Assignment 5
Assignment 5 Hendrik Gauger
Exercise 6
Distribution of Y
| x | -2 | 0 | 2 |
|---|---|---|---|
| P( Yt = x ) | 1/4 | 1/2 | 1/4 |
Distribution of (Yt, Yt+1)^T
| Yt/Yt+1 | -2 | 0 | 2 | Sum |
|---|---|---|---|---|
| -2 | 1/8 | 1/8 | 0 | 1/4 |
| 0 | 1/8 | 1/4 | 1/8 | 1/2 |
| 2 | 0 | 1/8 | 1/8 | 1/4 |
| Sum | 1/4 | 1/2 | 1/4 | 1 |
Distribution of (Yt, Yt+1, Yt+2)^T
| Yt+1 = -2 | ||||
|---|---|---|---|---|
| Yt / Yt+2 | -2 | 0 | 2 | Sum |
| -2 | 0.0625 | 0.0625 | 0 | 0.125 |
| 0 | 0.0625 | 0.0625 | 0 | 0.125 |
| 2 | 0 | 0 | 0 | 0 |
| Sum | 0.125 | 0.125 | 0 | 1/4 |
| Yt+1=0 | ||||
| Yt / Yt+2 | -2 | 0 | 2 | Sum |
| -2 | 0 | 0.0625 | 0.0625 | 0.125 |
| 0 | 0.0625 | 0.125 | 0.0625 | 0.25 |
| 2 | 0.0625 | 0.0625 | 0 | 0.125 |
| Sum | 0.125 | 0.25 | 0.125 | 1/2 |
| Yt+1=2 | ||||
| Yt/Yt+2 | -2 | 0 | 2 | Sum |
| -2 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0.0625 | 0.0625 | 0.125 |
| 2 | 0 | 0.0625 | 0.0625 | 0.125 |
| Sum | 0 | 0.125 | 0.125 | 0.25 |
| E0 | E1 | E2 | E4 | Y1 | Y2 | Y3 | E( Y3 I Y1,Y2) | error^2 |
|---|---|---|---|---|---|---|---|---|
| 1 | 1 | -1 | 1 | 2 | 0 | 0 | -1 | 1 |
| 1 | 1 | -1 | -1 | 2 | 0 | -2 | -1 | 1 |
| 1 | 1 | 1 | 1 | 2 | 2 | 2 | 1 | 1 |
| 1 | 1 | 1 | -1 | 2 | 2 | 0 | 1 | 1 |
| 1 | -1 | -1 | 1 | 0 | -2 | 0 | -1 | 1 |
| 1 | -1 | -1 | -1 | 0 | -2 | -2 | -1 | 1 |
| 1 | -1 | 1 | 1 | 0 | 0 | 2 | 0 | 4 |
| 1 | -1 | 1 | -1 | 0 | 0 | 0 | 0 | 0 |
| -1 | 1 | -1 | 1 | 0 | 0 | 0 | 0 | 0 |
| -1 | 1 | -1 | -1 | 0 | 0 | -2 | 0 | 4 |
| -1 | 1 | 1 | 1 | 0 | 2 | 2 | 1 | 1 |
| -1 | 1 | 1 | -1 | 0 | 2 | 0 | 1 | 1 |
| -1 | -1 | -1 | 1 | -2 | -2 | 0 | -1 | 1 |
| -1 | -1 | -1 | -1 | -2 | -2 | -2 | -1 | 1 |
| -1 | -1 | 1 | 1 | -2 | 0 | 2 | 1 | 1 |
| -1 | -1 | 1 | -1 | -2 | 0 | 0 | 1 | 1 |
Sum of error^2: 20
| E0 | E1 | E2 | E3 | Y1 | Y2 | Y3 | E(Y3 I Y1.Y2) | error^2 |
|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 2 | 0 | 0 | -(2/3) | 4/9 |
| 1 | 1 | 1 | -1 | 2 | 0 | -2 | -(2/3) | 16/9 |
| 1 | 1 | -1 | 1 | 2 | 2 | 2 | 2/3 | 16/9 |
| 1 | 1 | -1 | -1 | 2 | 2 | 0 | 2/3 | 4/9 |
| 1 | -1 | 1 | 1 | 0 | -2 | 0 | -(4/3) | 16/9 |
| 1 | -1 | 1 | -1 | 0 | -2 | -2 | -(4/3) | 4/9 |
| 1 | -1 | -1 | 1 | 0 | 0 | 2 | 0 | 4 |
| 1 | -1 | -1 | -1 | 0 | 0 | 0 | 0 | 0 |
| -1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| -1 | 1 | 1 | -1 | 0 | 0 | -2 | 0 | 4 |
| -1 | 1 | -1 | 1 | 0 | 2 | 2 | 4/3 | 4/9 |
| -1 | 1 | -1 | -1 | 0 | 2 | 0 | 4/3 | 16/9 |
| -1 | -1 | 1 | 1 | -2 | -2 | 0 | -(2/3) | 4/9 |
| -1 | -1 | 1 | -1 | -2 | -2 | -2 | -(2/3) | 16/9 |
| -1 | -1 | -1 | 1 | -2 | 0 | 2 | 2/3 | 16/9 |
| -1 | -1 | -1 | -1 | -2 | 0 | 0 | 2/3 | 4/9 |
Sum of error^2: 64/3
a)
(i) The best predictor for Y3 given Y1 = 2 and Y2 = 0 is: -1
(ii) The best linear predictor for Y3 given Y1 = 2 and Y2 = 0 is: -(2/3)
(iii) MSE of the best predictor: 20/16 = 1.25
(iv) MSE of the best linear predictor: 64/3/16 = 4/3 = 1.333
b)
The partial autocorrelation at lag 2 matches the slope of Y1 for the linear predictor of Y3.
=> Autocorrelation of -(1/3)
| Condition | Corr( Y[t-2],Y[t] I Y[t-1] ) | Sign |
|---|---|---|
| Y[t-1] = -2 | 0 | 0 |
| Y[t-1] = 0 | negative | - |
| Y[t-1] = 2 | 0 | 0 |
Only the correlation conditioned to Y[t-1] = 0 has the same sign as the partial autocorrelation at lag 2 and therefore matches it.