Basic Expected Value Problem (probability)

[shawn87411]

E[X]=2
Var(X)=3
Find E[4+4x+x^2]

I'm just confused what its asking. The expected value of this function is 2 so the average of it is 2 and the variance is how much it varies which is 3? Every example I have for expected values is related to an example such as cards, not just a polynomial

 

[Dick]

Var(X)=E(X^2-E(X)^2). Just solve that for E(X^2). Then you can find E of the quadratic.

 

[shawn87411]

Var(X)=E(X^2-E(X)^2). Just solve that for E(X^2). Then you can find E of the quadratic.

Is the E(X^2-E(X)^2) = E(2^2-(4+4x+X^2)^2)?

 

[Dick]

No... Var(X)=3=E(X^2)-E(X)^2. E(X)=2. What's E(X^2)?? E(4+4X+X^2)=E(4)+E(4X)+E(X^2). Right? Etc. Use the linearity properties of 'E'.

 

[shawn87411]

No... Var(X)=3=E(X^2)-E(X)^2. E(X)=2. What's E(X^2)?? E(4+4X+X^2)=E(4)+E(4X)+E(X^2). Right? Etc.

I think I'm mixing up the terms E(X^2) and e(X)^2. Which one is E[X]=2?

So E(X^2)=E(4)^2+E(4x)^2+(x^2)^2?

 

[Dick]

E(X)^2=4, since E(X)=2.

 

[shawn87411]

E(X)^2=4, since E(X)=2.

Right but what's the difference between E(X^2) and E(X)^2?

Is E(X^2)=E(4^2)+E(4x^2)+E((x^2)^2) with x=2?

 

[Dick]

No! E(X^2) is not the same as E(X)^2. They aren't directly related to each other. The only way you can find E(X^2) from the information you are given is to use Var(X)=3.

 

[shawn87411]

Var(X)+E(X)^2=E(X^2)
3+4=7=E(x^2)

Then use the fact that E(4+4X+X^2)=E(4)+E(4X)+E(X^2).

 

[Dick]

Var(X)+E(X)^2=E(X^2)
3+4=7=E(x^2)

Then use the fact that E(4+4X+X^2)=E(4)+E(4X)+E(X^2).

Yes, that's it.