# Math Help - Expected Values

1. ## Expected Values

Can anybody tell me why
E(f hat (x) - f(x))^2
= E(f hat (x) - f(x))^2 + var f hat (x)
Thanks.

2. I think your missing $E\hat f$ somewhere.
What you have here, is variance of something equal to zero, hence it's a constant.
Are you trying to show that the MSE = Bias squared + variance?

3. Thats exactly what im trying to do.
I think the next step is split it into
E(hat f -E(hat f) + E(hat f) -f)^2
a = hat f -E(hat f)
b = E(hat f) -f
and then expand (a+b)^2
but after that Im getting nowhere.
Thanks

4. Originally Posted by markrvr
Can anybody tell me why
E(f hat (x) - f(x))^2
= E(f hat (x) - f(x))^2 + var f hat (x)
Thanks.

you didn't read what I wrote.
This ...
E(f hat (x) - f(x))^2 = E(f hat (x) - f(x))^2 + var f hat (x)
implies that var f hat (x)=0.
which mean f hat (x) is constant

Please write what you really want to prove.

5. I am trying to show that the MSE = Bias squared + variance

6. Originally Posted by markrvr
I am trying to show that the MSE = Bias squared + variance
I guessed that two days ago, but what you wrote does not make sense.
You're missing the paramter $\theta$.

After you correctly state MSE, all you do is add and subtract
$E(\theta)$ inside of the MSE and you expand.

7. Yea I did it like this
E(hat f -E(hat f) + E(hat f) -f)^2
and expanded but I'm having trouble showing that the middle term of the expanded square equals 0.

8. Originally Posted by markrvr
Yea I did it like this
E(hat f -E(hat f) + E(hat f) -f)^2
and expanded but I'm having trouble showing that the middle term of the expanded square equals 0.
BECAUSE that is NOT MSE