1. ## 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!!

2. ## Re: Var + e

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.

3. ## 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?

4. ## 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

5. ## Re: Var + e

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 $E(Z)~?$

6. ## 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.