Best Linear Unbiased Prediction (BLUP) of Random Effects in the Normal Linear Mixed Effects Model. *Modified notes from Dr. Dan Nettleton from ISU

Transcription

1 Best Linear Unbiased Prediction (BLUP) of Random Effects in the Normal Linear Mixed Effects Model *Modified notes from Dr. Dan Nettleton from ISU

2 Suppose intelligence quotients (IQs) for a population of students are normally distributed with a mean µ and variance σ 2 u. IQ ~ N(µ,σ u 2 ) µ

3 Suppose an IQ test was given to an i.i.d sample of such students. Suppose that, given the IQ of a student (something hard to measure), the test score for that student is normally distributed with a mean equal to the student s IQ and a variance of σ 2 and is independent of the test score of any other student. score IQ ~ N(IQ,σ 2 ) IQ

4 Consider our linear mixed effects model where [ u e Y = X β + Zu + e ] ([ 0 N 0 ] [ G 0, 0 R ]) Note that this model coincides with u N(0, G), e N(0, R), independent of each other.

5 Given the data y, what is our best guess for the unobserved vector u? (The random student effects). Because u is a random vector rather than a fixed parameter, we talk about predicting u rather than estimating u. We seek a Best Linear Unbiased Predictor (BLUP) for u, which we will denote by û.

6 To be a BLUP, we require û to be a linear function of y, 2. û to be unbiased for u so that E(û u) = 0, and 3. Var(û u) to be no larger than the Var(v u), where v is any other linear and unbiased predictor.

7 The BLUP of u is û = GZ Σ 1 (y X ˆβ Σ ) And for the usual case in which G and Σ = ZG Z + R are unknown, we replace the matrices by estimates and approximate the BLUP of u by û = ĜZ ˆΣ 1 (y X ˆβ ˆΣ)

8 Let s return to the IQ example... Suppose it is known that σ2 u σ 2 =9 If the we sample 100 students and their sample mean IQ was 100, what is the best prediction of the IQ of a student who scored 130 on the test?

9 We will assume u 1,..., u 100 iid N(0,σ 2 u ) independent of e 1,..., e 100 iid N(0,σ 2 ). If we let µ + u i denote the IQ of student i, then IQs of the students are N(µ,σ 2 u), as stated at the beginning. If we let y i = µ + u i + e i denote the test score of student i, then y i (µ + u i ) N(µ + u i,σ 2 ), as stated at the beginning.

10 For this case, we have n = 100 Y = X β + Zu + e where X =1 n, β = µ, Z = I n, G = σ 2 ui n, R = σ 2 I n Then, and Σ = ZG Z + R = (σ 2 u + σ 2 )I n. GZ Σ 1 = σ2 u σ 2 u + σ 2 I n

11 And the BLUP for u is û = GZ Σ 1 (y X ˆβ Σ ) = The i th element of this vector is û i = σ2 u σ 2 u + σ 2(y i ȳ ) σ2 u σu 2 + σ2(y 1ȳ ) Thus, the BLUP for µ + u i (the IQ of student i) is σ2 u ˆµ+û i = ȳ + σu 2 + σ 2(y i ȳ ) = σ2 u σ2 σu 2 + σ 2y i+ σu 2 + σ 2ȳ

12 Note that the BLUP is a weighted average of the individual score and the overall mean score. σ 2 u σ 2 u + σ 2y i + σ2 σ 2 u + σ 2ȳ If there is relatively high variability among student scores (compared to variability within a student), then more weight is put on the individual score.

13 Let s return to the IQ example... Suppose it is known that σ2 u σ 2 =9 If we sample 100 students and their sample mean IQ was 100, what is the best prediction of the IQ of a student who scored 130 on the test? σ 2 u σ 2 u + σ 2 = σ2 u σ 2 σu 2 σ + 1 = = 0.9 We would predict the IQ of a student who scored 130 on the test to be somewhat shrunk toward the mean as 0.9(130) + 0.1(100) = 127

14 Example: Gene Expression Earlier in the semester, we introduced random effects using a gene expression example where there were 10 randomly chosen lines and 3 replicates within each line for a given gene. Y ij = µ + L i + ɛ ij for i = 1, 2,..., 10 and j = 1, 2, 3 with L i iid N(0, σ 2 L ) and ɛ ij iid N(0, σ 2 )

15 Example: Gene Expression Fit the random effects model for gene 1 and save the blups in a data set using the ODS output statement ods output SolutionR=blups; proc mixed data=gene1; class Line; model Expression=; random Line/solution; /* <---- */ run; ods output close;

16 Example: Gene Expression The grand mean is data blups; set blups; LineBlup = Estimate; keep Line LineBlup; proc print data=blups; run; Obs Line LineBlup

17 Example: Gene Expression Get the line means and compare to blups. ods output summary=means; proc means data=gene1; by Line; var Expression; run; ods output close; data means; set means; keep Line Expression_Mean Expression_N; run; data both; merge means blups; run; proc print data=both; run;

18 Example: Gene Expression Expression_ Expression_ Obs Line N Mean LineBlup Line means that are above the overall mean Ȳ.. = 4.10 have BLUPS that are brought down a bit (those that are below the overall mean have BLUPS that are brought up a bit). This is shrinkage toward the mean.

19 Example: Gene Expression proc sgplot data=both; scatter x=expression_mean y=lineblup; lineparm x=0 y=0 slope=1; refline / axis=x; refline / axis=y; run;

20 Example: Gene Expression We usually check the normality of the residuals (i.e. given the BLUPS, or conditioning on the BLUPS), but we could also check the normality of the random L i effects using the BLUPS, though I don t think this is done in practice very often.

21 Example: Gene Expression proc rank data=blups normal=blom out=diag; var LineBlup; ranks rankvalue; run; proc sgplot data=diag; scatter x=rankvalue y=lineblup; xaxis label="normal Quantiles"; run;