Math Problem Statement

Let X and Y be continuous random variables with joint density function

fx,y(x, y)=3x/ 4 for 0 < x < 2 and 0<y<2-x

What is P(X > 1)?

Solution

To find the probability P(X>1)P(X > 1) for the continuous random variables XX and YY with the given joint density function, we need to integrate the joint density over the region where X>1X > 1.

Step-by-Step Solution:

  1. Understand the Joint Density Function:

    The joint density function is given by: fX,Y(x,y)=3x4,for 0<x<2 and 0<y<2x.f_{X,Y}(x, y) = \frac{3x}{4}, \quad \text{for } 0 < x < 2 \text{ and } 0 < y < 2 - x.

  2. Set up the Probability P(X>1)P(X > 1):

    We need to integrate the joint density function over the region where X>1X > 1. Since XX ranges from 1 to 2, and for each value of xx, yy ranges from 0 to 2x2 - x, we have: P(X>1)=1202xfX,Y(x,y)dydx.P(X > 1) = \int_{1}^{2} \int_{0}^{2-x} f_{X,Y}(x, y) \, dy \, dx.

  3. Substitute the Joint Density Function:

    Substituting fX,Y(x,y)=3x4f_{X,Y}(x, y) = \frac{3x}{4}, the probability becomes: P(X>1)=1202x3x4dydx.P(X > 1) = \int_{1}^{2} \int_{0}^{2-x} \frac{3x}{4} \, dy \, dx.

  4. Integrate with Respect to yy First:

    P(X>1)=12(3x402xdy)dx.P(X > 1) = \int_{1}^{2} \left( \frac{3x}{4} \int_{0}^{2-x} dy \right) dx.

    The inner integral with respect to yy is: 02xdy=(2x)0=2x.\int_{0}^{2-x} dy = (2 - x) - 0 = 2 - x.

    So, the probability becomes: P(X>1)=123x4(2x)dx.P(X > 1) = \int_{1}^{2} \frac{3x}{4} (2 - x) \, dx.

  5. Simplify and Integrate with Respect to xx:

    Distribute 3x4\frac{3x}{4}: P(X>1)=12(3x243x24)dx=12(3x23x24)dx.P(X > 1) = \int_{1}^{2} \left( \frac{3x \cdot 2}{4} - \frac{3x^2}{4} \right) dx = \int_{1}^{2} \left( \frac{3x}{2} - \frac{3x^2}{4} \right) dx.

    Separate the integrals: P(X>1)=3212xdx3412x2dx.P(X > 1) = \frac{3}{2} \int_{1}^{2} x \, dx - \frac{3}{4} \int_{1}^{2} x^2 \, dx.

    Compute each integral:

    • xdx=x22\int x \, dx = \frac{x^2}{2}
    • x2dx=x33\int x^2 \, dx = \frac{x^3}{3}

    Plugging in the limits from 1 to 2:

    [ P(X > 1) = \frac{3}{2} \left[ \frac{x^2}{2} \right]{1}^{2} - \frac{3}{4} \left[ \frac{x^3}{3} \right]{1}^{2}. ]

    Evaluate the limits: [x22]12=222122=4212=212=32.\left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}.

    [x33]12=233133=8313=73.\left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.

    Therefore: P(X>1)=32323473.P(X > 1) = \frac{3}{2} \cdot \frac{3}{2} - \frac{3}{4} \cdot \frac{7}{3}.

    Simplify the terms: P(X>1)=9474=24=12.P(X > 1) = \frac{9}{4} - \frac{7}{4} = \frac{2}{4} = \frac{1}{2}.

Final Answer:

P(X>1)=12.P(X > 1) = \frac{1}{2}.

Would you like any further details or have any questions?

Relative Questions:

  1. How do you find the marginal density function of XX from a joint density function?
  2. What is the meaning of a joint density function in probability theory?
  3. How do you compute the probability P(X1)P(X \leq 1) for this problem?
  4. How do you verify if a joint density function is valid?
  5. What is the difference between continuous and discrete random variables?

Tip: To find a marginal distribution from a joint density, integrate the joint density over all possible values of the other variable.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Joint Density Function
Integral Calculus

Formulas

Joint density function
Integral of a continuous function

Theorems

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Suitable Grade Level

Advanced Mathematics