Question

The joint probability density function of the random variables X, Y, and Z is (e-(x+y+z) f(x, y, z) 0 < x, 0 < y, 0 <z elsewhere (a) (3 pts) Verify that the joint density function is a valid density function. (b) (3 pts) Find the joint marginal density function of X and Y alone (by integrating over 2). (C) (4 pts) Find the marginal density functions for X and Y. (d) (3 pts) What are P(1 < X < 2.5) and P(1 < Y < 2.5)? (e) (2 pts) Determine if X and Y are independent.

Answer 1

a)
this is valid density function as
f(x,y,z) >= 0 and

f(1, y, z)d.odydz = 1

b)

f(x,y) = [ f(x, y, z)dz = |e-(x+y++)dz = 0+(+y)

c)

f(x) = | f(x, y)dy = / e-Cz+vdy = e-> VO

similarly

f(y) = e^(-y)

d)

P(1 < X < 2.5)

=

72.5 e-1dc = e-1-e-2.5 = 0.2858

since distribution is identical

P(1 < Y < 2.5) is also 0.2858

e)

yes, as f(x,y) = f(x)f(y)

Please give me a thumbs-up if this helps you out. Thank you! :)