Unit 3: Joint Distributions and Probability Modeling Fundamentals and Applications

REVIEW OF CONTINUOUS PROBABILITY DISTRIBUTIONS

  • Summary of Key Distributions: Important continuous probability distributions include the uniform, normal, Student's T, exponential, beta, gamma, and Weibull distributions.

  • PDF and CDF: Probability Density Functions (PDFs) were defined for all these distributions. In several cases, explicit formulas for the Cumulative Distribution Function (CDF) were provided.

  • Parameters: Distributions vary by the number of parameters they utilize:

    • One parameter: Bernoulli, geometric, and Poisson distributions.

    • Two parameters: Binomial and negative binomial distributions.

  • The Normal Distribution:

    • The Standard Normal Distribution is a special case denoted as N(0,1)N(0, 1).

    • Transformation to Standard Normal (Standardizing): If a random variable XX follows a normal distribution XN(μ,σ2)X \sim N(\mu, \sigma^2) for μR\mu \in \mathbb{R} and \sigma > 0, then the random variable Z=XμσZ = \frac{X - \mu}{\sigma} holds the distribution ZN(0,1)Z \sim N(0, 1).

    • Inverse Transformation: If XN(0,1)X \sim N(0, 1), then the random variable Z=μ+σXZ = \mu + \sigma X follows the distribution ZN(μ,σ2)Z \sim N(\mu, \sigma^2).

  • Importance of Support: When defining the PDF of any continuous distribution, the support must be explicitly stated. The support consists of all values where the PDF is positive.

    • Beta Distribution Support: The interval [0,1][0, 1].

    • Gamma Distribution Support: The interval (0,)(0, \infty).

INTRODUCTION TO JOINT DISTRIBUTIONS

  • Rationale: Real-world data is usually multidimensional. To analyze such data, joint probability distributions are required to model the probability of value tuples.

  • Key Definitions:

    • Joint Probability Mass (Density) Function: A function of several variables providing the probability (or density) of tuples of random variables.

    • Joint Probability Measure: A probability measure for a tuple (pair, triple, etc.) of random variables.

    • Joint Cumulative Distribution Function (CDF): A function of several variables giving the accumulated probability across a range of values for tuples of random variables.

  • Sample Spaces and Cross-Products: Given two sample spaces Ω1\Omega_1 and Ω2\Omega_2, the cross-product is defined as Ω1×Ω2={(x,y)xΩ1,yΩ2}\Omega_1 \times \Omega_2 = \{ (x, y) \mid x \in \Omega_1, y \in \Omega_2 \}, consisting of all pairs (x,y)(x, y).

    • Example: If Ω1={1,2,3}\Omega_1 = \{ 1, 2, 3 \} and Ω2={a,b}\Omega_2 = \{ a, b \}, the cross-product is Ω1×Ω2={(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}\Omega_1 \times \Omega_2 = \{ (1, a), (1, b), (2, a), (2, b), (3, a), (3, b) \}.

    • If a probability measure PP assigns P(x)=16P(x) = \frac{1}{6} for each xΩ1×Ω2x \in \Omega_1 \times \Omega_2, then each pair is equally likely.

DISCRETE JOINT DISTRIBUTIONS

  • Joint Probability Mass Function (PMF): For sample spaces Ω1,Ω2\Omega_1, \Omega_2 with a probability measure PP on Ω1×Ω2\Omega_1 \times \Omega_2, consider discrete random variables X:Ω1{x1,x2,x3,}RX: \Omega_1 \rightarrow \{ x_1, x_2, x_3, \dots \} \subseteq \mathbb{R} and Y:Ω2{y1,y2,y3,}RY: \Omega_2 \rightarrow \{ y_1, y_2, y_3, \dots \} \subseteq \mathbb{R}. A function f:{x1,x2,}×{y1,y2,}[0,1]f: \{ x_1, x_2, \dots \} \times \{ y_1, y_2, \dots \} \rightarrow [0, 1] is a joint PMF if it satisfies:

    1. P(ω1,ω2Ω1×Ω2X(ω1)=xi,Y(ω2)=yk)=P(X=xi,Y=yk)=f(xi,yk)P(\omega_1, \omega_2 \in \Omega_1 \times \Omega_2 \mid X(\omega_1) = x_i, Y(\omega_2) = y_k) = P(X = x_i, Y = y_k) = f(x_i, y_k) for all i,kNi, k \in \mathbb{N}.

    2. i=1k=1f(xi,yk)=1\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} f(x_i, y_k) = 1.

  • Support: The set of all points where f(x_i, y_k) > 0.

  • Joint CDF (Discrete): For fixed xΩ1,yΩ2x \in \Omega_1, y \in \Omega_2:     FX,Y(x,y)=i:xixk:ykyfX,Y(xi,yk)F_{X,Y}(x, y) = \sum_{\forall i: x_i \le x} \sum_{\forall k: y_k \le y} f_{X,Y}(x_i, y_k)

CONTINUOUS JOINT DISTRIBUTIONS

  • Joint Probability Density Function (PDF): For sample spaces Ω1,Ω2\Omega_1, \Omega_2 with piecewise continuous random variables X,Y:ΩRX, Y: \Omega \rightarrow \mathbb{R}, a piecewise continuous function f:R×R[0,1]f: \mathbb{R} \times \mathbb{R} \rightarrow [0, 1] is a joint PDF if it satisfies:

    1. P(Xx,Yy)=xyf(t,s)dsdt=yxf(t,s)dtdsP(X \le x, Y \le y) = \int_{-\infty}^x \int_{-\infty}^y f(t, s) \,ds \,dt = \int_{-\infty}^y \int_{-\infty}^x f(t, s) \,dt \,ds for all x,yRx, y \in \mathbb{R}.

    2. f(t,s)dsdt=1\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t, s) \,ds \,dt = 1.

  • Support: The set of all points where f(x, y) > 0 for x,yRx, y \in \mathbb{R}.

  • Joint CDF (Continuous):     FX,Y(x,y)=xyf(t,s)dsdtF_{X,Y}(x, y) = \int_{-\infty}^x \int_{-\infty}^y f(t, s) \,ds \,dt

  • Geometric Interpretation: While a one-dimensional integral measures area under a curve, the joint integral measures the volume below the function ff and above the rectangle defined by (,x]×(,y](-\infty, x] \times (-\infty, y] for fixed x,yRx, y \in \mathbb{R}.

  • Fubini’s Theorem: This theorem proves that for these types of functions, the order of integration does not affect the value of the integral.

EXAMPLES OF JOINT DISTRIBUTIONS

Example 3.1: Rolling a Fair Die
  • Variables:

    • X=1X = 1 for the event (1,2)(1, 2), otherwise 00.

    • Y=1Y = 1 for an odd result (1,3,5)(1, 3, 5), otherwise 00.

  • Mapping Outcomes to (x,y)(x, y):

    • Outcome 1: (1,1)(1, 1)

    • Outcome 2: (1,0)(1, 0)

    • Outcome 3: (0,1)(0, 1)

    • Outcome 4: (0,0)(0, 0)

    • Outcome 5: (0,1)(0, 1)

    • Outcome 6: (0,0)(0, 0)

  • Inverse Mapping (Events):

    • (0,0){4,6}(0, 0) \rightarrow \{ 4, 6 \}

    • (0,1){3,5}(0, 1) \rightarrow \{ 3, 5 \}

    • (1,0){2}(1, 0) \rightarrow \{ 2 \}

    • (1,1){1}(1, 1) \rightarrow \{ 1 \}

  • Joint PMF Table:

    • fX,Y(0,0)=26=13f_{X,Y}(0, 0) = \frac{2}{6} = \frac{1}{3}

    • fX,Y(0,1)=26=13f_{X,Y}(0, 1) = \frac{2}{6} = \frac{1}{3}

    • fX,Y(1,0)=16f_{X,Y}(1, 0) = \frac{1}{6}

    • fX,Y(1,1)=16f_{X,Y}(1, 1) = \frac{1}{6}

Example 3.2: Drawing Marbles Without Replacement
  • Setup: Urn with 3 Green (G) and 6 Blue (B) marbles. Two are drawn without replacement.

  • Outcome Mapping:

    • GG(2,0)GG \rightarrow (2, 0)

    • GB(1,1)GB \rightarrow (1, 1)

    • BG(1,1)BG \rightarrow (1, 1)

    • BB(0,2)BB \rightarrow (0, 2)

  • Probability Calculations:

    • P(GG)=39×28=672P(GG) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}

    • P(GB)=39×68=1872P(GB) = \frac{3}{9} \times \frac{6}{8} = \frac{18}{72}

    • P(BG)=69×38=1872P(BG) = \frac{6}{9} \times \frac{3}{8} = \frac{18}{72}

    • P(BB)=69×58=3072P(BB) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72}

  • Joint PMF Results:

    • fX,Y(2,0)=672=112f_{X,Y}(2, 0) = \frac{6}{72} = \frac{1}{12}

    • fX,Y(1,1)=18+1872=3672=12f_{X,Y}(1, 1) = \frac{18+18}{72} = \frac{36}{72} = \frac{1}{2}

    • fX,Y(0,2)=3072=512f_{X,Y}(0, 2) = \frac{30}{72} = \frac{5}{12}

MULTIVARIATE HYPER-GEOMETRIC DISTRIBUTION

  • Context: A generalization of sampling without replacement from a population with multiple categories (colors).

  • Parameters:

    • KK: Number of categories (colors).

    • nkn_k: Number of items of the kthk^{th} color.

    • nn: Total number of items drawn.

    • XkX_k: Random variable counting the number of items of color kk in the sample.

  • Support Constraints:

    • x1+x2++xK=nx_1 + x_2 + \dots + x_K = n

    • 0xkmin(n,nk)0 \le x_k \le \min(n, n_k) for k=1,2,,Kk = 1, 2, \dots, K.

  • Joint PMF Formula:     fX1,,XK(x1,,xK)=(n1x1)(n2x2)(nKxK)(nkn)f_{X_1, \dots, X_K}(x_1, \dots, x_K) = \frac{\binom{n_1}{x_1} \binom{n_2}{x_2} \dots \binom{n_K}{x_K}}{\binom{\sum n_k}{n}}

  • Revisiting Example 3.2: With K=2K=2, n=2n=2, n1=3n_1=3 (green), and n2=6n_2=6 (blue):     fX1,X2(x1,x2)=(3x1)(6x2)(92)f_{X_1, X_2}(x_1, x_2) = \frac{\binom{3}{x_1} \binom{6}{x_2}}{\binom{9}{2}}     for x1+x2=2x_1 + x_2 = 2.

BIVARIATE UNIFORM DISTRIBUTION

  • Definition: A generalization of the univariate uniform distribution over a rectangular region [a,b]×[c,d][a, b] \times [c, d].

  • Joint PDF:     fX,Y(x,y)=1(ba)(dc)f_{X,Y}(x, y) = \frac{1}{(b-a)(d-c)} if (x,y)[a,b]×[c,d](x, y) \in [a, b] \times [c, d], and zero otherwise.

  • Measure Principle: The constant value of the PDF is the reciprocal of the area (measure) of the support.

  • Example Calculation over Non-Rectangular Region:

    • Given support [0,4]×[0,2][0, 4] \times [0, 2], then fX,Y(x,y)=1(40)(20)=18f_{X,Y}(x, y) = \frac{1}{(4-0)(2-0)} = \frac{1}{8}.

    • To find P(X+3Y3)P(X + 3Y \le 3), integrate over the triangular region with vertices (0,0),(0,1),(3,0)(0, 0), (0, 1), (3, 0).

    • Parametrize: 0y10 \le y \le 1 and 0x33y0 \le x \le 3 - 3y.

    • P=01033y18dxdy=0118(33y)dy=38[yy22]01=38(12)=316P = \int_0^1 \int_0^{3-3y} \frac{1}{8} \,dx \,dy = \int_0^1 \frac{1}{8}(3 - 3y) \,dy = \frac{3}{8} \left[ y - \frac{y^2}{2} \right]_0^1 = \frac{3}{8}\left(\frac{1}{2}\right) = \frac{3}{16}.

MULTIVARIATE NORMAL DISTRIBUTION

  • Bivariate Case Definition: Defined by a mean vector μ=(μ1,μ2)T\mu = (\mu_1, \mu_2)^T and a covariance matrix Σ=(σ12amp;σ12σ12amp;σ22)\Sigma = \begin{pmatrix} \sigma_1^2 & \sigma_{12} \\ \sigma_{12} & σ_2^2 \end{pmatrix}.

  • Determinant and Inverse:

    • det(Σ)=σ12σ22σ122\det(\Sigma) = \sigma_1^2 \sigma_2^2 - \sigma_{12}^2

    • Σ1=1det(Σ)(σ22amp;σ12σ12amp;σ12)\Sigma^{-1} = \frac{1}{\det(\Sigma)} \begin{pmatrix} \sigma_2^2 & -\sigma_{12} \\ -\sigma_{12} & \sigma_1^2 \end{pmatrix}

  • Joint PDF:     fX,Y(x,y)=12πdet(Σ)e12(xμ)TΣ1(xμ)f_{X,Y}(x, y) = \frac{1}{2\pi \sqrt{\det(\Sigma)}} e^{-\frac{1}{2} (\mathbf{x} - \mu)^T \Sigma^{-1} (\mathbf{x} - \mu)}

  • Covariance Effects:

    • Identity Covariance (Σ=(1amp;00amp;1)\Sigma = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}): Circular contours, no relationship between variables.

    • Positive Covariance (σ12=0.85\sigma_{12} = 0.85): Elliptical contours with positive slope. Higher values of one variable suggest higher values of the other.

    • Negative Covariance (σ12=0.85\sigma_{12} = -0.85): Elliptical contours with negative slope. Higher values of one variable suggest lower values of the other.

MARGINAL DISTRIBUTIONS

  • Concept: Determining the probability distribution of a single variable within a multivariate system by "summing out" or "integrating out" the other variables.

Discrete Marginal Distributions
  • Marginal PMF of X: fX(x)=j=1fX,Y(x,yj)f_X(x) = \sum_{j=1}^{\infty} f_{X,Y}(x, y_j).

  • Marginal PMF of Y: fY(y)=i=1fX,Y(xi,y)f_Y(y) = \sum_{i=1}^{\infty} f_{X,Y}(x_i, y).

  • Tabular Method: In a two-way table, the marginals are found by summing the rows or columns.

Continuous Marginal Distributions
  • Marginal PDF of X: fX(x)=fX,Y(x,y)dyf_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x, y) \,dy.

  • Marginal PDF of Y: fY(y)=fX,Y(x,y)dxf_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x, y) \,dx.

  • Example 3.3: Given fX,Y(x,y)=e(x+y)f_{X,Y}(x, y) = e^{-(x+y)} for x, y > 0:

    • fX(x)=0e(x+y)dy=ex0eydy=ex1=exf_X(x) = \int_0^{\infty} e^{-(x+y)} \,dy = e^{-x} \int_0^{\infty} e^{-y} \,dy = e^{-x} \cdot 1 = e^{-x}.

    • Both marginals follow an Exponential distribution with parameter 1.

  • Example 3.4: Given fX,Y(x,y)=x+y3f_{X,Y}(x, y) = \frac{x+y}{3} for x[0,1],y[0,2]x \in [0, 1], y \in [0, 2]:

    • fX(x)=02x+y3dy=[xy3+y26]02=2x3+46=23(x+1)f_X(x) = \int_0^2 \frac{x+y}{3} \,dy = \left[ \frac{xy}{3} + \frac{y^2}{6} \right]_0^2 = \frac{2x}{3} + \frac{4}{6} = \frac{2}{3}(x + 1).

    • fY(y)=01x+y3dx=[x26+xy3]01=16+y3=16(2y+1)f_Y(y) = \int_0^1 \frac{x+y}{3} \,dx = \left[ \frac{x^2}{6} + \frac{xy}{3} \right]_0^1 = \frac{1}{6} + \frac{y}{3} = \frac{1}{6}(2y + 1).

INDEPENDENT RANDOM VARIABLES

  • Definiton (Discrete): XX and YY are independent if fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x, y) = f_X(x) f_Y(y) for all possible pairs (x,y)(x, y).

  • Definition (Continuous): XX and YY are independent if fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x, y) = f_X(x) f_Y(y) for all x,yRx, y \in \mathbb{R}.

  • Standard Normal Case: If XN(0,1)X \sim N(0, 1) and YN(2,1)Y \sim N(2, 1) are independent, their joint PDF is:     fX,Y(x,y)=12πex2+(y2)22f_{X,Y}(x, y) = \frac{1}{2\pi} e^{-\frac{x^2 + (y-2)^2}{2}}

  • Diagonal Covariance Matrix: If the covariance matrix of a multivariate normal distribution is diagonal (σ12=0\sigma_{12} = 0), then the variables are independent.

  • Testing for Independence:

    • To prove independence, the product rule must hold for every pair.

    • To prove dependence, the product rule must fail for at least one pair.

  • Real-World Application: Global Warming Survey (N=500N=500 Participants):

    • Education levels (None, Undergraduate, Graduate) and Belief (Yes, No).

    • Modeled as multi-noulli distributions.

    • Expected counts under independence are calculated as Ei,k=NP(X=xi)P(Y=yk)E_{i,k} = N \cdot P(X=x_i) \cdot P(Y=y_k).

    • The χ2\chi^2 (Chi-square) distribution is used quantitatively to model departures from independence.

CONDITIONAL DISTRIBUTIONS

  • Core Formula: Based on the event probability P(AB)=P(AB)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

  • Discrete Conditional PMF: fXY(xy)=fX,Y(x,y)fY(y)f_{X|Y}(x \mid y) = \frac{f_{X,Y}(x, y)}{f_Y(y)} for f_Y(y) > 0.

  • Continuous Conditional PDF: fXY(xy)=fX,Y(x,y)fY(y)f_{X|Y}(x \mid y) = \frac{f_{X,Y}(x, y)}{f_Y(y)} for f_Y(y) > 0.

  • Deriving Joint from Conditional: fX,Y(x,y)=fYX(yx)fX(x)f_{X,Y}(x, y) = f_{Y|X}(y \mid x) f_X(x).

Example 3.9: Composite Distribution
  • Given: XUniform(1,2)X \sim \text{Uniform}(1, 2) and YX=xUniform(0,x)Y \mid X=x \sim \text{Uniform}(0, x).

  • Joint PDF:

    • fX(x)=1f_X(x) = 1 for 1x21 \le x \le 2.

    • fYX(yx)=1xf_{Y|X}(y \mid x) = \frac{1}{x} for 0yx0 \le y \le x.

    • fX,Y(x,y)=1xf_{X,Y}(x, y) = \frac{1}{x} for 0yx0 \le y \le x and 1x21 \le x \le 2.

  • Marginal of Y:

    • For 0y10 \le y \le 1: fY(y)=121xdx=ln(2)ln(1)=ln(2)f_Y(y) = \int_1^2 \frac{1}{x} \,dx = \ln(2) - \ln(1) = \ln(2).

    • For 1 < y \le 2: fY(y)=y21xdx=ln(2)ln(y)f_Y(y) = \int_y^2 \frac{1}{x} \,dx = \ln(2) - \ln(y).

Example 3.10: Ecological Predation Data
  • Data Provided:

    • P(Eagle)=0.2P(\text{Eagle}) = 0.2

    • P(Rabbit)=0.5P(\text{Rabbit}) = 0.5

    • P(EagleRabbit)=0.05P(\text{Eagle} \cap \text{Rabbit}) = 0.05

  • Completed Joint PMF Table:

    • fE,R(1,1)=0.05f_{E,R}(1, 1) = 0.05 (Eagle and Rabbit)

    • fE,R(1,0)=0.20.05=0.15f_{E,R}(1, 0) = 0.2 - 0.05 = 0.15 (Eagle, no Rabbit)

    • fE,R(0,1)=0.50.05=0.45f_{E,R}(0, 1) = 0.5 - 0.05 = 0.45 (Rabbit, no Eagle)

    • fE,R(0,0)=1(0.05+0.15+0.45)=0.35f_{E,R}(0, 0) = 1 - (0.05 + 0.15 + 0.45) = 0.35 (Neither)

  • Selected Conditionals:

    • P(E=1R=1)=0.050.5=0.1P(E=1 \mid R=1) = \frac{0.05}{0.5} = 0.1

    • P(E=1R=0)=0.150.5=0.3P(E=1 \mid R=0) = \frac{0.15}{0.5} = 0.3

    • P(R=1E=1)=0.050.2=0.25P(R=1 \mid E=1) = \frac{0.05}{0.2} = 0.25

    • P(R=1E=0)=0.450.8=0.5625P(R=1 \mid E=0) = \frac{0.45}{0.8} = 0.5625