Suppose f(x,y)=(1/8)(6-x-y) for 0<x<2 and 2<y<4. If all else is the same, then why can’t x...

Suppose f(x,y)=(1/8)(6-x-y) for 0<x<2 and 2<y<4.
1. If all else is the same, then why can’t x be defined on the range [0,3]?
2. Find p(0.5 < X < 1, 2 < Y < 3)
3. Find f_X(x) and f_Y(y)
4. Are X and Y independent? Why or why not?
5. Find p(0.5 < X < 1) and p(2 < Y < 3)
6. Find p(Y<3|X=1)
7. Find p(Y<3|0.5<X<1)

Solutions

Expert Solution

We would be looking at the first 4 parts here as:

a) Let x be defined on [0,3]

Then, the total probability is computed here as:

As the total probability has to be 1, which is not the case with X range [0,3], therefore x cannot have that range here.

b) The probability here is computed as:

therefore 0.171875 is the required probability here.

c) The marginal density functions for X and Y here are obtained as:

This is the required marginal density function for X here.

This is the required marginal density function for Y here.

d) We have here:

This is not equal to the joint PDF for X, Y

Therefore X and Y are not independent variables here.

orchestra answered 1 year ago

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