I'm stuck, can anyone help me with this "easy" statistics?

X Y p(x,y)

0 0 0,20

0 1 0,25

1 0 0,30

1 1 0,25

Var (X+Y) = ?

E (Y(X=0) = ?

Thanks a lot!!

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- Jan 3rd 2013, 06:26 AMemme1Var + e
I'm stuck, can anyone help me with this "easy" statistics?

X Y p(x,y)

0 0 0,20

0 1 0,25

1 0 0,30

1 1 0,25

Var (X+Y) = ?

E (Y(X=0) = ?

Thanks a lot!! - Jan 3rd 2013, 06:38 AMPlatoRe: Var + e
- Jan 3rd 2013, 06:40 AMHallsofIvyRe: Var + e
x and y, separately, can be 0 or 1 so x+ y can be 0, 1, or 2.

What must x and y be so that x+ y= 0? What is the probability of that?

What must x and y be so that x+ y= 1? What is the probability of that?

What must x and y be so that x+ y= 2? What is the probability of that? - Jan 3rd 2013, 06:48 AMemme1Re: Var + e
I did this...

z pz(z)

0 0,20

1 0,55

2 0,25

Var(Z) = 0^2*0,20 + 1^2*0,55 + 2^2*0,25

But it doesn't feel right - Jan 3rd 2013, 07:08 AMPlatoRe: Var + e
- Jan 3rd 2013, 08:01 AMHallsofIvyRe: Var + e
Your probability distribution for z= x+ y, 0,20 that z= 0, 0,55 that z= 1, and 0,25 that z= 2, is correct. Now, the expected value is 0(0,20)+ 1(0,55)+ 2(0,25)= 1,10+ 0,50= 1,60. The variance, then, is 0,20(0- 1,60)^2+ 0,55(1- 1,60)^2+ 0,25(2- 1,60)^2. You used the values, not the variation from the expected value.