(first) in this problem i want to know why integral range defined like that (0 t
ID: 3340165 • Letter: #
Question
(first)
in this problem
i want to know why integral range defined like that (0 to y)
why define zero to y not zero to 1
(second)
why dy's integral range defined x to 1?
is it possible 1 to y?
(third)
why dx's integral range defined y to 0?
is it possible y to 1/2?
i want to understand why range definded like that
0, otherwise. Example 5.7-1 Suppose that the joint PDF for random variables X and Y is given by xy(x,y)"(0, otherwise. 2y Determine(c) (n(x) and f,(y,and the probability that Solution: We first determine the value of constant c //fry(x,y)drdy. = 1 j[Icdr|dy=1. cy 0Explanation / Answer
1. The range of X and Y are dependent on each other. So,
while determining the value of c, we have to integrate over
the ranges of X and Y. We should have one dependent and
other one as independent range. So we can keep the range of
Y as independent i.e. 0 to 1 and range of X as dependent i.e.
0 to Y. Also, we can keep range of X as independent i.e. 0 to 1
and range of Y as dependent i.e. X to 1.
2. While computing the marginal density of X and as the range
of Y is dependent on X, we should have the integral of Y as
X to 1. It can't be 1 to y, as we are integrating over Y.
3. While computing marginal density of Y, we have to integrate
over X and the range of X, being dependent on Y, we should
have the integral from 0 to Y. It can't be from y to 1/2, as for
x > 1/2, the integral value will not be there.
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