User:Timothee Flutre/Notebook/Postdoc/2011/06/28

From OpenWetWare

< User:Timothee Flutre | Notebook | Postdoc | 2011 | 06

Revision as of 11:03, 21 November 2012 by Timothee Flutre (Talk | contribs)

(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)

Jump to: navigation, search

Project name

Main project page
Next entry

Linear regression by ordinary least squares

Data: let's assume that we obtained data from $N$ individuals. We note $y_1,\ldots,y_N$ the (quantitative) phenotypes (e.g. expression level at a given gene), and $g_1,\ldots,g_N$ the genotypes at a given SNP. We want to assess their linear relationship.

Model: to start with, we use a simple linear regression (univariate phenotype, single predictor).

$\forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta g_i + \epsilon_i \text{ with } \epsilon_i \sim \mathcal{N}(0,\sigma^2)$

In vector-matrix notation:

$\vec{y} = X B + \vec{e}$ with $\vec{e} \sim \mathcal{N}_N(0,\sigma^2 I_N)$ and $B T = [μβ]$

The parameters of the model are: $Θ = {μ,β,σ}$

Use only summary statistics: most importantly, we want the following estimates: $\hat{\beta}$ , $\hat{\sigma}$ and $se(\hat{\beta})$ (its standard error). In the case where we don't have access to the original data (e.g. because genotypes are confidential) but only to some summary statistics (see below), it is still possible to calculate the estimates.

The well-known ordinary-least-square (OLS) estimator of $B$ is:

$\hat{B} = (X^T X)^{-1} X^T \vec{y}$

$\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \left( \begin{bmatrix} 1 & \ldots & 1 \\ g_1 & \ldots & g_N \end{bmatrix} \begin{bmatrix} 1 & g_1 \\ \vdots & \vdots \\ 1 & g_N \end{bmatrix} \right)^{-1} \begin{bmatrix} 1 & \ldots & 1 \\ g_1 & \ldots & g_N \end{bmatrix} \begin{bmatrix} y_1 \\ \vdots \\ y_N \end{bmatrix}$

$\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \begin{bmatrix} N & \sum_i g_i \\ \sum_i g_i & \sum_i g_i^2 \end{bmatrix}^{-1} \begin{bmatrix} \sum_i y_i \\ \sum_i g_i y_i \end{bmatrix}$

$\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \frac{1}{N \sum_i g_i^2 - (\sum_i g_i)^2} \begin{bmatrix} \sum_i g_i^2 & - \sum_i g_i \\ - \sum_i g_i & N \end{bmatrix} \begin{bmatrix} \sum_i y_i \\ \sum_i g_i y_i \end{bmatrix}$

$\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \frac{1}{N \sum_i g_i^2 - (\sum_i g_i)^2} \begin{bmatrix} \sum_i g_i^2 \sum_i y_i - \sum_i g_i \sum_i g_i y_i \\ - \sum_i g_i \sum_i y_i + N \sum_i g_i y_i \end{bmatrix}$

Let's now define 4 summary statistics, very easy to compute:

$\bar{y} = \frac{1}{N} \sum_{i=1}^N y_i$

$\bar{g} = \frac{1}{N} \sum_{i=1}^N g_i$

$\vec{g}^T \vec{g} = \sum_{i=1}^N g_i^2$

$\vec{g}^T \vec{y} = \sum_{i=1}^N g_i y_i$

This allows to obtain the estimate of the effect size only by having the summary statistics available:

$\hat{\beta} = \frac{\vec{g}^T \vec{y} - N \bar{g} \bar{y}}{\vec{g}^T \vec{g} - N \bar{g}^2}$

By the way, in this case (i.e. simple linear regression, a single predictor), it's easy to see that:

$\hat{\beta} = \frac{Cov[\vec{y},\vec{g}]}{Var[\vec{g}]}$

The same works for the estimate of the standard deviation of the errors:

$\hat{\sigma}^2 = \frac{1}{N-r}(\vec{y} - X\hat{B})^T(\vec{y} - X\hat{B})$

We can also benefit from this for the standard error of the parameters:

$V(\hat{B}) = \hat{\sigma}^2 (X^T X)^{-1}$

$V(\hat{B}) = \hat{\sigma}^2 \frac{1}{N \vec{g}^T \vec{g} - N^2 \bar{g}^2} \begin{bmatrix} \vec{g}^T\vec{g} & -N\bar{g} \\ -N\bar{g} & N \end{bmatrix}$

$V(\hat{\beta}) = \frac{\hat{\sigma}^2}{\vec{g}^T\vec{g} - N\bar{g}^2}$

which corresponds to:

$V(\hat{\beta}) = \frac{1}{N} \frac{Var[\vec{e}]}{Var[\vec{g}]}$

Simulation with a given PVE: when testing an inference model, the first step is usually to simulate data. However, how do we choose the parameters? In our case, the model is $y = μ + β g + ε$ . Therefore, the variance of $y$ can be decomposed like this:

$V (y) = V (μ + β g + ε) = V (μ) + V (β g) + V (ε) = β 2 V (g) + σ 2$

The most intuitive way to simulate data is therefore to fix the proportion of variance in $y$ explained by the genotype, for instance $P V E = 60%$ , as well as the standard deviation of the errors, typically $σ = 1$ . From this, we can calculate the corresponding effect size $β$ of the genotype:

$PVE = \frac{V(\beta g)}{V(y)}$

Therefore: $\beta = \pm \sigma \sqrt{\frac{PVE}{(1 - PVE) * V(g)}}$

Note that $g$ is the random variable corresponding to the genotype encoded in allele dose, such that it is equal to 0, 1 or 2 copies of the minor allele. For our simulation, we will fix the minor allele frequency $f$ (eg. $f = 0.3$ ) and we will assume Hardy-Weinberg equilibrium. Then $g$ is distributed according to a binomial distribution with 2 trials for which the probability of success is $f$ . As a consequence, its variance is $V (g) = 2 f (1 - f)$ .

Here is some R code implementing all this:

set.seed(1859)
N <- 100 # sample size
mu <- 4
pve <- 0.6
sigma <- 1
maf <- 0.3 # minor allele frequency
beta <- sigma * sqrt(pve / ((1 - pve) * 2 * maf * (1 - maf))) # 1.89
g <- rbinom(n=N, size=2, prob=maf) # assuming Hardy-Weinberg equilibrium
y <- mu + beta * g + rnorm(n=N, mean=0, sd=sigma)
ols <- lm(y ~ g)
summary(ols) # muhat=4.1+-0.13, betahat=1.6+-0.16, R2=0.49
sqrt((1/(N-2) * sum(ols$residuals^2))) # sigmahat=0.99
plot(x=0, type="n", xlim=range(g), ylim=range(y),
     xlab="genotypes (allele counts)", ylab="phenotypes",
     main="Simple linear regression")
for(i in unique(g))
  points(x=jitter(g[g == i]), y=y[g == i], col=i+1, pch=19)
abline(a=coefficients(ols)[1], b=coefficients(ols)[2])

Several predictors: let's now imagine that we also know the gender of the N sampled individuals. We hence want to account for that in our estimate of the effect of the genotype. In matrix notation, we still have the same model, $\vec{y} = XB + \vec{e}$ with $\vec{y}$ an Nx1 vector, X an Nx3 matrix with 1's in the first column, the genotypes in the second and the genders in the third, B a 3x1 vector and $\vec{e}$ an Nx1 vector following a multivariate Normal distribution centered on 0 and with covariance matrix $σ 2 I N$ .

As above, we want $\hat{B}$ , $\hat{\sigma}$ and $V(\hat{B})$ . To efficiently get them, we start with the singular value decomposition of X:

$X = U D V T$

This allows us to get the Moore-Penrose pseudoinverse matrix of X:

$X + = (X T X) - 1 X T = V D - 1 U T$

From this, we get the OLS estimate of the effect sizes:

$\hat{B} = X^+ \vec{y}$

Then it's straightforward to get the residuals:

$\hat{\vec{e}} = \vec{y} - X \hat{B}$

With them we can calculate the estimate of the error variance:

$\hat{\sigma}^2 = \frac{1}{N-3} \hat{\vec{e}}^T \hat{\vec{e}}$

And finally the standard errors of the estimates of the effect sizes:

$V(\hat{B}) = \hat{\sigma}^2 V D^{-2} V^T$

We can check this with some R code:

## simulate the data
set.seed(1859)
N <- 100
mu <- 4
pve.g <- 0.4 # genotype
pve.c <- 0.2 # other covariate, eg. gender
sigma <- 1
maf <- 0.3
sex.ratio <- 0.5
beta.g <- sigma * sqrt((1 / (2 * maf * (1 - maf))) * (pve.g / (1 - pve.g - pve.c))) # 1.543
beta.c <- beta.g * sqrt((pve.c / pve.g) * (2 * maf * (1 - maf) / sex.ratio * (1 - sex.ratio))) # 0.707
x.g <- rbinom(n=N, size=2, prob=maf)
x.c <- rbinom(n=N, size=1, prob=sex.ratio)
y <- mu + beta.g * x.g + beta.c * x.c + rnorm(n=N, mean=0, sd=sigma)
ols <- lm(y ~ x.g + x.c)
summary(ols) # muhat=3.9+-0.17, beta.g.hat=1.6+-0.17, beta.c.hat=0.58+-0.21, R2=0.51
sqrt((1/(N-3)) * sum(ols$residuals^2)) # sigma.hat = 1.058

## perform the OLS analysis with the SVD of X
X <- cbind(rep(1,N), x.g, x.c)
Xp <- svd(x=X)
B.hat <- Xp$v %*% diag(1/Xp$d) %*% t(Xp$u) %*% y
E.hat <- y - X %*% B.hat
sigma.hat <- as.numeric(sqrt((1/(N-3)) * t(E.hat) %*% E.hat)) # 1.058
var.theta.hat <- sigma.hat^2 * Xp$v %*% diag((1/Xp$d)^2) %*% t(Xp$v)
sqrt(diag(var.theta.hat)) # 0.168 0.175 0.212

Such an analysis can also be done easily in a custom C/C++ program thanks to the GSL (here).

User:Timothee Flutre/Notebook/Postdoc/2011/06/28

From OpenWetWare

Linear regression by ordinary least squares

Views

Personal tools

Navigation

research

Search

Toolbox