Basic Expected Value Problem (probability)

• shawn87411
In summary, 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?
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

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.

mathman said:
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.

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

Precisely.

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.

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.

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

I don't 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.

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)?

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.

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 don't say Var(X)=3 and E(X^2)=4 and plug that in?

Last edited:
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.

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

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

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.

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:
E(4) is 4, yes.

So E(4)+4E(X)+Var(X)+E(X)^2 = 4 + 4*2 + 3 + 2^2 = 19

1. What is a basic expected value problem?

A basic expected value problem involves calculating the average outcome or value of a random event or experiment. It is a way of predicting the most likely outcome based on probabilities.

2. How do you calculate expected value?

To calculate expected value, you multiply each possible outcome by its probability and then add all the results together. The formula is: E(x) = Σ(x * P(x)), where x is the outcome and P(x) is the probability of that outcome.

3. What is the significance of expected value in probability?

Expected value is important in probability because it helps us make informed decisions based on the likelihood of different outcomes. It can also be used to evaluate the fairness of a game or situation.

4. Can expected value be negative?

Yes, expected value can be negative. This means that the average outcome or value of the event or experiment is less than zero. A negative expected value indicates that the event is not likely to be profitable or beneficial.

5. How is expected value used in real-life situations?

Expected value can be used to make decisions in various real-life situations, such as insurance, investing, and gambling. For example, insurance companies use expected value to determine premiums, investors use it to assess the potential return of an investment, and gamblers use it to decide which games offer the best odds.

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