Show that the expectation is zero

In summary: Therefore, we can conclude that the expression is minimized when \phi(X) = E(Y|X). In summary, we can see that in order to minimize E[Y - \phi(X)]^{2}, we need to choose \phi(X) = E(Y|X).
  • #1
island-boy
99
0
In one of the proofs I'm reading, I need to show that the
[tex]\phi(X)[/tex]
that will minimize [tex]E[Y - \phi(X)]^{2}[/tex] is
[tex]\phi(X) = E(Y|X)[/tex]

the proof is shown to proceed as follows:
[tex]E[Y - \phi(X)]^{2}[/tex]
[tex]=E[(Y - E(Y|X)) + (E(Y|X) - \phi(X))]^{2}[/tex]

expanding, we have

[tex]=E[(Y - E(Y|X))^{2} + 2(Y - E(Y|X))(E(Y|X) - \phi(X)) + (E(Y|X) - \phi(X))^{2}][/tex]
[tex]=E[(Y - E(Y|X))^{2}] + 2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] + E[(E(Y|X) - \phi(X))]^{2}][/tex]

the proof then states that the middle term is equal to zero. How is this true?
I got
[tex]2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] [/tex]
[tex]=2E[YE(Y|X) - (E(Y|X))^{2} - Y\phi(X) + E(Y|X)\phi(X)][/tex]
[tex]=2E[YE(Y|X)] - 2E[(E(Y|X))^{2}] - 2E[Y\phi(X)] + 2E[E(Y|X)\phi(X)][/tex]
[tex]=2E(Y|X)E(Y) - 2(E(Y|X))^{2} - 2E[Y\phi(X)] + 2E(Y|X)E(\phi(X))[/tex]

I'm not sure on how to proceed next.
I'm thinking that the first 2 terms will cancel as will the last two...but how exactly I'm not sure.

help please. thanks.
 
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  • #2
The key observation here is that E(Y|X) is a function of X and that E(Y) and E(\phi(X)) are constants with respect to X. Therefore, the middle term in the expansion can be written as:2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] = 2E[YE(Y|X)] - 2E[(E(Y|X))^{2}] - 2E[Y\phi(X)] + 2E[E(Y|X)\phi(X)]= 2E(Y|X)E(Y) - 2(E(Y|X))^{2} - 2E(Y)E(\phi(X)) + 2E(Y|X)E(\phi(X))Since E(Y) and E(\phi(X)) are both constants, they can be taken out of the expectations. Thus, we have:2E[(Y - E(Y|X))(E(Y|X) - \phi(X))] = 2E(Y|X)E(Y) - 2(E(Y|X))^{2} - 2E(Y)E(\phi(X)) + 2E(Y|X)E(\phi(X))= 2E(Y|X)[E(Y) - E(\phi(X))] - 2(E(Y|X))^{2}Now, since the goal is to minimize the expression E[Y - \phi(X)]^{2}, the best choice for \phi(X) is the one that minimizes the right hand side of this equation. This is achieved when E(Y) - E(\phi(X)) = 0, or equivalently, when \phi(X) = E(Y|X).
 

1. What does it mean to show that the expectation is zero?

To show that the expectation is zero means to prove that the average or expected value of a random variable is equal to zero. This can be done through mathematical calculations or statistical analysis.

2. Why is it important to show that the expectation is zero?

Showing that the expectation is zero is important because it indicates that there is no systematic bias or preference in the data. This is a crucial assumption in many statistical analyses and can affect the validity of the results.

3. How can you prove that the expectation is zero?

The expectation can be proven to be zero by using mathematical equations and properties of probability distributions. This can involve calculating the mean, variance, and other statistical measures to show that the expected value is equal to zero.

4. What does it mean if the expectation is not zero?

If the expectation is not zero, it means that there is a systematic bias or preference in the data. This can be caused by various factors such as measurement errors, sampling bias, or underlying patterns in the data.

5. Is it possible for the expectation to be zero for all values of a random variable?

Yes, it is possible for the expectation to be zero for all values of a random variable. This can occur if the distribution of the data is symmetrical around zero or if the data follows a specific pattern that results in an expected value of zero.

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