# Var + e

• Jan 3rd 2013, 06:26 AM
emme1
Var + 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 AM
Plato
Re: Var + e
Quote:

Originally Posted by emme1
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[FONT=sans-serif][COLOR=#000000](X=0)

Well, what kind of help? What had you done so far on each of them?
What don't you understand on each?

If you are expecting someone here to do them for you, that is not going to happen.
• Jan 3rd 2013, 06:40 AM
HallsofIvy
Re: 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 AM
emme1
Re: 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 AM
Plato
Re: Var + e
Quote:

Originally Posted by emme1
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

That is not right.

Did you find $E(Z)~?$.

How is $\text{Var}(Z)$ defined in terms of $E(Z)~?$
• Jan 3rd 2013, 08:01 AM
HallsofIvy
Re: 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.