2

If $X$ has $0$ mean and second moment $\sigma^2_X<\infty$ and is observed with some error $Z$, which is White noise independent of $X$ with second moment $\sigma^2_Z$ (again finite) i.e $Y = X + Z$

How do I find the best linear predictor of $X$ and its mean-squared error based on $n$ observations of $Y_1, Y_2, ... , Y_n$ ?

If my understanding is correct, I need to find the best linear predictor of $X$ in terms of $Y_n$ ie $X =a_0(Y-Z)_n + a_1(Y-Z)_{n-1} + ... + a_n(Y-Z)_{1}$

Where would I proceed from here? Would the MSE would be $|X-[a_0(Y-Z)_n + a_1(Y-Z)_{n-1} + ... + a_n(Y-Z)_{1}]|^2$

elbarto
  • 3,356
  • Definitions again... What are a predictor of X to you, and the associated MSE? Numbers, functions, something else? You seem to be using these words randomly... – Did Oct 18 '14 at 14:36
  • @Did a linear predictor is any linear combination of values which we can use to forecast a system. The Best linear predictor is one which we have a minimum mean squared error.

    So if I were to find a linear combination forecasting $X$ would in terms of $Y$, would I not re-write the system as $ X = Y-Z$ and then go from there?

    Where have I made an error? I am going by the definition provided in Brockwell & Davis' book for time series and forecasting

     – elbarto Oct 19 '14 at 04:41
  • 1
    In other words you are looking for $(a_1,a_2,\ldots,a_n)$ achieving the minimum of the mean-squared error $$R(a_1,a_2,\ldots,a_n)=E((a_1Y_1+a_2Y_2+\cdots+a_nY_n-X)^2).$$ Now that you are back on the right track, can you proceed? – Did Oct 19 '14 at 06:44

0 Answers0