Question

Consider the triangle with vertices (0,0), (1,0), (0,1). Suppose that (X, Y) is a uniformly chosen random point from the interior of this triangle.

a) Find the marginal density functions of X and Y
b)Calculate the expectation E[X] and E[Y]
c)Calculate the expectation E[XY]

Answer 1

a) To find the marginal density function of x and y, we need to integrate the joint density function over the other variable. The joint density function is given by:

f(x,y) = 1, 0 < x < 1, 0 < y < 1-x, 0 otherwise

To find the marginal density function of x, we integrate f(x,y) over y from 0 to 1-x:

f(x) = ∫[0,1-x] f(x,y) dy = ∫[0,1-x] 1 dy = 1-x, 0 < x < 1

Similarly, to find the marginal density function of y, we integrate f(x,y) over x from 0 to 1-y:

f(y) = ∫[0,1-y] f(x,y) dx = ∫[0,1-y] 1 dx = 1-y, 0 < y < 1

b) To find the expectation of x and y, we integrate x*f(x) and y*f(y), respectively, over their ranges:

E[x] = ∫[0,1] x*f(x) dx = ∫[0,1] x(1-x) dx = 1/3

E[y] = ∫[0,1] y*f(y) dy = ∫[0,1] y(1-y) dy = 1/3

c) To find the expectation of xy, we integrate x*y*f(x,y) over the triangle:

E[xy] = ∫[0,1]∫[0,1-x] x*y*f(x,y) dydx = ∫[0,1]∫[0,1-x] x*y dxdy = 1/24

Therefore, the answers are:

a) f(x) = 1-x, 0 < x < 1; f(y) = 1-y, 0 < y < 1

b) E[x] = 1/3, E[y] = 1/3

c) E[xy] = 1/24

Explanation: