Continuous Probability Distributions and Mathematical Statistics

Overview of Continuous Probability Distributions

• The primary continuous probability distributions utilized in mathematical statistics include:
  • Normal Distribution
  • Standard Normal Distribution
  • Chi-Square ($\chi^2$) Distribution
  • Student's t-Distribution
  • Fisher-Snedecor F-Distribution

Normal Distribution

• Definition: The normal distribution is the most widely used continuous probability distribution and is characterized by a symmetric bell-shaped curve.
• Parameters:
  • Mean: $\mu$, where $-\infty < \mu < \infty$.
  • Standard Deviation: $\sigma$, where $\sigma > 0$ (and variance $\sigma^2 > 0$).
• Probability Density Function (PDF):
  • A continuous random variable $Y$ follows a normal probability distribution if and only if its density function is given by:         $f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(y-\mu)^2 / (2\sigma^2)}, \quad -\infty < y < \infty$
• Expected Value and Variance:
  • $E(Y) = \mu$
  • $V(Y) = \sigma^2$
• Distribution Notation: $Y \sim N(\mu, \sigma^2)$
• Moment Generating Function (MGF):
  • $m(t) = e^{\mu t + \frac{1}{2}t^2 \sigma^2}$

• Area Under the Normal Curve:
  • The probability $P(a \le Y \le b)$ corresponds to the area under the normal density function between $y = a$ and $y = b$, requiring the evaluation of the integral:         $P(a \le Y \le b) = \int_a^b \frac{1}{\sigma \sqrt{2\pi}} e^{-(y-\mu)^2 / (2\sigma^2)}\,dy$

Standard Normal Distribution

• Definition: A standard normal distribution is a specific normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$ (variance $\sigma^2 = 1$).
• Transformation Formula (Standardization):
  • Any normal random variable $Y \sim N(\mu, \sigma^2)$ can be transformed to a standard normal random variable $Z \sim N(0, 1)$ using the relationship:         $Z = \frac{Y - \mu}{\sigma}$
  • The variable $Z$ represents the distance from the mean of a normal distribution expressed in units of standard deviation.
• Moment Generating Function (MGF) of Standard Normal Variable $Z$:
  • $M_Z(t) = \exp\left\{\frac{1}{2}t^2\right\} = e^{\frac{1}{2}t^2}, \quad -\infty < t < \infty$

Worked Example: Example 4.9

• Problem Statement: The achievement scores for a college entrance examination are normally distributed with mean $\mu = 75$ and standard deviation $\sigma = 10$. What fraction of the scores lies between $80$ and $90$?
• Model: Let $Y$ denote the score, where $Y \sim N(75, 10^2)$. We wish to find $P(80 \le Y \le 90)$.
• Standardization Procedure:
  • For $y_1 = 80$:         $z_1 = \frac{80 - 75}{10} = 0.5$
  • For $y_2 = 90$:         $z_2 = \frac{90 - 75}{10} = 1.5$
• Probability Calculation:
  • $P(80 \le Y \le 90) = P(0.5 \le Z \le 1.5)$

• Method 1 (Using Right-Hand Tail Areas $A(z)$):
  • Let $A(z) = P(Z \ge z)$.
  • From Table 4 (Standard Normal Curve Areas), $A(0.5) = 0.3085$ and $A(1.5) = 0.0668$.
  • $A = A(0.5) - A(1.5) = 0.3085 - 0.0668 = 0.2417$
• Method 2 (Using Cumulative Distribution Functions $P(Z \le z)$):
  • $P(Z \le 1.5) = 0.9332$
  • $P(Z \le 0.5) = 0.6915$
  • $P(0.5 \le Z \le 1.5) = 0.9332 - 0.6915 = 0.2417$
• Conclusion: $24.17\%$ (or a fraction of $0.2417$) of the achievement scores lie between $80$ and $90$.

Practice Exercises for Standard Normal Distribution

• Exercise 4.58: Use Table 4 to find the following probabilities for a standard normal random variable $Z$:
  • a) $P(0 \le Z \le 1.2)$
  • b) $P(-0.9 \le Z \le 0)$
  • c) $P(0.3 \le Z \le 1.56)$
• Exercise 4.59: If $Z$ is a standard normal random variable, find the value $z_0$ such that:
  • a) $P(Z > z_0) = 0.5$
  • b) $P(Z < z_0) = 0.8643$
  • c) $P(-z_0 < Z < z_0) = 0.90$
  • d) $P(-z_0 < Z < z_0) = 0.99$

Chi-Square ($\chi^2$) Distribution

• Definition: A continuous random variable $X$ is said to have a chi-square distribution with $r$ degrees of freedom if its probability density function is given by:     $f(x) = \begin{cases} \frac{1}{\Gamma(r/2) 2^{r/2}} x^{r/2 - 1} e^{-x/2}, & 0 < x < \infty \\ 0, & \text{elsewhere} \end{cases}$
• Parameter:
  • Degrees of Freedom: $r$ (or $k$), where $r$ is a positive integer representing the parameter of the distribution.
• Moment Generating Function (MGF):
  • $M(t) = (1 - 2t)^{-r/2}, \quad t < \frac{1}{2}$
• Mean and Variance:
  • Mean: $\mu = E(X) = r$
  • Variance: $\sigma^2 = V(X) = 2r$
• Notation: $X \sim \chi^2(r)$ or $X \sim \chi^2(k)$.

• Tabulation and Critical Values:
  • The percentage points of the chi-square distribution are tabulated in Table 6.
  • The symbol $\chi^2_\alpha(r)$ denotes the value such that the area to its right under the chi-square density curve with $r$ degrees of freedom equals $\alpha$:         $P(\chi^2 > \chi^2_\alpha(r)) = \alpha$

Worked Examples and Exercises for Chi-Square Distribution

• Example 3.3.3:
  • Given PDF:$f(x) = \begin{cases} \frac{1}{4} x e^{-x/2}, & 0 < x < \infty \\ 0, & \text{elsewhere} \end{cases}$
  • Identification: Matching with the general chi-square PDF formula:         $\frac{1}{\Gamma(r/2) 2^{r/2}} x^{r/2 - 1} e^{-x/2}$         Here $r/2 - 1 = 1 \implies r/2 = 2 \implies r = 4$. Thus, $X \sim \chi^2(4)$.
  • Properties:
    • Mean: $\mu = r = 4$
    • Variance: $\sigma^2 = 2r = 8$
    • MGF: $M(t) = (1 - 2t)^{-2}$ for $t < \frac{1}{2}$
• Example 3.3.5:
  • Let $X \sim \chi^2(10)$.
  • Finding $P(3.25 \le X \le 20.5)$:$P(3.25 \le X \le 20.5) = P(X \le 20.5) - P(X \le 3.5) = 0.975 - 0.025 = 0.95$
  • Finding $a$ such that $P(a < X) = 0.05$:$P(a < X) = 0.05 \implies P(X \le a) = 0.95$         Using Table 6 with $r = 10$, $a = 18.3$.
• Exercise 3.3.1:
  • If $M(t) = (1 - 2t)^{-6}$ for $t < \frac{1}{2}$ is the MGF of $X$, then $r/2 = 6 \implies r = 12$, meaning $X \sim \chi^2(12)$. Find $P(X < 5.23)$.
• Exercise 3.3.2:
  • If $X \sim \chi^2(5)$, determine constants $c$ and $d$ such that $P(c < X < d) = 0.95$ and $P(X < c) = 0.025$

Student's t-Distribution

• Derivation and Construction:
  • Let $W$ be a standard normal random variable, $W \sim N(0, 1)$.
  • Let $V$ be a chi-square random variable with $r$ degrees of freedom, $V \sim \chi^2(r)$.
  • Assume $W$ and $V$ are independent random variables.
• Joint Probability Density Function of $W$ and $V$:
  • Since $W$ and $V$ are independent, their joint PDF $h(w, v)$ is the product of their individual PDFs:         $h(w, v) = \begin{cases} \frac{1}{\sqrt{2\pi}} e^{-w^2/2} \times \frac{1}{\Gamma(r/2) 2^{r/2}} v^{r/2 - 1} e^{-v/2}, & -\infty < w < \infty, \; 0 < v < \infty \\ 0, & \text{elsewhere} \end{cases}$
• Definition of Student's t Variable:
  • The random variable $T$ defined by:         $T = \frac{W}{\sqrt{V/r}}$         has a Student's t-distribution with $r$ degrees of freedom.
• Parameters and Properties:
  • Degrees of Freedom: $\nu = r$.
  • Mean: $E(T) = 0$.
  • Variance: $V(T) = \frac{r}{r-2}$ for r > 2$.\n* **Shape and Properties:**\n * Symmetric bell-shaped curve centered at zero.\n * Exhibits heavier tails (more dispersion) than the standard normal distribution.\n * As the degrees of freedom r \to \infty$, the t-distribution approaches the standard normal distribution$N(0,1).\n\n\n\n## Exercises for Student's t-Distribution\n\n* **Exercise 3.6.1:** Let T$have a t-distribution with$10$degrees of freedom ($\nu = 10$). Find$P(|T| > 2.228).\n * *Solution:* P(|T| > 2.228) = P(T > 2.228) + P(T < -2.228) = 2 \times P(T > 2.228)$. From Table 5 with$\nu = 10$,$t_{0.025} = 2.228$. Thus,$P(|T| > 2.228) = 2(0.025) = 0.05.\n* **Exercise 3.6.2:** Let T$have a t-distribution with$14$degrees of freedom ($\nu = 14$). Determine$b$so that$P(-b < T < b) = 0.90\n\n# Fisher-Snedecor F-Distribution\n\n* **Derivation and Construction:**\n * Let U$and$V$be independent chi-square random variables with$r_1$and$r_2$degrees of freedom, respectively ($U \sim \chi^2(r_1)$and$V \sim \chi^2(r_2)).\n* **Joint Probability Density Function of U$and$V:**\n * h(u, v) = \begin{cases} \frac{1}{\Gamma(r_1/2)\Gamma(r_2/2) 2^{(r_1+r_2)/2}} u^{r_1/2 - 1} v^{r_2/2 - 1} e^{-(u+v)/2}, & 0 < u, v < \infty \ 0, & \text{elsewhere} \end{cases}\n* **Definition of F Variable:**\n * The ratio of two independent chi-square random variables, each divided by its respective degrees of freedom, defines an F-distributed random variable:\n        F = \frac{U / r_1}{V / r_2}\n* **Parameters:**\n * Numerator degrees of freedom: d_1 = \nu_1 = r_1\n * Denominator degrees of freedom: d_2 = \nu_2 = r_2\n* **Mean:**\n * E(F) = \frac{r_2}{r_2 - 2}, \quad \text{for } r_2 > 2\n\n\n\n* **Inversion Property of F-Distribution:**\n * If X \sim F(\nu_1, \nu_2)$, then its reciprocal$Y = \frac{1}{X} \sim F(\nu_2, \nu_1).\n* **Notation for Critical Values:**\n * F_{\alpha, \nu_1, \nu_2} represents the critical value such that:\n        P(F \ge F_{\alpha, \nu_1, \nu_2}) = \alpha\n\n## Exercises for F-Distribution\n\n* **Exercise 3.6.9:** If F$has an F-distribution with parameters$r_1 = 5$and$r_2 = 10$, find$a$and$b$so that$P(F \le a) = 0.05$and$P(F \le b) = 0.95$, and accordingly,$P(a < F < b) = 0.90$$