Math Problem Statement

Suppose X and Y take values in {0,1} with joint mass function p(x,y). Write p(0,0) = a, p(0,1) = b, p(1,0) = c, p(1,1) = d. Find necessary and sufficient conditions on a, b, c, d for X, Y to be (a) uncorrelated (b) independent.

Solution

The given problem describes two random variables XX and YY taking values in {0,1}\{0,1\}, with the joint probability mass function p(x,y)p(x,y). It defines the probabilities:

p(0,0)=a,p(0,1)=b,p(1,0)=c,p(1,1)=dp(0,0) = a, \quad p(0,1) = b, \quad p(1,0) = c, \quad p(1,1) = d

(a) Uncorrelated Condition:

Two random variables XX and YY are uncorrelated if their covariance is zero. The covariance is given by:

[ \text{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] ]

Step 1: Compute E[X]\mathbb{E}[X] and E[Y]\mathbb{E}[Y]

E[X]=0(a+b)+1(c+d)=c+d\mathbb{E}[X] = 0 \cdot (a + b) + 1 \cdot (c + d) = c + d E[Y]=0(a+c)+1(b+d)=b+d\mathbb{E}[Y] = 0 \cdot (a + c) + 1 \cdot (b + d) = b + d

Step 2: Compute E[XY]\mathbb{E}[XY]

E[XY]=00a+01b+10c+11d=d\mathbb{E}[XY] = 0 \cdot 0 \cdot a + 0 \cdot 1 \cdot b + 1 \cdot 0 \cdot c + 1 \cdot 1 \cdot d = d

Step 3: Covariance formula

Cov(X,Y)=d(c+d)(b+d)\text{Cov}(X,Y) = d - (c+d)(b+d)

Thus, for XX and YY to be uncorrelated, the covariance must be zero:

d(c+d)(b+d)=0d - (c+d)(b+d) = 0

(b) Independent Condition:

Two random variables XX and YY are independent if the joint probability factorizes into the product of the marginal probabilities. This means:

p(x,y)=pX(x)pY(y)p(x,y) = p_X(x)p_Y(y)

Step 1: Marginal probabilities

  • Marginal probability of XX: pX(0)=a+b,pX(1)=c+dp_X(0) = a + b, \quad p_X(1) = c + d
  • Marginal probability of YY: pY(0)=a+c,pY(1)=b+dp_Y(0) = a + c, \quad p_Y(1) = b + d

Step 2: Factorization condition

For independence, we need: p(0,0)=pX(0)pY(0),p(0,1)=pX(0)pY(1),p(1,0)=pX(1)pY(0),p(1,1)=pX(1)pY(1)p(0,0) = p_X(0)p_Y(0), \quad p(0,1) = p_X(0)p_Y(1), \quad p(1,0) = p_X(1)p_Y(0), \quad p(1,1) = p_X(1)p_Y(1)

This gives the following system of equations: a=(a+b)(a+c),b=(a+b)(b+d),c=(c+d)(a+c),d=(c+d)(b+d)a = (a+b)(a+c), \quad b = (a+b)(b+d), \quad c = (c+d)(a+c), \quad d = (c+d)(b+d)

Summary:

  • For XX and YY to be uncorrelated, the condition is: d=(c+d)(b+d)d = (c+d)(b+d)
  • For XX and YY to be independent, the conditions are the system of four equations: a=(a+b)(a+c),b=(a+b)(b+d),c=(c+d)(a+c),d=(c+d)(b+d)a = (a+b)(a+c), \quad b = (a+b)(b+d), \quad c = (c+d)(a+c), \quad d = (c+d)(b+d)

Let me know if you want more details or have further questions!

Related Questions:

  1. How do you compute the marginal probabilities from a joint probability table?
  2. What is the difference between uncorrelated and independent random variables?
  3. Can two random variables be uncorrelated but not independent?
  4. How does the covariance between two random variables affect their correlation?
  5. What are the properties of joint probability mass functions for discrete variables?

Tip: Independence implies uncorrelation, but the reverse is not always true. Be mindful when interpreting relationships between variables.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Covariance
Joint Probability
Independence

Formulas

Cov(X,Y) = E[XY] - E[X]E[Y]
E[X] = Σ x * p(x)
E[Y] = Σ y * p(y)
Independence condition: p(x, y) = p_X(x)p_Y(y)

Theorems

Uncorrelated Variables: Cov(X,Y) = 0
Independence of Variables: Joint distribution equals the product of marginals

Suitable Grade Level

College-level (Advanced Probability, Statistics)