91Ó°ÊÓ

Joint distribution of X and Y is uniform over a set A in the xy plane

let X and Y be two random variables.

The probability of simultaneous occurrence of the event x and y is called as joint pdf of X and Y.it can be given by\(f\left( {x,y} \right)\)

The marginal pdf of X and Y can be given as:

\({f_1}\left( x \right) = \int\limits_y {f\left( {x,y} \right)} dy\)

\({f_2}\left( y \right) = \int\limits_x {f\left( {x,y} \right)} dx\)

The conditional pdf of x and y can be defined as:

\({g_1}\left( {x|y} \right) = \frac{{f\left( {x,y} \right)}}{{{f_2}\left( y \right)}}\)

\({g_2}\left( {y|x} \right) = \frac{{f\left( {x,y} \right)}}{{{f_1}\left( x \right)}}\)

By using multiplication rule:

\(\begin{array}{c}f\left( {x,y} \right) = {g_1}\left( {x|y} \right){f_2}\left( y \right)\\ = {g_2}\left( {y|x} \right){f_1}\left( y \right)\end{array}\)

Suppose that the joint distribution of x and y is uniform over the set A in the xy-plane independent of the two random variables X and Y

The random variables are said to be independent ,if and only if for each y and x we have

\(\begin{array}{l}{\rm P}\left( {X = x|Y = y} \right)\\ = {\rm P}\left( {X = x} \right)\end{array}\)

A be the circle with radius 1 and center at the origin.

Area of the circle can be given by the equation\({x^2} + {y^2} = 1\)

From the given equation we can say that x and y are dependent which gives the value of the radius.

Hence x and y are not independent.

The joint density function \({x^2} + {y^2} = r\)can not be factored is not contant, hence it is assumed to not be independent.

A be the circle with radius 1 and center at the point (3,5)

area of the circle can be given by the equation\({x^2} + {y^2} = {1^2}\)

we can say that x and y are dependent on each other which gives the value for the radius.

Hence we conclude that x and y are not dependent.

The joint density function \({x^2} + {y^2} = r\)can not be factored is not constant, hence it is assumed to not be independent

A be the square with the vertices at the four points(1,1) (1,-1) (-1,-1) and (-1,1)

It is given that the joint probability distribution of the four points above is resulting into uniform distribution.

The joint pdf of X and Y is positive over a rectangle with sides parallel to the axes.

Since the probability density function of the uniform distribution is constant we can factor the pdf

Hence it can be concluded that x and y are independent.

A be the rectangle with vertices at the points(0,0) (0,3) (1,3) (1,0)

It is given that the joint pdf of these 4 points results into uniform distribution.

The joint pdf of x and y is positive over a rectangle with sides parallel to the axes.

Since the pdf of the uniform distribution is constant we can factor this pdf

Therefore x and y are independent

A be the square with vertices at the points(0,0) (1,1) (0,2) (-1,1)

It is given that the joint pdf of these 4 points results into uniform distribution

The joint pdf of x and y is positive over a rectangle with sides parallel to the axes.

Since the pdf of the uniform distribution is not constant we can not factor this pdf

Therefore x and y are not independent