Math Problem Statement

Consider the joint probability density function of continuous random variables X and Y: f(x, y) = { kxy^2, 0 < x < 2, 0 < y < 1 { 0, otherwise (a) Find the value of k. (b) Find the marginal probability density functions of X and Y. (c) Find an expression for the conditional probability density function f_{X|Y}(x|y) for 0 < y < 1.

Solution

To solve the problem related to the joint probability density function f(x,y)f(x, y), let us address each part systematically:

(a) Find the value of kk:

The total probability for a joint probability density function must integrate to 1 over the entire range. Thus, we solve: 0102kxy2dxdy=1.\int_{0}^{1} \int_{0}^{2} kxy^2 \, dx \, dy = 1.

Solution:

  1. Perform the inner integral with respect to xx: 02kxy2dx=ky202xdx=ky2[x22]02=ky22=2ky2.\int_{0}^{2} kxy^2 \, dx = ky^2 \int_{0}^{2} x \, dx = ky^2 \left[ \frac{x^2}{2} \right]_{0}^{2} = ky^2 \cdot 2 = 2ky^2.

  2. Perform the outer integral with respect to yy: 012ky2dy=2k01y2dy=2k[y33]01=2k13=2k3.\int_{0}^{1} 2ky^2 \, dy = 2k \int_{0}^{1} y^2 \, dy = 2k \left[ \frac{y^3}{3} \right]_{0}^{1} = 2k \cdot \frac{1}{3} = \frac{2k}{3}.

  3. Set this equal to 1 and solve for kk: 2k3=1    k=32.\frac{2k}{3} = 1 \implies k = \frac{3}{2}.

Thus, k=32k = \frac{3}{2}.


(b) Find the marginal probability density functions of XX and YY:

fX(x)f_X(x):

The marginal density fX(x)f_X(x) is obtained by integrating the joint density over yy: fX(x)=01f(x,y)dy=0132xy2dy.f_X(x) = \int_{0}^{1} f(x, y) \, dy = \int_{0}^{1} \frac{3}{2} xy^2 \, dy.

  1. Compute the integral: 0132xy2dy=32x01y2dy=32x[y33]01=32x13=x2.\int_{0}^{1} \frac{3}{2} xy^2 \, dy = \frac{3}{2} x \int_{0}^{1} y^2 \, dy = \frac{3}{2} x \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{3}{2} x \cdot \frac{1}{3} = \frac{x}{2}.

Thus, fX(x)=x2f_X(x) = \frac{x}{2}, for 0<x<20 < x < 2.

fY(y)f_Y(y):

The marginal density fY(y)f_Y(y) is obtained by integrating the joint density over xx: fY(y)=02f(x,y)dx=0232xy2dx.f_Y(y) = \int_{0}^{2} f(x, y) \, dx = \int_{0}^{2} \frac{3}{2} xy^2 \, dx.

  1. Compute the integral: 0232xy2dx=32y202xdx=32y2[x22]02=32y22=3y2.\int_{0}^{2} \frac{3}{2} xy^2 \, dx = \frac{3}{2} y^2 \int_{0}^{2} x \, dx = \frac{3}{2} y^2 \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{3}{2} y^2 \cdot 2 = 3y^2.

Thus, fY(y)=3y2f_Y(y) = 3y^2, for 0<y<10 < y < 1.


(c) Find the conditional probability density function fXY(xy)f_{X|Y}(x|y):

The conditional density fXY(xy)f_{X|Y}(x|y) is given by: fXY(xy)=f(x,y)fY(y).f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)}.

  1. Substitute f(x,y)=32xy2f(x, y) = \frac{3}{2} xy^2 and fY(y)=3y2f_Y(y) = 3y^2: fXY(xy)=32xy23y2.f_{X|Y}(x|y) = \frac{\frac{3}{2} xy^2}{3y^2}.

  2. Simplify: fXY(xy)=12x,0<x<2,0<y<1.f_{X|Y}(x|y) = \frac{1}{2} x, \quad 0 < x < 2, \, 0 < y < 1.

Thus, fXY(xy)=12xf_{X|Y}(x|y) = \frac{1}{2} x.


Summary of Results:

  • (a) k=32k = \frac{3}{2}.
  • (b) Marginal densities:
    • fX(x)=x2,0<x<2f_X(x) = \frac{x}{2}, \, 0 < x < 2.
    • fY(y)=3y2,0<y<1f_Y(y) = 3y^2, \, 0 < y < 1.
  • (c) Conditional density:
    • fXY(xy)=12x,0<x<2,0<y<1f_{X|Y}(x|y) = \frac{1}{2} x, \, 0 < x < 2, \, 0 < y < 1.

Would you like further explanations or graphical visualizations of these results? Here are five related questions to expand your understanding:

  1. How would the range of xx and yy affect the validity of the joint probability density function?
  2. Can you verify that the marginal densities fX(x)f_X(x) and fY(y)f_Y(y) integrate to 1?
  3. How is the conditional density fXY(xy)f_{X|Y}(x|y) interpreted in terms of dependence between XX and YY?
  4. What is the expected value of XX and YY using their marginal densities?
  5. How would you extend this to find Var(X)\text{Var}(X) or Cov(X,Y)\text{Cov}(X, Y)?

Tip: Always ensure the joint density integrates to 1 across its range before proceeding with marginal or conditional calculations!

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Math Problem Analysis

Mathematical Concepts

Joint Probability Density Functions
Marginal Probability Density
Conditional Probability Density

Formulas

∫∫ f(x, y) dx dy = 1 (Normalization condition for PDF)
f_X(x) = ∫ f(x, y) dy (Marginal PDF of X)
f_Y(y) = ∫ f(x, y) dx (Marginal PDF of Y)
f_{X|Y}(x|y) = f(x, y) / f_Y(y) (Conditional PDF)

Theorems

Law of Total Probability
Normalization Condition of PDF

Suitable Grade Level

Undergraduate