Question

QUESTION :

Given: fx,y(x,y)=be^-(x+y) for 0<x<a and 0<y<Infinity and =0 elsewhere
a) Use Property 2 to determine the value of b that will make this a valid density function. Ans: b=1/(1-e^-a)
b) Use Property 3 to determine Fx,y(x,y) Ans: Fxy=(1-e^-x)(1-e^-y)/(1-e^-a)
c) Take the derivative of Fx,y(x,y) to show that it equals fx,y(x,y).

ANSWERS ARE GIVEN FOR A,B JUST PROVIDE THE STEPS

Properties of Joint Distribution Functions:

2) ??,?(∞, ∞) = ?

Example Given: ??,?(?, ?) = 0.2?(? − 1)?(? − 1) + 0.3?(? − 2)?(? − 1) + 0.5?(? − 3)?(? − 3) Then ??,?(+∞, +∞)

=

3) ? ≤ ??,?(?, ?) ≤ ?

Proof: This must be the valid range since 4a and 4b show that 0 and 1 are in the range and probability outside this range is impossible.

Answer 1