Answer :
Let's address each part of the question step by step.
### Part (a): Find \( k \).
To find the constant \( k \), we must use the fact that the total probability over the entire range of the joint probability density function (pdf) must equal 1. This means integrating the given function over the specified range.
The joint pdf is \( f_{X,Y}(x,y) = k(x-y) \) for \( 0 < y < x < 2 \).
1. The integration bounds for \( x \) are from \( y \) to 2.
2. The integration bounds for \( y \) are from 0 to 2.
So, we need to solve:
[tex]\[ \int_{0}^{2} \int_{y}^{2} k(x-y) \, dx \, dy = 1 \][/tex]
First, integrate with respect to \( x \):
[tex]\[ \int_{0}^{2} \left[ \int_{y}^{2} k(x-y) \, dx \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} \left[ k \int_{y}^{2} (x-y) \, dx \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} \left[ k \left( \frac{(x-y)^2}{2} \Bigg|_{y}^{2} \right) \right] dy \][/tex]
Evaluating the inner integral:
[tex]\[ = \int_{0}^{2} \left[ k \left( \frac{(2-y)^2}{2} - 0 \right) \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} k \frac{(2-y)^2}{2} \, dy \][/tex]
Next, simplify and integrate with respect to \( y \):
[tex]\[ = k \int_{0}^{2} \frac{(2-y)^2}{2} \, dy \][/tex]
[tex]\[ = k \frac{1}{2} \int_{0}^{2} (4 - 4y + y^2) \, dy \][/tex]
[tex]\[ = k \frac{1}{2} \left[ 4y - 2y^2 + \frac{y^3}{3} \Bigg|_{0}^{2} \right] \][/tex]
Evaluating this:
[tex]\[ = k \frac{1}{2} \left[ (8 - 8 + \frac{8}{3}) - ( 0 ) \right] \][/tex]
[tex]\[ = k \frac{1}{2} \cdot \frac{8}{3} \][/tex]
[tex]\[ = \frac{4k}{3} \][/tex]
Set this result equal to 1:
[tex]\[ \frac{4k}{3} = 1 \][/tex]
Solving for \( k \):
[tex]\[ k = \frac{3}{4} \][/tex]
### Part (b): Find \( P(\max \{X, Y\} \leq 1) \).
The event \(\{ \max \{X, Y\} \leq 1 \}\) indicates that both \( X \) and \( Y \) must be less than or equal to 1.
The range for the integration is \(0 < y < x < 1 \):
[tex]\[ P(\max\{X, Y\} \leq 1) = \int_{0}^{1} \int_{y}^{1} k(x-y) \, dx \, dy \][/tex]
Using \( k = \frac{3}{4} \):
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \int_{y}^{1} (x-y) \, dx \right] dy \][/tex]
Integrating with respect to \( x \):
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \frac{(x-y)^2}{2} \Bigg|_{y}^{1} \right] dy \][/tex]
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \frac{(1-y)^2}{2} \right] dy \][/tex]
[tex]\[ = \frac{3}{8} \int_{0}^{1} (1-y)^2 \, dy \][/tex]
[tex]\[ = \frac{3}{8} \left[ \frac{(1-y)^3}{3} \Bigg|_{0}^{1} \right] \][/tex]
[tex]\[ = \frac{3}{8} \cdot \frac{1}{3} \][/tex]
[tex]\[ = \frac{1}{8} \][/tex]
### Part (c): Find \(\widehat{x}_M(y)\), the minimum mean square error estimate of \( X \) given \( Y=y \), \( 0
The conditional expectation of \( X \) given \( Y = y \) is:
[tex]\[ \widehat{x}_M(y) = E[X|Y=y] \][/tex]
This can be found by:
[tex]\[ E[X|Y=y] = \frac{\int_{y}^{2} x f_{X|Y}(x|y) \, dx}{\int_{y}^{2} f_{X|Y}(x|y) \, dx} \][/tex]
Given \( f_{X,Y}(x,y) = \frac{3}{4}(x-y) \):
The conditional pdf \( f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \):
First, find the marginal pdf of \( Y \):
[tex]\[ f_Y(y) = \int_{y}^{2} \frac{3}{4}(x-y) \, dx \][/tex]
[tex]\[ = \frac{3}{4} \int_{y}^{2} (x-y) \, dx \][/tex]
[tex]\[ = \frac{3}{4} \left[ \frac{(x-y)^2}{2} \Bigg|_{y}^{2} \right] \][/tex]
[tex]\[ = \frac{3}{4} \left[ \frac{(2-y)^2}{2} - 0 \right] \][/tex]
[tex]\[ = \frac{3}{8} (2-y)^2 \][/tex]
Now, calculate the conditional expectation:
[tex]\[ E[X|Y=y] = \frac{\int_{y}^{2} x \cdot \frac{3}{4}(x-y) \, dx}{f_Y(y)} \][/tex]
[tex]\[ = \frac{\int_{y}^{2} x \cdot \frac{3}{4}(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
Simplify the numerator:
[tex]\[ = \frac{\int_{y}^{2} \frac{3}{4}x(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \int_{y}^{2} x(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \int_{y}^{2} (x^2 - xy) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{x^3}{3} - \frac{xy^2}{2} \Bigg|_{y}^{2} \right]}{\frac{3}{8}(2-y)^2} \][/tex]
Evaluating the integral at the limits:
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{8}{3} - 2y - 0 \right]}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{8}{3} - 2y \right]}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{6 - 3y}{(2-y)^2} \][/tex]
Thus, \(\widehat{x}_M(y) = \frac{6 - 3y}{(2-y)^2}\).
So, the final answers are:
(a) \( k = \frac{3}{4} \)
(b) \( P(\max \{X, Y\} \leq 1) = \frac{1}{8} \)
(c) [tex]\( \widehat{x}_M(y) = \frac{6 - 3y}{(2-y)^2} \)[/tex]
### Part (a): Find \( k \).
To find the constant \( k \), we must use the fact that the total probability over the entire range of the joint probability density function (pdf) must equal 1. This means integrating the given function over the specified range.
The joint pdf is \( f_{X,Y}(x,y) = k(x-y) \) for \( 0 < y < x < 2 \).
1. The integration bounds for \( x \) are from \( y \) to 2.
2. The integration bounds for \( y \) are from 0 to 2.
So, we need to solve:
[tex]\[ \int_{0}^{2} \int_{y}^{2} k(x-y) \, dx \, dy = 1 \][/tex]
First, integrate with respect to \( x \):
[tex]\[ \int_{0}^{2} \left[ \int_{y}^{2} k(x-y) \, dx \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} \left[ k \int_{y}^{2} (x-y) \, dx \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} \left[ k \left( \frac{(x-y)^2}{2} \Bigg|_{y}^{2} \right) \right] dy \][/tex]
Evaluating the inner integral:
[tex]\[ = \int_{0}^{2} \left[ k \left( \frac{(2-y)^2}{2} - 0 \right) \right] dy \][/tex]
[tex]\[ = \int_{0}^{2} k \frac{(2-y)^2}{2} \, dy \][/tex]
Next, simplify and integrate with respect to \( y \):
[tex]\[ = k \int_{0}^{2} \frac{(2-y)^2}{2} \, dy \][/tex]
[tex]\[ = k \frac{1}{2} \int_{0}^{2} (4 - 4y + y^2) \, dy \][/tex]
[tex]\[ = k \frac{1}{2} \left[ 4y - 2y^2 + \frac{y^3}{3} \Bigg|_{0}^{2} \right] \][/tex]
Evaluating this:
[tex]\[ = k \frac{1}{2} \left[ (8 - 8 + \frac{8}{3}) - ( 0 ) \right] \][/tex]
[tex]\[ = k \frac{1}{2} \cdot \frac{8}{3} \][/tex]
[tex]\[ = \frac{4k}{3} \][/tex]
Set this result equal to 1:
[tex]\[ \frac{4k}{3} = 1 \][/tex]
Solving for \( k \):
[tex]\[ k = \frac{3}{4} \][/tex]
### Part (b): Find \( P(\max \{X, Y\} \leq 1) \).
The event \(\{ \max \{X, Y\} \leq 1 \}\) indicates that both \( X \) and \( Y \) must be less than or equal to 1.
The range for the integration is \(0 < y < x < 1 \):
[tex]\[ P(\max\{X, Y\} \leq 1) = \int_{0}^{1} \int_{y}^{1} k(x-y) \, dx \, dy \][/tex]
Using \( k = \frac{3}{4} \):
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \int_{y}^{1} (x-y) \, dx \right] dy \][/tex]
Integrating with respect to \( x \):
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \frac{(x-y)^2}{2} \Bigg|_{y}^{1} \right] dy \][/tex]
[tex]\[ = \frac{3}{4} \int_{0}^{1} \left[ \frac{(1-y)^2}{2} \right] dy \][/tex]
[tex]\[ = \frac{3}{8} \int_{0}^{1} (1-y)^2 \, dy \][/tex]
[tex]\[ = \frac{3}{8} \left[ \frac{(1-y)^3}{3} \Bigg|_{0}^{1} \right] \][/tex]
[tex]\[ = \frac{3}{8} \cdot \frac{1}{3} \][/tex]
[tex]\[ = \frac{1}{8} \][/tex]
### Part (c): Find \(\widehat{x}_M(y)\), the minimum mean square error estimate of \( X \) given \( Y=y \), \( 0
The conditional expectation of \( X \) given \( Y = y \) is:
[tex]\[ \widehat{x}_M(y) = E[X|Y=y] \][/tex]
This can be found by:
[tex]\[ E[X|Y=y] = \frac{\int_{y}^{2} x f_{X|Y}(x|y) \, dx}{\int_{y}^{2} f_{X|Y}(x|y) \, dx} \][/tex]
Given \( f_{X,Y}(x,y) = \frac{3}{4}(x-y) \):
The conditional pdf \( f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \):
First, find the marginal pdf of \( Y \):
[tex]\[ f_Y(y) = \int_{y}^{2} \frac{3}{4}(x-y) \, dx \][/tex]
[tex]\[ = \frac{3}{4} \int_{y}^{2} (x-y) \, dx \][/tex]
[tex]\[ = \frac{3}{4} \left[ \frac{(x-y)^2}{2} \Bigg|_{y}^{2} \right] \][/tex]
[tex]\[ = \frac{3}{4} \left[ \frac{(2-y)^2}{2} - 0 \right] \][/tex]
[tex]\[ = \frac{3}{8} (2-y)^2 \][/tex]
Now, calculate the conditional expectation:
[tex]\[ E[X|Y=y] = \frac{\int_{y}^{2} x \cdot \frac{3}{4}(x-y) \, dx}{f_Y(y)} \][/tex]
[tex]\[ = \frac{\int_{y}^{2} x \cdot \frac{3}{4}(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
Simplify the numerator:
[tex]\[ = \frac{\int_{y}^{2} \frac{3}{4}x(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \int_{y}^{2} x(x-y) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \int_{y}^{2} (x^2 - xy) \, dx}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{x^3}{3} - \frac{xy^2}{2} \Bigg|_{y}^{2} \right]}{\frac{3}{8}(2-y)^2} \][/tex]
Evaluating the integral at the limits:
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{8}{3} - 2y - 0 \right]}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{\frac{3}{4} \left[ \frac{8}{3} - 2y \right]}{\frac{3}{8}(2-y)^2} \][/tex]
[tex]\[ = \frac{6 - 3y}{(2-y)^2} \][/tex]
Thus, \(\widehat{x}_M(y) = \frac{6 - 3y}{(2-y)^2}\).
So, the final answers are:
(a) \( k = \frac{3}{4} \)
(b) \( P(\max \{X, Y\} \leq 1) = \frac{1}{8} \)
(c) [tex]\( \widehat{x}_M(y) = \frac{6 - 3y}{(2-y)^2} \)[/tex]
