Basic Expected Value Problem (probability)

1. Feb 25, 2008

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

2. Feb 25, 2008

mathman

The word variance seems to confuse you. By definition (almost) var(X)=E(X^2)-(E(X))^2.

Also as you should know, E(A+B+C)=E(A)+E(B)+E(C)

You should be able to work it out.

3. Feb 25, 2008

shawn87411

So Var(x) = 3 = E(2^2-(E(4)+E(4x)+E(x^2))?

Sorry I had a bad flu the past week so I missed my monday and wednesday lecture and our book is pretty lackluster

4. Feb 25, 2008

AsianSensationK

Precisely.

5. Feb 26, 2008

No, in your working, the brackets are in the wrong place, and you've ended up saying X = 2 in the first part, rather than E(X) = 2.

Use the equations mathman said, rearrange your Var(X) equation to get E(X^2) = Var(X) + E(X)^2, and from there, its just plugging in what you already know.

6. Feb 26, 2008

shawn87411

So I have 7=E(4)+E(4x)+E(x^2) but I'm confused what i do now. Do I plug in E(x)=2 so E(4x)=8 and E(x^2)=E(4)? if so where do you go from there, i'm just confused what i'm solving for

7. Feb 26, 2008

I dont know where your getting this 7 from?

E(4)+E(4X)+E(X^2), as you've stated, is E(4) + 4E(X) + E(X^2), which is E(4) + 4E(X) + Var(X) + E(X)^2, all of which you have values for. Again, you've said E(X^2) = E(4), here your saying X = 2, but X is a random variable, you cannot assume a value for X.

8. Feb 26, 2008

shawn87411

E(X^2) = Var(X) + E(X)^2, with Var(X)=3 and E(X)^2=4 so 4+3 gives the 7.

So do you not solve anything and it stays at 7=E(4)+4E(X)+E(X^2)?

9. Feb 26, 2008

All your doing is working out the value of E[4+4X+X^2], and your using the fact that E(X^2) = 7 to help you, your not equating anything to 7.

10. Feb 26, 2008

shawn87411

Your just trying to find the expected value of 4+4x+x^2, so all I have to say is E(4+4x+x6x^2)=E(4)+4E(x)+Var(X)+E(X^2)? Thats it? You dont say Var(X)=3 and E(X^2)=4 and plug that in?

Last edited: Feb 26, 2008
11. Feb 26, 2008

You've got to plug in the numerical values you know for E(4), E(X), Var(X) and E(X^2) to get a numerical answer.

12. Feb 26, 2008

shawn87411

E(4)=4
E(X)=2
Var(X)=3
E(X^2)=4

E(4)+4E(X)+4E(X^2)=13?

13. Feb 26, 2008

E(4+4X+X^2) = E(4) + 4E(X) + Var(X) + E(X)^2, as we discussed, I'm not sure where you've got your above equation from.

14. Feb 26, 2008

shawn87411

Alright, I'm really confused. You said I know the values for E(4), E(X), Var(X) and E(X^2) and those need to be plugged in. You also said

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

Its given that E(X)=2, Var(X)=3. This gives E(X)^2=4 and I assumed E(4)=4? Plugging those in gives 4+(4*2)+3+4 which gives 19, is that the right approach? Forgot to multiply by 4 for that one term, typed the above one wrong, sorry. (I really appreciate the help)

Last edited: Feb 26, 2008
15. Feb 26, 2008