Find out the joint distribution, Civil Engineering

Find out the joint distribution:

Let XI and X2 be two independent random variables each distributed uniformly in the interval [ 0, a ], where a > 0 is a constant. Find out the joint distribution of

Yl = Xl + X2 and Y2 = X1 - X2.

Instead, in vector notation, what is the distribution of Y = XA.

where

x = (X1,X2),Y= (Y1,Y2), A = 584_Find out the joint distribution.png? Find also the marginal distributions of Y1 and Y2. ?

Solution:

The joint pdf of X is

fx(x) = 1/a2, (x1,x2)? R(x)

= 0 otherwise.

Where

R(x) = {(x1,x2):0 ≤ x1 ≤ a, 0 ≤  x2 ≤ a}

The Jacobian of the transformation is

347_Find out the joint distribution1.png

Hence the pdf of Y is

fy(y) = 1/2a2, (y1,y2)? R(y)

= 0 otherwise.

where R ( y ) is the transformed region R ( x ) under the transformation Y = XA. The range of variation of Yl is clearly [ 0,2a ] and that of Y2 is [ - a, + a ]. However Yl and Y2 are not independent.

Since the inverse transformation is

X1= ½ (Y1 + Y2), X2 = ½ (Y1 - Y2) and 0≤ x1, x2 ≤ a,

the region R ( y ) is given by

R(y) = {( Y1 + Y2) : 0 ≤ Y1 + Y2 ≤ 2a, 0≤ Y1 - Y2 ≤2a},

The Relation between R ( x ) and R ( y ) is illustrated in Figure 2.

1951_Find out the joint distribution2.png

Figure: Relation between R ( x ) and R ( y ).

Note that the variables xl and x2 are independent and the region R ( x ) is such that for Xl - xl, the variation X2 does not depend on xl, but the region R ( Y ) is not of that type and the transformed variables Yl and Y2 are not independent.

The variable Yl varies in the interval [ 0, 2a]and for a fixed yl, if 0≤ y1≤ a, then y2 takes on values -y1≤y2≤ y1, while, if a< y1≤ 2a then y2 varies in the interval

-(2a-y1) <.y2 ≤ (2a - y1)

Integrating fy ( y ) with respect to y2, the marginal pdf of y2 is obtained as follows

fY1(y1) = 2283_Find out the joint distribution3.png 1/2a2 dy2 = y1/a2, for 0 ≤ y1 ≤ a

462_Find out the joint distribution4.png 1/2a2 dy2  = 2a-y1/a2, for a< y1 ≤ 2a

= 0 otherwise.

In a similar manner, we note that for a given Y2, if -a ≤ y2 ≤ 0 then

-y2 ≤ y1 ≤ 2a-y2, and if 0≤ y2 ≤ a then y2 ≤ y1 ≤ 2a - y2

Hence,

fY2(y2) = 119_Find out the joint distribution5.png1/2a2 dy1 = a+y2/a2, -a ≤ y2 ≤ 0

960_Find out the joint distribution6.png 1/2a2 dy1 = a-y2/a2 , 0< y2 ≤ a

= 0 otherwise.

Remarks:

The forms of pdf the marginal distributions In Example 5 are shown in Figure 3. Due to their triangular shape of pdf's, the distributions are called triangular distributions.

2222_Find out the joint distribution7.png

 

Figure: The forms of the marginal distributions of YI and Y2

Posted Date: 1/30/2013 7:16:52 AM | Location : United States







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