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

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 Revision as of 14:35, 30 July 2012 (view source) (→Simple linear regression: add regression with multiple predictors)← Previous diff Revision as of 20:57, 14 August 2012 (view source) (→Linear regression by ordinary least squares: improv simul with PVE + expand variance geno)Next diff → Line 76: Line 76: - * '''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 (linear regression: $y = \mu + \beta g + \epsilon$), it is frequent to fix the proportion of variance in $y$ explained by $\beta g$: + * '''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 $y = \mu + \beta g + \epsilon$). Therefore, the variance of $y$ can be decomposed like this: - $PVE = \frac{V(\beta g)}{V(y)} = \frac{V(\beta g)}{V(\beta g) + V(\epsilon)}$ with $V(\beta g) = \frac{1}{N}\sum_{n=1}^N (\beta g_n - \bar{\beta g})^2$ and $V(\epsilon) = \sigma^2$ + $V(y) = V(\mu + \beta g + \epsilon) = V(\mu) + V(\beta g) + V(\epsilon) = \beta^2 V(g) + \sigma^2$ - This way, by also fixing $\beta$, it is easy to calculate the corresponding $\sigma$: + The most intuitive way to simulate data is therefore to fix the proportion of variance in $y$ explained by the genotype, for instance $PVE=60%$, as well as the standard deviation of the errors, typically $\sigma=1$. From this, we can calculate the corresponding effect size $\beta$ of the genotype: - $\sigma = \sqrt{\frac{1}{N}\sum_{n=1}^N (\beta g_n - \bar{\beta g})^2 \frac{1 - PVE}{PVE}}$ + $PVE = \frac{V(\beta g)}{V(y)}$ - Here is some R code implementing this: + 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 $V(g)=2f(1-f)$. + + Here is some R code implementing all this: set.seed(1859) set.seed(1859) - N <- 100 + N <- 100 # sample size - mu <- 5 + mu <- 4 - g <- sample(x=0:2, size=N, replace=TRUE, prob=c(0.5, 0.3, 0.2)) # MAF=0.2 + pve <- 0.6 - beta <- 0.5 + sigma <- 1 - pve <- 0.8 + maf <- 0.3 - beta.g.bar <- mean(beta * g) + beta <- sigma * sqrt(pve / ((1 - pve) * 2 * maf * (1 - maf))) # 1.88 - sigma <- sqrt((1/N) * sum((beta * g - beta.g.bar)^2) * (1-pve) / pve) # 0.18 + g <- sample(x=0:2, size=N, replace=TRUE, prob=c(maf^2, 2*maf*(1-maf), (1-maf)^2)) y <- mu + beta * g + rnorm(n=N, mean=0, sd=sigma) y <- mu + beta * g + rnorm(n=N, mean=0, sd=sigma) + ols <- lm(y ~ g) + summary(ols) # muhat=3.5, betahat=2.1, R2=0.64 + sqrt(mean(ols$residuals^2)) # sigmahat = 0.98 plot(x=0, type="n", xlim=range(g), ylim=range(y), plot(x=0, type="n", xlim=range(g), ylim=range(y), - xlab="genotypes (allele dose)", ylab="phenotypes", + xlab="genotypes", ylab="phenotypes", main="Simple linear regression") main="Simple linear regression") for(i in unique(g)) for(i in unique(g)) points(x=jitter(g[g == i]), y=y[g == i], col=i+1, pch=19) points(x=jitter(g[g == i]), y=y[g == i], col=i+1, pch=19) - ols <- lm(y ~ g) - summary(ols) # muhat=5.01, betahat=0.46, R2=0.779 abline(a=coefficients(ols)[1], b=coefficients(ols)[2]) abline(a=coefficients(ols)[1], b=coefficients(ols)[2]) ## Revision as of 20:57, 14 August 2012 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 (eg. 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). $\forall n \in {1,\ldots,N}, \; y_n = \mu + \beta g_n + \epsilon_n \text{ with } \epsilon_n \sim N(0,\sigma^2)$ In matrix notation: y = Xθ + ε with ε˜NN(0,σ2IN) and θT = (μ,β) • Use only summary statistics: most importantly, we want the following estimates: $\hat{\beta}$, $se(\hat{\beta})$ (its standard error) and $\hat{\sigma}$. In the case where we don't have access to the original data (eg. because genotypes are confidential) but only to some summary statistics (see below), it is still possible to calculate the estimates. Here is the ordinary-least-square (OLS) estimator of θ: $\hat{\theta} = (X^T X)^{-1} X^T Y$ $\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \left( \begin{bmatrix} 1 & \ldots & 1 \\ g_1 & \ldots & g_N \end{bmatrix} \begin{bmatrix} 1 & g_1 \\ \vdots & \vdots \\ 1 & g_N \end{bmatrix} \right)^{-1} \begin{bmatrix} 1 & \ldots & 1 \\ g_1 & \ldots & g_N \end{bmatrix} \begin{bmatrix} y_1 \\ \vdots \\ y_N \end{bmatrix}$ $\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \begin{bmatrix} N & \sum_n g_n \\ \sum_n g_n & \sum_n g_n^2 \end{bmatrix}^{-1} \begin{bmatrix} \sum_n y_n \\ \sum_n g_n y_n \end{bmatrix}$ $\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \frac{1}{N \sum_n g_n^2 - (\sum_n g_n)^2} \begin{bmatrix} \sum_n g_n^2 & - \sum_n g_n \\ - \sum_n g_n & N \end{bmatrix} \begin{bmatrix} \sum_n y_n \\ \sum_n g_n y_n \end{bmatrix}$ $\begin{bmatrix} \hat{\mu} \\ \hat{\beta} \end{bmatrix} = \frac{1}{N \sum_n g_n^2 - (\sum_n g_n)^2} \begin{bmatrix} \sum_n g_n^2 \sum_n y_n - \sum_n g_n \sum_n g_n y_n \\ - \sum_n g_n \sum_n y_n + N \sum_n g_n y_n \end{bmatrix}$ Let's now define 4 summary statistics, very easy to compute: $\bar{y} = \frac{1}{N} \sum_{n=1}^N y_n$ $\bar{g} = \frac{1}{N} \sum_{n=1}^N g_n$ $g^T g = \sum_{n=1}^N g_n^2$ $g^T y = \sum_{n=1}^N g_n y_n$ This allows to obtain the estimate of the effect size only by having the summary statistics available: $\hat{\beta} = \frac{g^T y - N \bar{g} \bar{y}}{g^T g - N \bar{g}^2}$ The same works for the estimate of the standard deviation of the errors: $\hat{\sigma}^2 = \frac{1}{N-r}(y - X\hat{\theta})^T(y - X\hat{\theta})$ We can also benefit from this for the standard error of the parameters: $V(\hat{\theta}) = \hat{\sigma}^2 (X^T X)^{-1}$ $V(\hat{\theta}) = \hat{\sigma}^2 \frac{1}{N g^T g - N^2 \bar{g}^2} \begin{bmatrix} g^Tg & -N\bar{g} \\ -N\bar{g} & N \end{bmatrix}$ $V(\hat{\beta}) = \frac{\hat{\sigma}^2}{g^Tg - N\bar{g}^2}$ • 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 + βg + ε). Therefore, the variance of y can be decomposed like this: V(y) = V(μ + βg + ε) = V(μ) + Vg) + V(ε) = β2V(g) + σ2 The most intuitive way to simulate data is therefore to fix the proportion of variance in y explained by the genotype, for instance PVE = 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) = 2f(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 beta <- sigma * sqrt(pve / ((1 - pve) * 2 * maf * (1 - maf))) # 1.88 g <- sample(x=0:2, size=N, replace=TRUE, prob=c(maf^2, 2*maf*(1-maf), (1-maf)^2)) y <- mu + beta * g + rnorm(n=N, mean=0, sd=sigma) ols <- lm(y ~ g) summary(ols) # muhat=3.5, betahat=2.1, R2=0.64 sqrt(mean(ols$residuals^2)) # sigmahat = 0.98
plot(x=0, type="n", xlim=range(g), ylim=range(y),
xlab="genotypes", 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 genotypic effect. In matrix notation, we still have the same model Y = XB + E with 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 E an Nx1 vector following a multivariate Normal distribution centered on 0 and with covariance matrix σ2IN.

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 = UDVT

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

X + = (XTX) − 1XT

X + = VD − 1UT

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

$\hat{B} = X^+ Y$

Then it's straightforward to get the residuals:

$\hat{E} = Y - X \hat{B}$

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

$\hat{\sigma} = \sqrt{\frac{1}{N-3} \hat{E}^T \hat{E}}$

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 <- 5
Xg <- sample(x=0:2, size=N, replace=TRUE, prob=c(0.5, 0.3, 0.2)) # genotypes
beta.g <- 0.5
Xc <- sample(x=0:1, size=N, replace=TRUE, prob=c(0.7, 0.3)) # gender
beta.c <- 0.3
pve <- 0.8
betas.gc.bar <- mean(beta.g * Xg + beta.c * Xc) # 0.405
sigma <- sqrt((1/N) * sum((beta.g * Xg + beta.c * Xc - betas.gc.bar)^2) *
(1-pve) / pve) # 0.2
y <- mu + beta.g * Xg + beta.c * Xc + rnorm(n=N, mean=0, sd=sigma)

## perform the OLS analysis with the SVD of X
X <- cbind(rep(1,N), Xg, Xc)
Xp <- svd(x=X)
B.hat <- Xp$v %*% diag(1/Xp$d) %*% t(Xp$u) %*% y E.hat <- y - X %*% B.hat sigma.hat <- as.numeric(sqrt((1/(N-3)) * t(E.hat) %*% E.hat)) # 0.211 var.theta.hat <- sigma.hat^2 * Xp$v %*% diag((1/Xp$d)^2) %*% t(Xp$v)
sqrt(diag(var.theta.hat)) # 0.0304 0.0290 0.0463

## check all this
ols <- lm(y ~ Xg + Xc)
summary(ols) # muhat=4.99+-0.03, beta.g.hat=0.52+--.-29, beta.c.hat=0.24+-0.046, R2=0.789



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