(from an actuarial exam) the random variables x and y have joint probability mass function p(x, y) for x = 0, 1 and y = 0, 1, 2. suppose that 3p(1, 1) = p(1, 2) and p(1, 1) maximizes the variance of xy . calculate the probability that x or y is zero.

0.4696

Step-by-step explanation:

Step 1: we say,

p(0,1) + p(0,2) + p(1,1) + p(1,0) + 3p(1,2) + p(0,0) = 1

P(Y=0) + p(X=0) - p(X=0, Y=0) + 4p(1,1) = 1.

Step2: let p(1,1)= w.

Therefore, 3w= p(1,2)

E(XY)= 2p(1,2) + p(1,1)

= 7w.

E[(XY)^2] = 4p(1,2) + p(1,1)

=13w.

variance(XY) = E[(XY)^2] - [E( XY)]^2

= -49w+ 13w

therefore, 0,1 are elements of w.

Step 3: calculate the derivative

13= 2x × 49

X = 13÷98

X= 0.1326

X=0, Y=0 = 1

= 1 - [0.1326 × (-4)]

= 0.4696.

it's both, now can you answer my most recent question?