3.6 Simulation Results

In this section, we carry out simulations to check the performance of the proposed OPG method and MAVE method. Some comparisons of the methods are made. Let $ {{\cal S}}(B) $ be the space spanned by the column vectors in $ B$. To describe the error of estimation, we define the distance from $ b$ to $ {{\cal S}}(B) $ as $ d(b, B) = b^{\top }(I - BB^{\top }) b $, where the columns of $ B$ form an orthogonal standard basis of the space. It is easy to see that $ 0\le d(b, B) \le 1 $. If $ d(b, B) = 0 $ then $ b \in {{\cal S}}(B) $; if $ d(b, B) = 1 $ then $ b \perp {{\cal
S}}(B) $.

Example 3.6.1. Consider the following model

$\displaystyle y = X^{\top } \beta_1 (X^{\top } \beta_2)^2 + (X^{\top } \beta_3 )(X^{\top }\beta_4) + 0.5 \varepsilon,$     (3.39)

where $ X \sim N(0, I_{10}) $ and $ \varepsilon \sim N(0,1) $ and they are independent. In model (3.39), the coefficients $ \beta_1 = (1, 2, 3, 4, 0, 0, 0, 0, 0, 0)^{\top }/\sqrt{30} $, $ \beta_2 = (-2, 1, -4, 3, 1,$ $ 2,0,0,0,0)^{\top }$ $ /\sqrt{35} $, $ \beta_3 = (0, 0, 0, 0, 2,-1,2,1,2,1)^{\top }/\sqrt{15} $, $ \beta_4 = (0, 0, 0, 0, 0,0,$ $ -1,-1,1,1)^{\top }/2 $ and there are four e.d.r. directions. Let $ B_0 =
(\beta_1,\beta_2,\beta_3,\beta_4)$. In our simulations, the SIR method and the ADE method perform quite poorly for this model. Next, we use this model to check the OPG method and the MAVE method.


Table: Average distance $ d(\hat \beta_k, B_0)$ for model (3.39) using different methods
$ d(\hat \beta_k, B_0)$ Freq. of Est. No. of
n Methods $ k = 1$ $ k=2$ $ k=3$ $ k=4$ e.d.r directions
pHd .2769 .2992 .4544 .5818 $ f_1$=0 $ f_2$=10 $ f_3$=23
OPG .1524 .2438 .3444 .4886 $ f_4$=78 $ f_5$=44 $ f_6$=32
100 MAVE .1364 .1870 .2165 .3395 $ f_7$=11 $ f_8$=1 $ f_9$=1
rMAVE .1137 .1397 .1848 .3356 $ f_{10}$=0
pHd .1684 .1892 .3917 .6006 $ f_1$=0 $ f_2$=0 $ f_3$=5
OPG .0713 .1013 .1349 .2604 $ f_4$=121 $ f_5$=50 $ f_6$=16
200 MAVE .0710 .0810 .0752 .1093 $ f_7$=8 $ f_8$=0 $ f_9$=0
rMAVE .0469 .0464 .0437 .0609 $ f_{10}$=0
pHd .0961 .1151 .3559 .6020 $ f_1$=0 $ f_2$=0 $ f_3$=0
OPG .0286 .0388 .0448 .0565 $ f_4$=188 $ f_5$=16 $ f_6$=6
400 MAVE .0300 .0344 .0292 .0303 $ f_7$=0 $ f_8$=0 $ f_9$=0
rMAVE .0170 .0119 .0116 .0115 $ f_{10}$=0


With sample sizes $ n=100$, $ 200$ and $ 400$, 200 independent samples are drawn. The average distance from the estimated e.d.r. directions to $ {{\cal S}}(B_0 ) $ is calculated for the pHd method (Li, 1992), the OPG method, the MAVE method and the rMAVE method. The results are listed in Table 3.1. The proposed OPG and MAVE methods work quite well. The results also show that the MAVE method is better than the OPG method, while the rMAVE method shows significant improvement over the MAVE method. Our method for the estimation of the number of e.d.r. directions also works quite well.

Example 3.6.2. We next consider the nonlinear time series model

$\displaystyle y_t = -1 +0.4 \beta_1^{\top } X_{t-1} - \cos(\frac\pi2
\beta_2^{\top } X_{t-1})
+ \exp\{-(\beta_3^{\top }X_{t-1})^2\} + 0.2 \varepsilon_t,$     (3.40)

where $ X_{t-1} = (y_{t-1}, \cdots, y_{t-6})^{\top } $, $ \varepsilon$ are i.i.d. $ N(0, 1)$, $ \beta_1 = (1, 0, 0, 2, 0,
0)^{\top }/\sqrt{5} $, $ \beta_2 = (0, 0, 2, 0, 0,
1)^{\top }/\sqrt{5} $ and $ \beta_3 = (-2, 2, -2, 1, -1,
1)^{\top }/\sqrt{15} $. A typical data set sample with size 1000 was drawn from this model. The points $ (y_t, y_{t-k}),\ t =
1,\dots, 1000 $, and $ k = 1, \cdots, 6 $ are plotted in Figures 3.7 (a)-(f). As there is no discernible symmetry, the SIR method will not be appropriate.

Figure 3.7: A typical data set from model (3.40). (a)-(f) are $ y_t $ plotted against $ y_{t-k},\ k= 1,\cdots , 6,$ respectively.
\includegraphics[width=1.2\defpicwidth]{d_gpoint.ps}

Now, we use the OPG method and the MAVE method. The simulation results are listed in Table 3.2. Both methods have quite small estimation errors. As expected, the rMAVE method works better than the MAVE method, and the MAVE method outperforms the OPG method. The number of the e.d.r. directions is also estimated correctly most of the time for suitable sample size.


Table: Mean of the distance $ d(\hat \beta_k, B_0)$ for model (3.40) using different methods
    $ d(\hat \beta_k, B_0)$ for k = Freq. of Est. No. of
$ n$ Method 1 2 3 e.d.r. directions
  pHd .1582 .2742 .3817 $ f_1$=3 $ f_2$=73
  OPG .0427 .1202 .2803 $ f_3$=94 $ f_4$=25
100 MAVE .0295 .1201 .2924 $ f_5$=4 $ f_6$=1
  rMAVE .0096 .0712 .2003    
  pHd .1565 .2656 .3690 $ f_1$=0 $ f_2$=34
  OPG .0117 .0613 .1170 $ f_3$=160 $ f_4$=5
200 MAVE .0059 .0399 .1209 $ f_5$=1 $ f_6$=0
  rMAVE .0030 .0224 .0632    
  pHd .1619 .2681 .3710 $ f_1$=0 $ f_2$=11
  OPG .0076 .0364 .0809 $ f_3$=185 $ f_4$=4
300 MAVE .0040 .0274 .0666 $ f_5$=0 $ f_6$=0
  rMAVE .0017 .0106 .0262    


Example 3.6.3. The multi-index model (3.6). For simplicity, we discuss only the generalized partially linear single-index models. This model was proposed by Xia, Tong, and Li (1999), which has been found adequate for quite a lot of real data sets. The model can be written as

$\displaystyle y = \theta_0^{\top } X + {\sl g}(\beta_0^{\top }X) + \varepsilon,$     (3.41)

where $ \theta_0 \perp \beta_0 $ and $ \Vert\beta_0 \Vert = 1 $. Following the idea of the MAVE method, we may estimate $ \theta_0
$ and $ \beta_0$ by minimizing
    $\displaystyle \sum_{j=1}^n \sum_{i=1}^n\Big [ y_i - \theta^{\top } X_i -
a_j - b_j \beta^{\top } (X_i - X_j)\Big ]^2
w_{ij},$  
    $\displaystyle \textrm{subject to}\ \ \theta\perp \beta.$ (3.42)

Suppose that $ \tilde \theta $ and $ \tilde \beta $ constitute the minimum point. Then we have the estimates
$\displaystyle \hat \theta = \tilde \theta - (\tilde \theta^{\top }\tilde
\beta)\tilde \beta \quad \textrm{and}\quad
\hat \beta = \tilde \beta .$      

To illustrate the performance of the above algorithm, we further consider the following model

$\displaystyle y = 3x_2 + 2x_3 + (x_1 - 2x_2 + 3x_3)^2 + \varepsilon,$     (3.43)

where $ x_1, x_2, x_3$ and $ \varepsilon$ are i.i.d. $ \sim N(0, 1)$. In this model, $ \theta_0 = (0, 3, 2)^{\top } $ and $ \beta_0 = (1, -2, 3)^{\top }/\sqrt{14} = (0.2673, -0.5345,
0.8018)^{\top } $. Let $ \theta_1 = \theta_0/\Vert\theta_0\Vert = (0,
0.8321, 0.5547)^{\top } $ and $ B_0 = (\beta_0, \theta_1) $. We can estimate the model using the MAVE method without assuming its specific form. The simulation results listed in Table 3.3 suggest that the estimation is quite successful. If we further assume that it is a generalized partially linear single-index model and estimate $ \theta_1 $ and $ \beta_0$ by minimizing (3.42), the estimation is much better as shown in Table 3.3.


Table: Mean of the estimated directions and average distance $ d(\hat \beta_0, B_0) $ or $ d(\hat \theta_1, B_0) $ in square brackets for model (3.43)
$ n$ no model specification model specified
$ \beta_0$ (0.2703 -0.5147 0.8136) [.000329] (0.2678 -0.5346 0.8014) [.000052]
50 $ \theta_1 $ (0.0264 0.8487 0.5281) [.013229] (0.0108 0.8319 0.5513) [.003946]
$ \beta_0$ (0.2679 -0.5307 0.8041) [.000052] (0.2665 -0.5346 0.8020) [.000019]
100 $ \theta_1 $ (0.0035 0.8341 0.5516) [.002244] (0.0014 0.8318 0.5540) [.001142]


Example 3.6.4. The varying-coefficient model (3.7). For simplicity, here we only discuss the single-index coefficient linear model proposed by Xia and Li (1999). The model can be written as

$\displaystyle y_t = c_0(\beta_0^{\top } X_{t-1}) + c_1(\beta_0^{\top }X_{t-1})
y_{t-1} + \cdots + c_p(\beta_0^{\top }X_{t-1})y_{t-p} +
\varepsilon_t,$     (3.44)

where $ X_{t-1} = (y_{t-1}, \cdots, y_{t-p})^{\top } $.

To see the performance of the MAVE method discussed in Section 3.3.2, we further consider the following model

$\displaystyle y_t = 0.4\sin(x_t) + 0.5 \Phi(x_t+0.6)y_{t-4} + 0.6\exp(-x_t^2/4)y_{t-5} + 0.2 \varepsilon_t,$     (3.45)

where $ x_t = 2(0.5y_{t-1} + y_{t-2} + 1.5y_{t-3})$, $ t = 0,\pm
1, \pm 2,\cdots $, and $ \varepsilon_t $ are i.i.d. $ \sim N(0, 1)$. In this model, $ \beta_0 = (1, 2, 3)^{\top }/\sqrt{13} =
(0.2774\ 0.5547\ 0.8321)^{\top } $. Model (3.45) is a combination of the TAR model and the EXPAR model (cf. Tong (1990)). Table 3.4 shows the simulation results, which also suggest that the estimation is satisfactory.


Table 3.4: Mean of the estimated directions and the standard deviation for model (3.45)
$ n$ Estimated direction s.d.
$ n$ = 100 (0.2794 0.5306 0.7895) [0.07896 0.0817 0.0646]
$ n$ = 200 (0.2637 0.5243 0.8052) [0.06310 0.0468 0.0303]