Find out the joint distribution:
Let X_{I} and X_{2} 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
Y_{l} = X_{l} + X_{2} and Y_{2} = X_{1} - X_{2}.
Instead, in vector notation, what is the distribution of Y = XA.
where
x = (X_{1},X_{2}),Y= (Y_{1},Y_{2}), A = ? Find also the marginal distributions of Y_{1} and Y_{2}. ?
Solution:
The joint pdf of X is
f_{x}(x) = 1/a^{2}, (x_{1},x_{2})? R(x)
= 0 otherwise.
Where
R(x) = {(x_{1},x_{2}):0 ≤ x_{1} ≤ a, 0 ≤ x_{2} ≤ a}
The Jacobian of the transformation is
Hence the pdf of Y is
f_{y}(y) = 1/2a^{2}, (y_{1},y_{2})? R(y)
= 0 otherwise.
where R ( y ) is the transformed region R ( x ) under the transformation Y = XA. The range of variation of Y_{l} is clearly [ 0,2a ] and that of Y_{2} is [ - a, + a ]. However Y_{l} and Y_{2} are not independent.
Since the inverse transformation is
X_{1}= ½ (Y_{1} + Y_{2}), X_{2} = ½ (Y_{1} - Y_{2}) and 0≤ x_{1}, x_{2} ≤ a,
the region R ( y ) is given by
R(y) = {( Y_{1} + Y_{2}) : 0 ≤ Y_{1} + Y_{2 }≤ 2a, 0≤ Y_{1} - Y_{2} ≤2a},
The Relation between R ( x ) and R ( y ) is illustrated in Figure 2.
Figure: Relation between R ( x ) and R ( y ).
Note that the variables x_{l} and x_{2} are independent and the region R ( x ) is such that for X_{l} - x_{l}, the variation X_{2} does not depend on x_{l}, but the region R ( Y ) is not of that type and the transformed variables Y_{l} and Y_{2} are not independent.
The variable Y_{l} varies in the interval [ 0, 2a]and for a fixed y_{l}, if 0≤ y_{1}≤ a, then y_{2} takes on values -y_{1}≤y_{2}≤ y_{1}, while, if a< y_{1}≤ 2a then y_{2} varies in the interval
-(2a-y_{1}) <.y_{2 }≤ (2a - y_{1})
Integrating f_{y} ( y ) with respect to y_{2}, the marginal pdf of y_{2} is obtained as follows
f_{Y1}(y1) = 1/2a^{2} dy_{2} = y_{1}/a^{2}, for 0 ≤ y1 ≤ a
1/2a^{2} dy_{2 } = 2a-y_{1}/a^{2}, for a< y_{1} ≤ 2a
= 0 otherwise.
In a similar manner, we note that for a given Y_{2}, if -a ≤ y_{2} ≤ 0 then
-y_{2} ≤ y_{1} ≤ 2a-y_{2}, and if 0≤ y_{2} ≤ a then y_{2} ≤ y_{1} ≤ 2a - y_{2}
Hence,
f_{Y2}(y_{2}) = 1/2a^{2} dy_{1} = a+y_{2}/a^{2}, -a ≤ y^{2 }≤ 0
1/2a^{2} dy_{1} = a-y_{2}/a^{2} , 0< y^{2} ≤ 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.
Figure: The forms of the marginal distributions of Y_{I} and Y_{2}