1. ## proving of equationa

hi! can anyone teach me how to prove the following?

E(X^2)= (E(x))^2 + Var X

and

Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)

2. Originally Posted by pinkiee
hi! can anyone teach me how to prove the following?

E(X^2)= (E(x))^2 + Var X

and

Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)

$\displaystyle V(x)=E[(X-u)^2]=E(X^2-2Xu+u^2)=E(X^2)-E(2Xu)+E(u^2)$

But u is a constant

$\displaystyle =E(X^2)-2uE(x)+u^2=E(X^2)-2u^2+u^2=E(X^2)-u^2$

But $\displaystyle E(X)=u$ so

$\displaystyle =E(X^2)-(E(X))^2$

3. but how do i swap the places of the terms? cuz i need to prove that E(X^2) is equal to (E(x))^2 + Var X

4. Originally Posted by pinkiee
but how do i swap the places of the terms? cuz i need to prove that E(X^2) is equal to (E(x))^2 + Var X
Surely you can re-arrange the proven result $\displaystyle Var(X) = E(X^2) - [E(X)]^2$ to make $\displaystyle E(X^2)$ the subject?!

In fact, Google gives tons of websites that more or less prove both of the results you want to prove. Did you try doing a Google search?

5. i tried googling, they gave me results of how to dervive var X
but i just cant seem to rearrange it to make E(X^2) the subject.
any mind helping out?

and how about Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)

6. $\displaystyle V(x)=E(X^2)-(E(X))^2$

Now you want to isolate one of those.....

y=x-z

Solve for x,

if you can do that, you can do the step above, which only involves one step of addition

7. ohh.. okie i try to work out that. thanks =)

how do i derive Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y) ?

8. Originally Posted by pinkiee
ohh.. okie i try to work out that. thanks =)

how do i derive Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y) ?
Hi - I would suggest you try it out for yourself. It is pretty straight fwd.

I will get you started. Cov(XV) = E((X-E(X)(Y-E(Y)) - This is the definition and your starting point
In this put X = aX+bZ and simplify the expression. Thats all there is to this prove

Hope you know that E(aX+bZ) = aE(X) + bE(Z)
If not try prove this first.

9. ok thanks! i gt it!

but i still can't expand E(X^2) such that i'll get (E(x))^2 + Var X

its not just rearranging it right? hmm so confused.

anyone can give me the 1st step of expanding E(X^2)?

10. Originally Posted by pinkiee
ok thanks! i gt it!

but i still can't expand E(X^2) such that i'll get (E(x))^2 + Var X

its not just rearranging it right? hmm so confused.

anyone can give me the 1st step of expanding E(X^2)?
Originally Posted by mr fantastic
Surely you can re-arrange the proven result $\displaystyle Var(X) = E(X^2) - [E(X)]^2$ to make $\displaystyle E(X^2)$ the subject?!
[snip]
This has nothing to do with expanding!

If you had A = B - C I would hope that at this level you could make B the subject. This is basic algebra. If you honestly cannot do this you have absolutely no business attempting questions like the one you posted.

11. u mean i just have to do this?

E(X^2)= (E(x))^2 + Var X

then rearrange to Var X = E(X^2) - (E(x))^2 ?

then derive that var x = E(X^2) - (E(x))^2 ?

12. Originally Posted by pinkiee
u mean i just have to do this?

E(X^2)= (E(x))^2 + Var X Mr F says: Yes. Read my bottom comment first.

then rearrange to Var X = E(X^2) - (E(x))^2 ?

then derive that var x = E(X^2) - (E(x))^2 ? Mr F says: This was proved in the very first reply you got. Did you read it?
..

13. yup, i read it alr.

i think i gt it now.

i thought i have to expand E(X^2) so that it gives me (E(x))^2 + Var X

thanks!! =D