Test of PHE

For using normal distribution to perform statistical tests or establish a confidence interval of P HE, we used Shapiro-Wilk statistic to test the normality of P HE. Table 6, Table 7, Table 10, Table 11, Table 14 and Table 15 also shows these p-values of using Shapiro-Wilk test under scenario I and scenario II. The histograms of PHE values under different situations are shown in Figure 3-14. Under scenario I, the normality of PHE doesn’t hold in most situations. But the normality of P HE holds in most situations under scenario II. In other words, although using normal distribution is not good , it isn’t too bad. Briefly, using normal distribution can be acceptable with a lower standard..

We first describe the information about the mean LOD-score curve under both scenario I and scenario II (Figure 15-26). The LOD-score in our simulation was found to be related to the number of families and the heritability of the trait due to the common disease gene, where the trait may be a phenotype or an endophenotype. When the heritability of the endophenotype due to the common disease gene is larger than the heritability of the phenotype due to the common disease gene, the endophenotype is useful to search the disease gene. We except that the heritability of the endophenotype due to the common disease gene isn’t smaller than the heritability of the phenotype due to the disease gene. These results were consistent to the results from other papers [Almasy and Blangero, 1998; Williams et al., 1999].

Table 8 and Table 9 contain results under scenario I. At the same time, Figure 15, Figure 16, Figure 17 and Figure 18 show the mean LOD-score curve under scenario I. Based on these figures, when the heritability of P due G was assumed to be 0.42 and the heritability of E due to G allowed being 0 and 0.15, we find that using endophenotype to search for the disease gene is worse than using phenotype because the mean LOD-score of P was higher than the mean LOD-score of E. That is, we don’t hope that these are endophenotypes. On the other hand, when the heritability of P due to G was assumed to be 0.42 and the heritability of E due to G was assumed to be 0.74, endophenotype-based genetic analysis is more likely to succeed than one in terms of search for the disease gene (i.e. the mean LOD-score of E is higher than the one of P ). Besides, when the heritability of P due to G was assumed to be 0.42 and the heritability of E due to G was assumed to be 0.42, the phenotype-based effect

and the endophenotype-based effect are same. Altogether, when the heritability of P due to G was assumed to be 0.42 and the heritability of E due to G was assumed to be 0.42 or 0.74, the endophenotype-based effect isn’t worse than the phenotype-based effect. As a result of above descriptions about mean LOD-score curve, based on Table A3 and Table A4, we view it endophenotype candidate if lower bound of 95% one-sided confidence interval is larger than 0.25 or 0.50. The criterion that lower bound of 95% one-sided confidence interval is larger than 0.50 can be seen as a stronger evidence and the criterion that lower bound of 95% one-sided confidence interval is larger than 0.25 is also a suitable frame of reference. With another viewpoint, using two cutpoints, 0.25 and 0.50, the power, that the probability of rejecting H0

when H1 holds, will exceed 0.7 or 0.8 except for the situation where the heritability of P due to G was assumed to be 0.42, the heritability of E due to G was assumed to be 0.42, ρ was assumed to be 0, and cutpoint is set as 0.50. It implies that endophenotype-based effect isn’t worse than the phenotype-based effect. If it is desired that there is a higher power such as 0.9, 0 may be an applicable cutpoint no matter ρ was either 0 or 0.5. But it also leads the result ,that endophenotype-based effect is worse than the phenotype-based effect, happen, such as the situation where the heritability of P due G was assumed to be 0.42 and the heritability of E due to G allowed being 0.15.

In scenario II, on account of disrupted P HE values with the heritability of P due to G3 and the heritability of E due to G2, the criteria under scenario I may become improper. Based on Table 12, Table 13, Table 16 and Table 17, we downscale the standard of these criteria for searching the endophenotype successfully. The criterion that lower bound of 95% one-sided confidence interval is larger than 0.25 is still a suitable one. But many useful endophenotypes will be missed. So, we find that the criterion that lower bound of 95% one-sided confidence interval is larger than 0 should be seen as the criterion that search the potential candidate of endophenotype. Furthermore, if we want to let the higher power be kept for the goal that endophenotype-based effect isn’t worse than the phenotype-based effect, considered cutpoint may be 0. However, if ρ was assumed to be 0.5, the chosen cutpoint, 0, is not sufficient because of the lower power.

In summary, three criteria are provided as follows. The first criterion that lower bound of

95% one-sided confidence interval is larger than 0 is the potential evidence for searching the endophenotype. The second criterion that lower bound of 95% one-sided confidence interval is larger than 0.25 is the moderate evidence for searching the endophenotype. And the third criterion that lower bound of 95% one-sided confidence interval is larger than 0.50 is the stronger evidence for searching the endophenotype. However, you can choose some different criteria depended on the different goals of different cases or use lower bound of 95% one-sided confidence interval directly as the evidence for searching the endophenotype.

In another aspect, using the viewpoint of "power", we try to construct some steps to help us determine the desired endophenotype. The process of our construction is as follows. At the first step, check if ρ is 0 because it brings different information about use of the PHE values.

If it doesn’t hold, we are careful with use of P HE values because there is a lower power of detecting the useful endophenotypes if ρ is 0.5 even when the cutpoint is set as 0. That is, the involvement of ρ 6= 0 leads much uncertainty to use PHE values. Furthermore, if ρ become larger, using the PHE values may loss much useful information of the endophenotypes. In other words, If the lower bound of 95% one-sided confidence interval isn’t larger than 0 when ρ is larger than 0, it doesn’t imply that the endophenotype is helpless. If ρ is 0, we will perform the second step.

At the second step, check if the lower bound of 95% one-sided confidence interval is larger than 0.25. If it holds, it implies two possibilities : (1) there is the single disease gene to lead a direct effect on phenotype and endophenotype such as Scenario I and endophenotype-based effect isn’t worse than the phenotype-based effect; (2) it implies that both the influences of other genes on phenotype and endophenotype can be small, relative to the influences of the shared genes on phenotype and endophenotype such as Scenario II and endophenotype-based effect is better than the phenotype-based effect. If the lower bound of 95% one-sided confidence interval isn’t larger than 0.25, we will proceed to perform the third step.

At the third step, check if the lower bound of 95% one-sided confidence interval is larger than 0. If it holds, there exists two possible situations : (1) there is the single disease gene to lead a direct effect on phenotype and endophenotype such as scenario I and endophenotype-based effect isn’t better than the phenotype-endophenotype-based effect. It is out of our desire; (2) the

influence of other genes of either phenotype or endophenotype can be large relatively to the influence of the shared genes of either phenotype or endophenotype respectively such as sce-nario II and endophenotype-based effect isn’t worse than the phenotype-based effect. If the lower bound of 95% one-sided confidence interval isn’t larger than 0 when ρ is 0, it means there is a high probability that it isn’t a useful endophonotype. In sum, using three steps is helpful to search a useful endophenotype.

5 DISCUSSION

Based on definition of an endophenotype proposed by Huang et al. [2005], we have at-tempted to provide criteria that can be used to validate an endophenotype. Huang et al.,2005 had shown that the proposed index, P HE, is useful in validating endophenotypes. In our re-port, we use P HE proposed by Huang et al. [2005] as the index for evaluating endophenotypes to provide more clear informations, three criteria and three steps, through the one-sided con-fidence interval or the statistical test. However, we can find that the more the total numbers of family members, the more efficiency of detecting a useful endophenotype.

As discussed in corresponding index for validating surrogate endpoints such as P T E, con-fidence intervals of P T E can be calculated using Fieller’s theorem [Buyse and Molenberghs, 1998], however, they are usually too wide to be useful. With our proposed theorem or corol-lary, we use Fieller’s theorem or delta method to calculate confidence intervals of P HE. Our simulation results show that the estimators of standard error of P HE values’ estimators are near “true” standard errors of these indices’ estimators. That is, they are quite reasonable to avoid too wide confidence interval to be useful. However, although they may be overestimated or underestimated , they are helpful to detect the useful endophenotype easily. This is be-cause that it tends to have a underestimator of standard error of P HE estimator for the good endophenotype and it leads the lower bound of 95% one-sided confidence interval to be easily larger than our set cutpoint. Otherwise, the lower bound of 95% one-sided confidence interval tend to be smaller than our set cutpoint for the useless endophenotype. In other words, it isn’t too serious for using these overestimated or underestimated estimators of standard error of

P HE values’ estimators to construct a reasonable one-sided confidence interval and to search a useful endophenotype.

Besides, our simulation results show that the multiple gene effect lowers P HE values to lead it confused for evaluating endophenotypes. We provide three criteria and three steps to help us understand the pattern of P HE values versus the relationship between endophenotype and phenotype. If you aren’t interested in the relationship between P HE values and the heritabilities caused by different genes, the second step can be omitted. However, among three steps, we need to check that ρ is 0. The SOLAR command "polygenic" can be used to calculate ρ . If ρ is near 0, we can view it 0 to use three criteria and three steps safely for searching a useful endophenotype. Furthermore, at the third step, we will face the situation that the influence of other genes of either phenotype or endophenotype can be large relatively to the influence of the shared genes of either phenotype or endophenotype respectively such as Scenario II. For the influence of other genes of phenotype or endophenotype, we can use linkage analysis to determine which heritability is relatively large. If the heritability of other genes of phenotype is relatively large to the heritability of the found disease gene of phenotype, it means that only using an endophenotype may be sufficient. We must to search more than one endophenotype to capture a complete feature of the specified phenotype. The following model can be tried to be considered.

P = αH+ γ_1HE1 + γ_2HE2 + τHZ + G + ,

where E1 is assumed to being a found endophenotype and E2 is assumed to being a new or interested endophenotype. And we calculate the P HE value, 1 − hE1E2

hN E

, directly and its lower bound of 95% one-sided confidence interval, where hE1E2 is the heritability calculated from the variance component analysis (24) including the endophenotypes, E1 and E2, with any other covariates. To avoid to get same information or to find similar endophonotypes, we also calculate the partial proportion of heritability explained (P P HE) by the endophenotype defined as

P P HE = 1− hE1E2

hE1

where hE1E2 is the heritability calculated from the variance component analysis (24) including the endophenotypes, E1 and E2, with any other covariates and hE1 is the heritability cal-culated from the variance component analysis (24) without including the endophenotype E2 with any other covariates. A good and new endophenotype is one that explains a large pro-portion of heritability given a found endophenotype E1, thus, the greater the P P HE value, the more likely E2 an desired endophenotype.

In the future, to make it clear for using the P HE values, especially when ρ 6= 0, we should simulate with ρ < 0 and ρ >> 0. The information of the P HE values involved with negative ρ is a loss of our report. However, the much higher ρ is considered to help us understand the efficiency of using the P HE values to detect a useful endophenotype clearly in a bad situation. If the power of using the P HE values to detect useful endophenotype candidates isn’t too low when ρ is a much larger value, P HE values will be very useful index to search a useful endophenotype to increase opportunities of finding susceptible disease genes.

Appendix:

By GEE [Zeger and Liang, 1992; Amos, 1994],

S_β^(t)³

The correlation parameter h may be estimated by simultaneously solving

S_β^(t)³

as given by Covariance after transformation in table I,

and W_r×r^(t) = Using Taylor’s expansion, we have

bh^(k)− h^(k)

According to above equation, we can obtain

Cov³

Besides, for simplicity, we can replace them with à _R

then ma-trices. Above equation can be written as

Cov³

obtain

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TABLE 6. Simulation results based on scenario I (1)

N o. of f amilies hP a h^a_E ρ^b h^c P HE^d s.e^e s.e(delta)^f s.e(F ieller)^g S.W − pvalue^h

200 0.42 0 0 0.405 −0.002 0.009 0.025 0.029 < 0.001

0.5 0.473 −0.201 0.138 0.215 0.271 < 0.001

0.15 0 0.337 0.202 0.079 0.128 0.154 < 0.001

0.5 0.269 0.322 0.158 0.151 0.234 0.039

0.42 0 0.183 0.562 0.138 0.107 0.204 0.698

0.5 0.075 0.816 0.149 0.087 0.118 < 0.001

0.74 0 0.053 0.875 0.125 0.084 0.094 < 0.001

0.5 0.028 0.937 0.093 0.075 0.088 < 0.001

ahP=heritability of P due to G; hE= heritability of E due to G

bρ =correlation between non-family deviations of E and P

ch=mean of heritability of P, conditional on E

dP HE=mean of proportion of heritability explained by endophenotype

es.e=standard deviation of proportion of heritability explained by endophenotype

fs.e(delta)=mean of estimator of s.e by delta method

gs.e(F ieller)=mean of (₂ ¹

×1.96× the range of confidence limits of PHE) by Fieller theorem

hS.W − pvalue=p value of using Shapiro-Wilk Test

TABLE 7. Simulation results based on scenario I (2)

N o. of f amilies hP a h^a_E ρ^b h^c P HE^d s.e^e s.e(delta)^f s.e(F ieller)^g S.W − pvalue^h

500 0.42 0 0 0.422 −0.0004 0.002 0.007 0.008 < 0.001

0.5 0.481 −0.173 0.071 0.117 0.122 < 0.001

0.15 0 0.339 0.189 0.042 0.074 0.076 0.001

0.5 0.282 0.331 0.081 0.084 0.088 0.282

0.42 0 0.187 0.552 0.084 0.066 0.068 0.012

0.5 0.076 0.817 0.092 0.050 0.052 0.003

0.74 0 0.048 0.889 0.079 0.048 0.049 < 0.001

0.5 0.017 0.959 0.053 0.045 0.046 < 0.001

ahP=heritability of P due to G; hE= heritability of E due to G

bρ =correlation between non-family deviations of E and P

ch=mean of heritability of P, conditional on E

dP HE=mean of proportion of heritability explained by endophenotype

es.e=standard deviation of proportion of heritability explained by endophenotype

fs.e(delta)=mean of estimator of s.e by delta method

gs.e(F ieller)=mean of (₂ ¹

×1.96× the range of confidence limits of PHE) by Fieller theorem

hS.W − pvalue=p value of using Shapiro-Wilk Test

TABLE 8. Simulation results based on scenario I (3)

delta method Fieller theorem

N o. of f amilies h^a_P h^a_E ρ^b D0.00^c D0.25^c D0.50^c D0.75^c F 0.00^d F 0.25^d F 0.50^d F 0.75^d

200 0.42 0 0 0 0 0 0 0 0 0 0

0.5 0 0 0 0 0.01 0.005 0 0

0.15 0 0.55 0 0 0 0.395 0.01 0.01 0.01

0.5 0.715 0.195 0.01 0 0.56 0.115 0.01 0

0.42 0 0.99 0.815 0.255 0 0.95 0.71 0.19 0

0.5 0.995 0.98 0.825 0.365 0.99 0.945 0.8 0.34

0.74 0 1 1 0.945 0.52 1 0.995 0.9 0.515

0.5 1 1 0.99 0.78 0.995 0.99 0.99 0.765

ahP=heritability of P due to G; hE= heritability of E due to G

bρ =correlation between non-family deviations of E and P

cDx=the porportion that (\P HE-1.645 ×s.e(P HE)\ _delta ^e) is larger than x;

dF x=the porportion that the lower 95% confidence limits at one side using Fieller theorem is larger than x;

es.e(P HE)\ _delta=the estimator of s.e by delta method

TABLE 9. Simulation results based on scenario I (4)

delta method Fieller theorem

N o. of f amilies h^a_P h^a_E ρ^b D0.00^c D0.25^c D0.50^c D0.75^c F 0.00^d F 0.25^d F 0.50^d F 0.75^d

500 0.42 0 0 0 0 0 0 0 0 0 0

0.5 0 0 0 0 0 0 0 0

0.15 0 0.935 0 0 0 0.89 0 0 0

0.5 0.975 0.28 0 0 0.945 0.24 0 0

0.42 0 1 0.985 0.26 0.005 1 0.98 0.22 0.005

0.5 1 1 0.995 0.4 1 1 0.985 0.39

0.74 0 1 1 1 0.74 1 1 1 0.725

0.5 1 1 1 0.97 1 1 1 0.965

ahP=heritability of P due to G; hE= heritability of E due to G

bρ =correlation between non-family deviations of E and P

cDx=the porportion that (\P HE-1.645 ×s.e(P HE)\ _delta ^e) is larger than x;

dF x=the porportion that the lower 95% confidence limits at one side using Fieller theorem is larger than x;;

es.e(P HE)\ _delta=the estimator of s.e by delta method

TABLE 10. Simulation results based on scenario II with P>E (1)

N o. of f amilies h(G1E)/h(G1P)^a h(G2E)/h(G3P)^b ρ^c h^c P HE^d s.e^e s.e(delta)^f s.e(F ieller)^f S.W − pvalue^g

200 0/0.42 0.3/0.17 0 0.580 −0.0009 0.0051 0.0116 0.0119 < 0.001

0.5 0.653 −0.138 0.065 0.101 0.106 < 0.001

0.15/0.42 0.25/0.17 0 0.530 0.093 0.040 0.077 0.080 0.300

0.5 0.581 −0.004 0.095 0.113 0.118 < 0.001

0.42/0.42 0.12/0.17 0 0.424 0.273 0.087 0.089 0.093 0.004

0.5 0.463 0.193 0.112 0.105 0.109 0.047

0.51/0.42 0.04/0.17 0 0.380 0.344 0.101 0.087 0.090 0.124

0.5 0.412 0.285 0.122 0.099 0.103 0.081

0.74/0.42 0.05/0.41 0 0.674 0.181 0.057 0.053 0.054 0.033

0.5 0.762 0.069 0.074 0.058 0.057 0.146

0.74/0.42 0.08/0.41 0 0.682 0.174 0.069 0.053 0.053 0.777

0.5 0.769 0.057 0.072 0.057 0.057 0.020

0.79/0.42 0.08/0.41 0 0.660 0.191 0.063 0.055 0.056 0.537

0.5 0.758 0.076 0.071 0.056 0.057 0.271

ah(G1_E)=heritability of E due to G1; h(G1_P)= heritability of P due to G1;

bh(G2E)=heritability of E due to G2; h(G3P)= heritability of P due to G3;

cρ =correlation between non-family deviations of E and P; h=mean of heritability of P, conditional on E

dP HE=mean of proportion of heritability explained by endophenotype

es.e=standard deviation of proportion of heritability explained by endophenotype

fs.e(delta)=mean of estimator of s.e by delta method; s.e(F ieller)=mean of (_2×1.96¹ × the range of confidence limits of PHE) by Fieller theorem

gS.W − pvalue=p value of using Shapiro-Wilk Test

TABLE 11. Simulation results based on scenario II with P>E (2)

N o. of f amilies h(G1E)/h(G1P)^a h(G2E)/h(G3P)^b ρ^c h^c P HE^d s.e^e s.e(delta)^f s.e(F ieller)^f S.W − pvalue^g

500 0/0.42 0.3/0.17 0 0.595 −0.0003 0.0017 0.0039 0.0039 < 0.001

0.5 0.659 −0.127 0.038 0.058 0.059 < 0.001

0.15/0.42 0.25/0.17 0 0.539 0.091 0.025 0.046 0.046 < 0.001

0.5 0.588 −0.003 0.054 0.069 0.070 0.108

0.42/0.42 0.12/0.17 0 0.432 0.267 0.051 0.055 0.056 0.367

0.5 0.471 0.202 0.068 0.063 0.064 0.186

0.51/0.42 0.04/0.17 0 0.388 0.344 0.053 0.053 0.054 0.084

0.5 0.418 0.287 0.073 0.060 0.061 0.170

0.74/0.42 0.05/0.41 0 0.672 0.185 0.038 0.034 0.034 0.805

0.5 0.762 0.074 0.044 0.035 0.035 0.394

0.74/0.42 0.08/0.41 0 0.681 0.175 0.038 0.033 0.034 0.495

0.5 0.770 0.067 0.044 0.035 0.035 0.206

0.79/0.42 0.08/0.41 0 0.664 0.192 0.041 0.034 0.034 0.681

0.5 0.755 0.075 0.048 0.036 0.036 0.034

ah(G1_E)=heritability of E due to G1; h(G1_P)= heritability of P due to G1;

bh(G2E)=heritability of E due to G2; h(G3P)= heritability of P due to G3;

cρ =correlation between non-family deviations of E and P; h=mean of heritability of P, conditional on E

dP HE=mean of proportion of heritability explained by endophenotype

es.e=standard deviation of proportion of heritability explained by endophenotype

fs.e(delta)=mean of estimator of s.e by delta method; s.e(F ieller)=mean of (_2×1.96¹ × the range of confidence limits of PHE) by Fieller theorem

gS.W − pvalue=p value of using Shapiro-Wilk Test

TABLE 12. Simulation results based on scenario II with P>E (3)

delta method Fieller theorem

N o. of f amilies h(G1E)/h(G1P)^a h(G2E)/h(G3P)^b ρ^c D0.00^d D0.25^d D0.50^d D0.75^d F 0.00^e F 0.25^e F 0.50^e F 0.75^e

200 0/0.42 0.3/0.17 0 0 0 0 0 0 0 0 0

0.5 0 0 0 0 0 0 0 0

0.15/0.42 0.25/0.17 0 0.275 0 0 0 0.2 0 0 0

0.5 0.04 0 0 0 0.02 0 0 0

0.42/0.42 0.12/0.17 0 0.9 0.09 0 0 0.85 0.065 0 0

0.5 0.585 0.025 0 0 0.515 0.025 0 0

0.51/0.42 0.04/0.17 0 0.95 0.35 0.01 0 0.93 0.29 0.01 0

0.5 0.805 0.2 0.01 0 0.755 0.16 0.01 0

0.74/0.42 0.05/0.41 0 0.945 0.01 0 0 0.925 0.01 0 0

0.5 0.41 0 0 0 0.38 0 0 0

0.74/0.42 0.08/0.41 0 0.87 0.01 0 0 0.855 0.01 0 0

0.5 0.35 0 0 0 0.335 0 0 0

0.79/0.42 0.02/0.41 0 0.925 0.01 0 0 0.91 0.01 0 0

0.5 0.415 0 0 0 0.39 0 0 0

ah(G1E)=heritability of E due to G1; h(G1P)= heritability of P due to G1;

bh(G2E)=heritability of E due to G2; h(G3P)= heritability of P due to G3;

cρ =correlation between non-family deviations of E and P;

dDx=the porportion that (\P HE-1.645 ×s.e(P HE)\ _delta ^f) is larger than x;

eF x=the porportion that the lower 95% confidence limits at one side using Fieller theorem is larger than x;

eF x=the porportion that the lower 95% confidence limits at one side using Fieller theorem is larger than x;

在文檔中 信賴區間與模擬研究-對於穩定表現型的遺傳解釋比例 (頁 33-83)