Math Problem Statement

The bivariate random variable (X, Y) takes the values (1, 1) with probability 0.25, (2, 1) with probability 0.25, and (1, 2) with probability 0.50. Find the median of X, the variance of Y, and the covariance of X and Y. Let F(x, y) be the distribution function of (X, Y). Find F(1, 2). Draw the distribution function of X and the probability function of Y.

Solution

The problem in the image involves a bivariate random variable (X,Y)(X, Y) with a given probability distribution. The random variable takes the following values:

  • (1,1)(1, 1) with probability 0.25
  • (2,1)(2, 1) with probability 0.25
  • (1,2)(1, 2) with probability 0.50

The task requires calculating:

  1. The median of XX
  2. The variance of YY
  3. The covariance of XX and YY
  4. The distribution function F(x,y)F(x, y)
  5. The value of F(1,2)F(1, 2)
  6. The distribution function of XX
  7. The probability function of YY

Let’s go through these calculations step-by-step:

1. Median of XX:

The median is the value where half of the data lies below and half above. We can organize the distribution of XX:

  • X=1X = 1 with total probability 0.25+0.50=0.750.25 + 0.50 = 0.75
  • X=2X = 2 with total probability 0.250.25

Thus, the median is X=1X = 1, since the cumulative probability reaches 0.75 for X=1X = 1, exceeding 0.5.

2. Variance of YY:

We need to calculate the expected value E(Y)E(Y) and then E(Y2)E(Y^2) to find the variance Var(Y)=E(Y2)[E(Y)]2\text{Var}(Y) = E(Y^2) - [E(Y)]^2.

  • P(Y=1)=0.25+0.25=0.50P(Y = 1) = 0.25 + 0.25 = 0.50
  • P(Y=2)=0.50P(Y = 2) = 0.50

Expected value of YY:

E(Y)=1(0.50)+2(0.50)=0.50+1=1.50E(Y) = 1(0.50) + 2(0.50) = 0.50 + 1 = 1.50

E(Y2)E(Y^2):

E(Y2)=12(0.50)+22(0.50)=0.50+4(0.50)=0.50+2=2.50E(Y^2) = 1^2(0.50) + 2^2(0.50) = 0.50 + 4(0.50) = 0.50 + 2 = 2.50

Variance of YY:

Var(Y)=E(Y2)[E(Y)]2=2.50(1.50)2=2.502.25=0.25\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = 2.50 - (1.50)^2 = 2.50 - 2.25 = 0.25

3. Covariance of XX and YY:

Covariance is given by: Cov(X,Y)=E(XY)E(X)E(Y)\text{Cov}(X, Y) = E(XY) - E(X)E(Y)

First, calculate E(XY)E(XY): E(XY)=(11)(0.25)+(21)(0.25)+(12)(0.50)=0.25+0.50+1=1.75E(XY) = (1 \cdot 1)(0.25) + (2 \cdot 1)(0.25) + (1 \cdot 2)(0.50) = 0.25 + 0.50 + 1 = 1.75

Now, calculate E(X)E(X): E(X)=1(0.75)+2(0.25)=0.75+0.50=1.25E(X) = 1(0.75) + 2(0.25) = 0.75 + 0.50 = 1.25

Now, use the previously calculated E(Y)=1.50E(Y) = 1.50.

Thus, Cov(X,Y)=1.75(1.25)(1.50)=1.751.875=0.125\text{Cov}(X, Y) = 1.75 - (1.25)(1.50) = 1.75 - 1.875 = -0.125

4. Distribution function F(x,y)F(x, y):

The distribution function F(x,y)F(x, y) is the probability that XxX \leq x and YyY \leq y. We will compute it for different values of xx and yy.

For F(1,2)F(1, 2), the probability that X1X \leq 1 and Y2Y \leq 2:

  • For X=1,Y=1X = 1, Y = 1 and X=1,Y=2X = 1, Y = 2, the combined probability is 0.25+0.50=0.750.25 + 0.50 = 0.75.

Thus, F(1,2)=0.75F(1, 2) = 0.75.

5. Distribution function of XX:

The marginal distribution of XX is:

  • P(X=1)=0.25+0.50=0.75P(X = 1) = 0.25 + 0.50 = 0.75
  • P(X=2)=0.25P(X = 2) = 0.25

6. Probability function of YY:

The marginal distribution of YY is:

  • P(Y=1)=0.25+0.25=0.50P(Y = 1) = 0.25 + 0.25 = 0.50
  • P(Y=2)=0.50P(Y = 2) = 0.50

Would you like any additional clarification on these solutions?

5 Related Questions:

  1. How is the covariance between two random variables interpreted in statistics?
  2. Can you explain the relationship between variance and standard deviation?
  3. What is the significance of the distribution function in probability theory?
  4. How would the distribution change if additional values were added for XX and YY?
  5. What are some common applications of bivariate distributions in real-life scenarios?

Tip:

Covariance helps determine whether two variables increase or decrease together. If it's positive, they tend to increase together; if negative, one tends to decrease when the other increases.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Random Variables
Bivariate Distributions
Covariance
Variance
Median

Formulas

E(Y) = Σ[y_i * P(Y = y_i)]
Var(Y) = E(Y^2) - [E(Y)]^2
Cov(X, Y) = E(XY) - E(X)E(Y)
F(x, y) = P(X ≤ x, Y ≤ y)

Theorems

Expected Value
Variance
Covariance
Cumulative Distribution Function

Suitable Grade Level

College-Level Statistics