \sim \text{Exp}(\lambda)$ with rate parameter \lambda > 0$.\n\n* **Probability Density Function (PDF)**:\n    f(x; \lambda) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0 \ 0 & \text{otherwise} \end{cases}\n\n    \n\n* **Cumulative Distribution Function (CDF)**:\n    F(x; \lambda) = \begin{cases} 0 & x < 0 \ 1 - e^{-\lambda x} & x \ge 0 \end{cases}\n\n* **Expectation, Variance, & MGF**:\n    E(X) = \frac{1}{\lambda}\n    V(X) = \frac{1}{\lambda^2}\n    M_X(t) = \left(1 - \frac{t}{\lambda}\right)^{-1}, \quad \text{for } t < \lambda\n\n* **Memoryless Property**:\n * For continuous component lifetime X \sim \text{Exp}(\lambda):\n        P(X \ge t + t_0 \mid X \ge t_0) = P(X \ge t)\n * *Substantive Meaning*: The distribution of remaining lifetime is independent of current age t_0$. An old component has the exact same probability of surviving an additional$t hours as a brand new component.\n\n* **Poisson Process Connection Proof**:\n * *Proposition*: Suppose event occurrences in any time interval follow a Poisson distribution with parameter at$($a$is the rate of events per unit time). Then elapsed time$T$between successive events is exponentially distributed with$\lambda = a$.
  • Proof:         $F(t) = P(T \le t) = 1 - P(T > t)$         The condition $T > t$ means 0 events occurred in time interval $t$. Let $X \sim \text{Pois}(at)$:         $P(T > t) = P(X = 0) = \frac{e^{-at}(at)^0}{0!} = e^{-at}$$F(t) = 1 - e^{-at}, \quad t \ge 0$         Taking the derivative yields the PDF:         $f(t) = F'(t) = a e^{-at}, \quad t \ge 0$         which is the PDF of $\text{Exp}(a)$.

• Worked Problems:

  • Cafeteria Arrivals (Poisson rate $30/\text{hr} \implies \lambda = 0.5/\text{min}$):
    1. Probability time between arrivals $> 3\,\text{minutes}$:             $P(T > 3) = e^{-0.5 \times 3} = e^{-1.5} \approx 0.2231$
    2. Expected time between arrivals:             $E(T) = \frac{1}{\lambda} = \frac{1}{0.5} = 2\,\text{minutes}$
  • Industrial Batteries System (Average life $8\,\text{years} \implies \lambda = 1/8$):
    • Probability a single battery lasts $> 8\,\text{years}$:             $p = P(X > 8) = e^{-\frac{1}{8} \times 8} = e^{-1} \approx 0.3679$
    • In a system of 6 installed batteries, probability at least 2 last $> 8\,\text{years}$ ($Y \sim \text{Bin}(6, 0.3679)$) :             $P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1) = 1 - (1-p)^6 - 6p(1-p)^5$$= 1 - (0.6321)^6 - 6(0.3679)(0.6321)^5 \approx 1 - 0.0625 - 0.2183 = 0.7192$
  • Car Battery Trip (Average life $10{,}000\,\text{km} \implies \lambda = 1/10000$):
    • Probability completing a $5000\,\text{km}$ trip without replacement:             $P(X \ge 5000) = e^{-\frac{1}{10000} \times 5000} = e^{-0.5} \approx 0.6065$
  • Battery Failure Threshold ($10\%$ fail within first $500\,\text{hours}$):
    • Finding $\lambda$:             $F(500) = 1 - e^{-500\lambda} = 0.10 \implies e^{-500\lambda} = 0.90$
    • Probability functioning after $2000\,\text{hours}$:             $P(X > 2000) = e^{-2000\lambda} = (e^{-500\lambda})^4 = (0.90)^4 = 0.6561$

The Chi-Squared Distribution

• Definition & Parameters:

  • The chi-squared distribution with $\nu$ degrees of freedom (df), denoted as $X \sim \chi^2(\nu)$ or $X \sim \chi^2_\nu$, is a gamma distribution with $\alpha = \frac{\nu}{2}$ and $\beta = 2$, where $\nu$ is a positive integer.

• Probability Density Function (PDF):     $f(x; \nu) = \begin{cases} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{(\nu/2)-1} e^{-x/2} & x \ge 0 \\ 0 & x < 0 \end{cases}$

• Expectation, Variance, & MGF:     $E(X) = \nu$$V(X) = 2\nu$$M_X(t) = (1 - 2t)^{-\nu/2}, \quad \text{for } t < \frac{1}{2}$

• Relationship with Standard Normal Distribution:

  • Proposition: If $Z \sim N(0,1)$, then $Z^2 \sim \chi^2(1)$.

• Chi-Squared Critical Values ($\chi^2_{\alpha, u}$):
  • The critical value $\chi^2_{\alpha, \nu}$ denotes the value on the horizontal axis such that an area of $\alpha$ lies to its right under the $\chi^2_\nu$ curve:         $P(X \ge \chi^2_{\alpha, \nu}) = \alpha$$P(X \le \chi^2_{\alpha, \nu}) = 1 - \alpha$

• Key Distribution Theorems:

  • Sum of Independent Chi-Squared Variables: If $X_i \sim \chi^2(\nu_i)$ ($i = 1, 2, \dots, n$) are independent, then:         $\sum_{i=1}^n X_i \sim \chi^2\left(\sum_{i=1}^n \nu_i\right)$
  • Gamma to Chi-Squared Conversion: If $2\alpha$ is a positive integer and $X \sim \Gamma(\alpha, \beta)$, then:         $\frac{2X}{\beta} \sim \chi^2(2\alpha)$

• Worked Problems & Applications:

  • Exponential to Chi-Squared Relation: If $X$ has exponential distribution with mean $2$, then $X \sim \text{Exp}(0.5) = \Gamma(1, 2)$. Comparing parameters with $\chi^2(\nu) = \Gamma(\nu/2, 2)$ gives $\nu/2 = 1 \implies \nu = 2$. Thus $X \sim \chi^2(2)$.
  • System of Electronic Components: Component life follows $\chi^2(10)$. Probability a component needs replacement within $16\,\text{hours}$:
    • From Table A.11, $\chi^2_{0.10, 10} = 15.987 \approx 16 \implies P(X > 16) = 0.10 \implies P(X \le 16) = 0.90$
    • For 4 independent components in a system, probability all 4 are replaced within $16\,\text{hours}$:             $P = (0.90)^4 = 0.6561$
  • Walmart Customer Arrivals (Poisson rate $150/\text{hr} = 2.5/\text{min}$):
    1. First customer arrives within $1\,\text{minute}$ ($T \sim \text{Exp}(2.5)$) :             $P(T \le 1) = 1 - e^{-2.5 \times 1} = 1 - e^{-2.5} \approx 0.9179$
    2. $12$ or more customers arrive in a 4-minute window ($Y \sim \text{Pois}(2.5 \times 4) = \text{Pois}(10)$) :             $P(Y \ge 12) = 1 - P(Y \le 11)$
    3. The 10th customer arrives between 6:07 am and 6:08 am ($t \in [7, 8]$ minutes). Elapsed time $X_{10} \sim \Gamma(10, \beta = 1/2.5 = 0.4)$.             Convert to $\chi^2$: $\frac{2X_{10}}{0.4} = 5X_{10} \sim \chi^2(20)$.             $P(7 \le X_{10} \le 8) = P(35 \le 5X_{10} \le 40) = P(34.170 \le \chi^2(20) \le 39.997)$             Given $\chi^2_{0.025, 20} = 34.170$ and $\chi^2_{0.005, 20} = 39.997$:             $P = 0.025 - 0.005 = 0.020$             In a week ($n=7$ days), probability this happens on exactly $2$ days ($D \sim \text{Bin}(7, 0.020)$) :             $P(D = 2) = \binom{7}{2} (0.020)^2 (0.980)^5 = 21 \times 0.0004 \times 0.9039 \approx 0.0076$
    4. Time until next customer arrival exceeds $2\,\text{minutes}$ ($T \sim \text{Exp}(2.5)$) :             $P(T > 2) = e^{-2.5 \times 2} = e^{-5} \approx 0.0067$
  • Critical Value Lookups:
    1. If $X \sim \chi^2(5)$ and $P(X \le x) = 0.96$, then $P(X \ge x) = 0.04 \implies x = \chi^2_{0.04, 5} = 11.64$
    2. If $\chi^2_{0.075, 5} = 10$, then for $X \sim \chi^2(5)$, $P(X \le 10) = 1 - 0.075 = 0.925$