Overview

Teaching: 70 min
Exercises: 50 min

Questions

• How can we apply linear regression in a high-dimensional setting?

• How can we benefit from the fact that we have many outcomes?

• How can we control for the fact that we do many tests?

Objectives

• Perform and critically analyse high-dimensional regression.

• Understand methods for shrinkage of noise parameters in high-dimensional regression.

• Perform multiple testing adjustment.

DNA methylation data

For the following few episodes, we will be working with human DNA methylation data from flow-sorted blood samples, described in data. DNA methylation assays measure, for each of many sites in the genome, the proportion of DNA that carries a methyl mark (a chemical modification that does not alter the DNA sequence). In this case, the methylation data come in the form of a matrix of normalised methylation levels (M-values), where negative values correspond to unmethylated DNA and positive values correspond to methylated DNA. Along with this, we have a number of sample phenotypes (eg, age in years, BMI).

Let’s read in the data for this episode:

library("here")
library("minfi")
methylation <- readRDS(here("data/methylation.rds"))

Note: the code that we used to download these data from its source is available here

methylation is actually a special Bioconductor SummarizedExperiment object that summarises lots of different information about the data. These objects are very useful for storing all of the information about a dataset in a high-throughput context. The structure of SummarizedExperiment objects is described in the vignettes on Bioconductor. Here, we show how to extract the information for analysis.

We can extract

• the dimensions of the dataset using dim(). Importantly, in these objects and data structures for computational biology in R generally, observations are stored as columns and features (in this case, sites in the genome) are stored as rows. This is in contrast to usual tabular data, where features or variables are stored as columns and observations are stored as rows;
• assays, (normalised methylation levels here), using assay();
• sample-level information using colData().

dim(methylation)

[1] 5000   37

You can see in this output that this object has a dim() of $5000 \times 37$, meaning it has 5000 features and 37 observations. To extract the matrix of methylation M-values, we can use the assay() function.

methyl_mat <- assay(methylation)

The distribution of these M-values looks like this:

hist(methyl_mat, xlab = "M-value")

Methylation levels are generally bimodally distributed.

You can see that there are two peaks in this distribution, corresponding to features which are largely unmethylated and methylated, respectively.

Similarly, we can examine the colData(), which represents the sample-level metadata we have relating to these data. In this case, the metadata, phenotypes, and groupings in the colData look like this for the first 6 samples:

head(colData(methylation))

In this episode, we will focus on the association between age and methylation. The following heatmap summarises age and methylation levels available in the methylation dataset:

library("ComplexHeatmap")

age <- methylation$Age

# sort methylation values by age 
order <- order(age)
age_ord <- age[order]
methyl_mat_ord <- methyl_mat[, order]

# plot heatmap
Heatmap(methyl_mat_ord,
        cluster_columns = FALSE,
        show_row_names = FALSE,
        show_column_names = FALSE,
        name = "M-value",
        row_title = "Feature", 
        column_title =  "Sample", 
        top_annotation = columnAnnotation(age = age_ord))

Heatmap of methylation values across all features.

Challenge 1

Why can we not just fit many linear regression models relating every combination of feature (colData and assays) and draw conclusions by associating all variables with significant model p-values?

Solution

There are a number of problems that this kind of approach presents. For example:

1. If we perform 5000 tests for each of 14 variables, even if there were no true associations in the data, we’d be likely to observe some strong spurious associations that arise just from random noise.
2. We may not have a representative sample for each of these covariates. For example, we may have very small sample sizes for some ethnicities, leading to spurious findings.
3. Without a research question in mind when creating a model, it’s not clear how we can interpret each model, and rationalising the results after the fact can be dangerous; it’s easy to make up a “story” that isn’t grounded in anything but the fact that we have significant findings.

In general, it is scientifically interesting to explore two modelling problems using the three types of data:

1. Predicting methylation levels using age as a predictor. In this case, we would have 5000 outcomes (methylation levels across the genome) and a single covariate (age).

2. Predicting age using methylation levels as predictors. In this case, we would have a single outcome (age) which will be predicted using 5000 covariates (methylation levels across the genome).

The examples in this episode will focus on the first type of problem, whilst the next episode will focus on the second.

Measuring DNA Methylation

DNA methylation is an epigenetic modification of DNA. Generally, we are interested in the proportion of methylation at many sites or regions in the genome. DNA methylation microarrays, as we are using here, measure DNA methylation using two-channel microarrays, where one channel captures signal from methylated DNA and the other captures unmethylated signal. These data can be summarised as “Beta values” ($\beta$ values), which is the ratio of the methylated signal to the total signal (methylated plus unmethylated). The $\beta$ value for site $i$ is calculated as

\[\beta_i = \frac{ m_i } { u_{i} + m_{i} }\]

where $m_i$ is the methylated signal for site $i$ and $u_i$ is the unmethylated signal for site $i$. $\beta$ values take on a value in the range $[0, 1]$, with 0 representing a completely unmethylated site and 1 representing a completely methylated site.

The M-values we use here are the $\log_2$ ratio of methylated versus unmethylated signal:

\[M_i = \log_2\left(\frac{m_i}{u_i}\right)\]

M-values are not bounded to an interval as Beta values are, and therefore can be easier to work with in statistical models.

Regression with many outcomes

In high-throughput studies, it is common to have one or more phenotypes or groupings that we want to relate to features of interest (eg, gene expression, DNA methylation levels). In general, we want to identify differences in the features of interest that are related to a phenotype or grouping of our samples. Identifying features of interest that vary along with phenotypes or groupings can allow us to understand how phenotypes arise or manifest. Analysis of this type is sometimes referred to using the term differential analysis.

For example, we might want to identify genes that are expressed at a higher level in mutant mice relative to wild-type mice to understand the effect of a mutation on cellular phenotypes. Alternatively, we might have samples from a set of patients, and wish to identify epigenetic features that are different in young patients relative to old patients, to help us understand how ageing manifests.

Using linear regression, it is possible to identify differences like these. However, high-dimensional data like the ones we’re working with require some special considerations. A first consideration, as we saw above, is that there are far too many features to fit each one-by-one as we might do when analysing low-dimensional datasets (for example using lm() on each feature and checking the linear model assumptions). A second consideration is that statistical approaches may behave slightly differently when applied to very high-dimensional data, compared to low-dimensional data. A third consideration is the speed at which we can actually compute statistics for data this large – methods optimised for low-dimensional data may be very slow when applied to high-dimensional data.

Ideally when performing regression, we want to identify cases like this, where there is a clear association, and we probably “don’t need” statistics:

A scatter plot of age and a feature of interest.

or equivalently for a discrete covariate:

A scatter plot of a grouping and a feature of interest.

However, often due to small differences and small sample sizes, the problem is more difficult:

A scatter plot of a grouping and a feature of interest.

And, of course, we often have an awful lot of features and need to prioritise a subset of them! We need a rigorous way to prioritise genes for further analysis.

Fitting a linear model

So, in the data we have read in, we have a matrix of methylation values $X$ and a vector of ages, $y$. One way to model this is to see if we can use age to predict the expected (average) methylation value for sample $j$ at a given locus $i$, which we can write as $X_{ij}$. We can write that model as:

where $\text{Age}_j$ is the age of sample $j$. In this model, $\beta_1$ represents the unit change in mean methylation level for each unit (year) change in age. For a specific CpG, we can fit this model and get more information from the model object. For illustration purposes, here we arbitrarily select the first CpG in the methyl_mat matrix (the one on its first row).

age <- methylation$Age
# methyl_mat[1, ] indicates that the 1st CpG will be used as outcome variable
lm_age_methyl1 <- lm(methyl_mat[1, ] ~ age)
lm_age_methyl1

Call:
lm(formula = methyl_mat[1, ] ~ age)

Coefficients:
(Intercept)          age  
   0.902334     0.008911  

We now have estimates for the expected methylation level when age equals 0 (the intercept) and the change in methylation level for a unit change in age (the slope). We could plot this linear model:

plot(age, methyl_mat[1, ], xlab = "Age", ylab = "Methylation level", pch = 16)
abline(lm_age_methyl1)

A scatter plot of age versus the methylation level for an arbitrarily selected CpG side (the one stored as the first column of methyl_mat). Each dot represents an individual. The black line represents the estimated linear model.

For this feature, we can see that there is no strong relationship between methylation and age. We could try to repeat this for every feature in our dataset; however, we have a lot of features! We need an approach that allows us to assess associations between all of these features and our outcome while addressing the three considerations we outlined previously. Before we introduce this approach, let’s go into detail about how we generally check whether the results of a linear model are statistically significant.

Hypothesis testing in linear regression

Using the linear model we defined above, we can ask questions based on the estimated value for the regression coefficients. For example, do individuals with different age have different methylation values for a given CpG? We usually do this via hypothesis testing. This framework compares the results that we observed (here, estimated linear model coefficients) to the results you would expect under a null hypothesis associated to our question. In the example above, a suitable null hypothesis would test whether the regression coefficient associated to age ($\beta_1$) is equal to zero or not. If $\beta_1$ is equal to zero, the linear model indicates that there is no linear relationship between age and the methylation level for the CpG (remember: as its name suggests, linear regression can only be used to model linear relationships between predictors and outcomes!). In other words, the answer to our question would be: no!

The output of a linear model typically returns the results associated with the null hypothesis described above (this may not always be the most realistic or useful null hypothesis, but it is the one we have by default!). To be more specific, the test compares our observed results with a set of hypothetical counter-examples of what we would expect to observe if we repeated the same experiment and analysis over and over again under the null hypothesis.

For this linear model, we can use tidy() from the broom package to extract detailed information about the coefficients and the associated hypothesis tests in this model:

library("broom")
tidy(lm_age_methyl1)

# A tibble: 2 × 5
  term        estimate std.error statistic p.value
  <chr>          <dbl>     <dbl>     <dbl>   <dbl>
1 (Intercept)  0.902      0.344      2.62   0.0129
2 age          0.00891    0.0100     0.888  0.381 

The standard errors (std.error) represent the statistical uncertainty in our regression coefficient estimates (often referred to as effect size). The test statistics and p-values (statistic and p.value) represent measures of how (un)likely it would be to observe results like this under the “null hypothesis”. For coefficient $k$ in a linear model (in our case, it would be the slope), the test statistic calculated in statistic above is a t-statistic given by:

$SE\left(\hat{\beta}_{k}\right)$ measures the uncertainty we have in our effect size estimate. Knowing what distribution these t-statistics follow under the null hypothesis allows us to determine how unlikely it would be for us to observe what we have under those circumstances, if we repeated the experiment and analysis over and over again. To demonstrate how the t-statistics are calculated, we can compute them “by hand”:

table_age_methyl1 <- tidy(lm_age_methyl1)
tvals <- table_age_methyl1$estimate / table_age_methyl1$std.error
all.equal(tvals, table_age_methyl1$statistic)

[1] TRUE

We can see that the t-statistic is just the ratio between the coefficient estimate and the standard error.

Calculating the p-values is a bit more tricky. Specifically, it is the proportion of the distribution of the test statistic under the null hypothesis that is as extreme or more extreme than the observed value of the test statistic. This is easy to observe visually, by plotting the theoretical distribution of the test statistic under the null hypothesis (see next call-out box for more details about it):

The p-value for a regression coefficient represents how often it'd be observed under the null.

The red-ish shaded region represents the portion of the distribution of the test statistic under the null hypothesis that is equal or greater to the value we observe for the intercept term. As our null hypothesis relates to a 2-tailed test (as the null hypothesis states that the regression coefficient is equal to zero, we would reject it if the regression coefficient is substantially larger or smaller than zero), the p-value for the test is twice the value of the shaded region. In this case, the shaded region is small relative to the total area of the null distribution; therefore, the p-value is small ($p=0.013$). The blue-ish shaded region represents the same measure for the slope term; this is larger, relative to the total area of the distribution, therefore the p-value is larger than the one for the intercept term ($p=0.381$). The p-value is a function of the test statistic: the ratio between the effect size we’re estimating and the uncertainty we have in that effect. A large effect with large uncertainty may not lead to a small p-value, and a small effect with small uncertainty may lead to a small p-value.

Calculating p-values from a linear model

Manually calculating the p-value for a linear model is a little bit more complex than calculating the t-statistic. The intuition posted above is definitely sufficient for most cases, but for completeness, here is how we do it:

Since the statistic in a linear model is a t-statistic, it follows a student t distribution under the null hypothesis, with degrees of freedom (a parameter of the student t-distribution) given by the number of observations minus the number of coefficients fitted, in this case $37 - 2 = 35$. We want to know what portion of the distribution function of the test statistic is as extreme as, or more extreme than, the value we observed. The function pt()(similar topnorm(), etc) can give us this information.

Since we’re not sure if the coefficient will be larger or smaller than zero, we want to do a 2-tailed test. Therefore we take the absolute value of the t-statistic, and look at the upper rather than lower tail. In the figure above the shaded areas are only looking at “half” of the t-distribution (which is symmetric around zero), therefore we multiply the shaded area by 2 in order to calculate the p-value.

Combining all of this gives us:

pvals <- 2 * pt(abs(tvals), df = lm_age_methyl1$df, lower.tail = FALSE)
all.equal(table_age_methyl1$p.value, pvals)

[1] TRUE

Challenge 2

In the model we fitted, the estimate for the intercept is 0.902 and its associated p-value is 0.0129. What does this mean?

Solution

The first coefficient in a linear model like this is the intercept, which measures the mean of the outcome (in this case, the methylation value for the first CpG) when age is zero. In this case, the intercept estimate is 0.902. However, this is not a particularly noteworthy finding as we do not have any observations with age zero (nor even any with age < 20!).

The reported p-value is associated to the following null hypothesis: the intercept ($\beta_0$ above) is equal to zero. Using the usual significance threshold of 0.05, we reject the null hypothesis as the p-value is smaller than 0.05. However, it is not really interesting if this intercept is zero or not, since we probably do not care what the methylation level is when age is zero. In fact, this question does not even make much sense! In this example, we are more interested in the regression coefficient associated to age, as that can tell us whether there is a linear relationship between age and methylation for the CpG.

Fitting a lot of linear models

In the linear model above, we are generally interested in the second regression coefficient (often referred to as slope) which measures the linear relationship between age and methylation levels. For the first CpG, here is its estimate:

coef_age_methyl1 <- tidy(lm_age_methyl1)[2, ]
coef_age_methyl1

# A tibble: 1 × 5
  term  estimate std.error statistic p.value
  <chr>    <dbl>     <dbl>     <dbl>   <dbl>
1 age    0.00891    0.0100     0.888   0.381

In this case, the p-value is equal to 0.381 and therefore we cannot reject the null hypothesis: there is no statistical evidence to suggest that the regression coefficient associated to age is not equal to zero.

Now, we could do this for every feature (CpG) in the dataset and rank the results based on their test statistic or associated p-value. However, fitting models in this way to 5000 features is not very computationally efficient, and it would also be laborious to do programmatically. There are ways to get around this, but first let us talk about what exactly we are doing when we look at significance tests in this context.

Sharing information across outcome variables

We are going to introduce an idea that allows us to take advantage of the fact that we carry out many tests at once on structured data. We can leverage this fact to share information between model parameters. The insight that we use to perform information pooling or sharing is derived from our knowledge about the structure of the data. For example, in a high-throughput experiment like a DNA methylation assay, we know that all of the features were measured simultaneously, using the same technique. This means that generally, we expect the base-level variability for each feature to be broadly similar.

This can enable us to get a better estimate of the uncertainty of model parameters than we could get if we consider each feature in isolation. So, to share information between features allows us to get more robust estimators. Remember that the t-statistic for coefficient $\beta_k$ in a linear model is the ratio between the coefficient estimate and its standard error:

It is clear that large effect sizes will likely lead to small p-values, as long as the standard error for the coefficent is not large. However, the standard error is affected by the amount of noise, as we saw earlier. If we have a small number of observations, it is common for the noise for some features to be extremely small simply by chance. This, in turn, causes small p-values for these features, which may give us unwarranted confidence in the level of certainty we have in the results (false positives).

There are many statistical methods in genomics that use this type of approach to get better estimates by pooling information between features that were measured simultaneously using the same techniques. Here we will focus on the package limma, which is an established software package used to fit linear models, originally for the gene expression micro-arrays that were common in the 2000s, but which is still in use in RNAseq experiments, among others. The authors of limma made some assumptions about the distributions that these follow, and pool information across genes to get a better estimate of the uncertainty in effect size estimates. It uses the idea that noise levels should be similar between features to moderate the estimates of the test statistic by shrinking the estimates of standard errors towards a common value. This results in a moderated t-statistic.

The process of running a model in limma is somewhat different to what you may have seen when running linear models. Here, we define a model matrix or design matrix, which is a way of representing the coefficients that should be fit in each linear model. These are used in similar ways in many different modelling libraries.

What is a model matrix?

R fits a regression model by choosing the vector of regression coefficients that minimises the differences between outcome values and predicted values using the covariates (or predictor variables). To get predicted values, we multiply the matrix of predictors by the coefficients. The latter matrix is called the model matrix (or design matrix). It has one row for each observation and one column for each predictor, plus one additional column of ones (the intercept column). Many R libraries contruct the model matrix behind the scenes, but limma does not. Usually, the model matrix can be extracted from a model fit using the function model.matrix(). Here is an example of a model matrix for the methylation model:

design_age <- model.matrix(lm_age_methyl1) # model matrix
head(design_age)

                    (Intercept) age
201868500150_R01C01           1  39
201868500150_R03C01           1  49
201868500150_R05C01           1  20
201868500150_R07C01           1  49
201868500150_R08C01           1  33
201868590193_R02C01           1  21

dim(design_age)

[1] 37  2

As you can see, the model matrix has the same number of rows as our methylation data has samples. It also has two columns - one for the intercept (similar to the linear model we fit above) and one for age. This happens “under the hood” when fitting a linear model with lm(), but here we have to specify it directly. The limma user manual has more detail on how to make model matrices for different types of experimental design, but here we are going to stick with this simple two-variable case.

We pass our methylation data to lmFit(), specifying the model matrix. Internally, this function runs lm() on each row of the data in an efficient way. The function eBayes(), when applied to the output of lmFit(), performs the pooled estimation of standard errors that results in the moderated t-statistics and resulting p-values.

library("limma")
design_age <- model.matrix(lm_age_methyl1) # model matrix
fit_age <- lmFit(methyl_mat, design = design_age)
fit_age <- eBayes(fit_age)

To obtain the results of the linear models, we can use the topTable() function. By default, this returns results for the first coefficient in the model. As we saw above when using lm(), and when we defined design_age above, the first coefficient relates to the intercept term, which we are not particularly interested in here; therefore we specify coef = 2. Further, topTable() by default only returns the top 10 results. To see all of the results in the data, we specify number = nrow(fit_age) to ensure that it returns a row for every row of the input matrix.

toptab_age <- topTable(fit_age, coef = 2, number = nrow(fit_age))
head(toptab_age)

                 logFC    AveExpr         t      P.Value   adj.P.Val        B
cg08446924 -0.02571353 -0.4185868 -6.039068 5.595675e-07 0.002797837 5.131574
cg06493994  0.01550941 -2.1057877  5.593988 2.239813e-06 0.005599533 3.747986
cg17661642  0.02266668 -2.0527722  5.358739 4.658336e-06 0.006048733 3.019698
cg05168977  0.02276336 -2.2918472  5.346500 4.838987e-06 0.006048733 2.981904
cg24549277  0.01975577 -1.7466088  4.939242 1.708355e-05 0.011508818 1.731821
cg04436528 -0.01943612  0.7033503 -4.917179 1.828563e-05 0.011508818 1.664608

The output of topTable includes the coefficient, here termed a log fold change logFC, the average level (aveExpr), the t-statistic t, the p-value (P.Value), and the adjusted p-value (adj.P.Val). We’ll cover what an adjusted p-value is very shortly. The table also includes B, which represents the log-odds that a feature is signficantly different, which we won’t cover here, but which will generally be a 1-1 transformation of the p-value. The coefficient estimates here are termed logFC for legacy reasons relating to how microarray experiments were traditionally performed. There are more details on this topic in many places, for example this tutorial by Kasper D. Hansen

Now we have estimates of effect sizes and p-values for the association between methylation level at each locus and age for our 37 samples. It’s useful to create a plot of effect size estimates (model coefficients) against p-values for each of these linear models, to visualise the magnitude of effects and the statistical significance of each. These plots are often called “volcano plots”, because they resemble an eruption.

plot(toptab_age$logFC, -log10(toptab_age$P.Value),
    xlab = "Effect size", ylab = bquote(-log[10](p-value)),
    pch = 19
)

Plotting p-values against effect sizes using limma; the results are similar to a standard linear model.

In this figure, every point represents a feature of interest. The x-axis represents the effect size observed for that feature in a linear model, while the y-axis is the $-\log_{10}(\text{p-value})$, where larger values indicate increasing statistical evidence of a non-zero effect size. A positive effect size represents increasing methylation with increasing age, and a negative effect size represents decreasing methylation with increasing age. Points higher on the y-axis represent features for which we think the results we observed would be very unlikely under the null hypothesis.

Since we want to identify features that have different methylation levels in different age groups, in an ideal case there would be clear separation between “null” and “non-null” features. However, usually we observe results as we do here: there is a continuum of effect sizes and p-values, with no clear separation between these two classes of features. While statistical methods exist to derive insights from continuous measures like these, it is often convenient to obtain a list of features which we are confident have non-zero effect sizes. This is made more difficult by the number of tests we perform.

Challenge 3

The effect size estimates are very small, and yet many of the p-values are well below a usual significance level of p < 0.05. Why is this?

Solution

Because age has a much larger range than methylation levels, the unit change in methylation level even for a strong relationship is very small!

As we mentioned, the p-value is a function of both the effect size estimate and the uncertainty (standard error) of that estimate. Because the uncertainty in our estimates is much smaller than the estimates themselves, the p-values are also small.

If we predicted age using methylation level, it is likely we would see much larger coefficients, though broadly similar p-values!

It is worthwhile considering what exactly the effect of the moderation or information sharing that limma performs has on our results. To do this, let us compare the effect sizes estimates and p-values from the two approaches.

Plot of effect sizes using limma vs. those using lm.

These are exactly identical! This is because limma does not perform any sharing of information when estimating effect sizes. This is in contrast to similar packages that apply shrinkage to the effect size estimates, like DESeq2. These often use information sharing to shrink or moderate the effect size estimates, in the case of DESeq2 by again sharing information between features about sample-to-sample variability. In contrast, let us look at the p-values from limma and R’s built-in lm() function:

Plot of p-values using limma vs. those using lm.

We can see that for the vast majority of features, the results are broadly similar. There seems to be a minor general tendency for limma to produce smaller p-values, but for several features, the p-values from limma are considerably larger than the p-values from lm(). This is because the information sharing tends to shrink large standard error estimates downwards and small estimates upwards. When the degree of statistical significance is due to an abnormally small standard error rather than a large effect, this effect results in this prominent reduction in statistical significance, which has been shown to perform well in case studies. The degree of shrinkage generally depends on the amount of pooled information and the strength of the evidence independent of pooling. For example, with very few samples and many features, information sharing has a larger effect, because there are a lot of genes that can be used to provide pooled estimates, and the evidence from the data that this is weighed against is relatively sparse. In contrast, when there are many samples and few features, there is not much opportunity to generate pooled estimates, and the evidence of the data can easily outweigh the pooling.

Shrinkage methods like these ones can be complex to implement and understand, but it is useful to develop an intuition about why these approaches may be more precise and sensitive than the naive approach of fitting a model to each feature separately.

Challenge 4

1. Try to run the same kind of linear model with smoking status as covariate instead of age, and making a volcano plot. Note: smoking status is stored as methylation$smoker.
2. We saw in the example in the lesson that this information sharing can lead to larger p-values. Why might this be preferable?

Solution

1. The following code runs the same type of model with smoking status:

   design_smoke <- model.matrix(~methylation$smoker)
   fit_smoke <- lmFit(methyl_mat, design = design_smoke)
   fit_smoke <- eBayes(fit_smoke)
   toptab_smoke <- topTable(fit_smoke, coef = 2, number = nrow(fit_smoke))
   plot(toptab_smoke$logFC, -log10(toptab_smoke$P.Value),
       xlab = "Effect size", ylab = bquote(-log[10](p)),
       pch = 19
   )
   

   

   A plot of significance against effect size for a regression of smoking against methylation.

2. Being a bit more conservative when identifying features can help to avoid false discoveries. Furthermore, when rejecting the null hypothesis is based more on a small standard error resulting from abnormally low levels of variability for a given feature, we might want to be a bit more conservative in our expectations.

Shrinkage

Shrinkage is an intuitive term for an effect of information sharing, and is something observed in a broad range of statistical models. Often, shrinkage is induced by a multilevel modelling approach or by Bayesian methods.

The general idea is that these models incorporate information about the structure of the data into account when fitting the parameters. We can share information between features because of our knowledge about the data structure; this generally requires careful consideration about how the data were generated and the relationships within.

An example people often use is estimating the effect of attendance on grades in several schools. We can assume that this effect is similar in different schools (but maybe not identical), so we can share information about the effect size between schools and shrink our estimates towards a common value.

For example in DESeq2, the authors used the observation that genes with similar expression counts in RNAseq data have similar dispersion, and a better estimate of these dispersion parameters makes estimates of fold changes much more stable. Similarly, in limma the authors made the assumption that in the absence of biological effects, we can often expect the technical variation in the measurement of the expression of each of the genes to be broadly similar. Again, better estimates of variability allow us to prioritise genes in a more reliable way.

There are many good resources to learn about this type of approach, including:

• a blog post by TJ Mahr
• a book by David Robinson
• a (relatively technical) book by Gelman and Hill

The problem of multiple tests

With such a large number of features, it would be useful to decide which features are “interesting” or “significant” for further study. However, if we were to apply a normal significance threshold of 0.05, it would be likely we end up with a lot of false positives. This is because a p-value threshold like this represents a $\frac{1}{20}$ chance that we observe results as extreme or more extreme under the null hypothesis (that there is no assocation between age and methylation level). If we carry out many more than 20 such tests, we can expect to see situations where, despite the null hypothesis being true, we observe observe signifiant p-values due to random chance. To demonstrate this, it is useful to see what happens if we permute (scramble) the age values and run the same test again:

set.seed(123) 
age_perm <- age[sample(ncol(methyl_mat), ncol(methyl_mat))]
design_age_perm <- model.matrix(~age_perm)

fit_age_perm <- lmFit(methyl_mat, design = design_age_perm)
fit_age_perm <- eBayes(fit_age_perm)
toptab_age_perm <- topTable(fit_age_perm, coef = 2, number = nrow(fit_age_perm))

plot(toptab_age_perm$logFC, -log10(toptab_age_perm$P.Value),
    xlab = "Effect size", ylab = bquote(-log[10](p)),
    pch = 19
)
abline(h = -log10(0.05), lty = "dashed", col = "red")

Plotting p-values against effect sizes for a randomised outcome shows we still observe 'significant' results.

Since we have generated a random sequence of ages, we have no reason to suspect that there is a true association between methylation levels and this sequence of random numbers. However, you can see that the p-value for many features is still lower than a traditional significance level of $p=0.05$. In fact, here 226 features are significant at p < 0.05. If we were to use this fixed threshold in a real experiment, it is likely that we would identify many features as associated with age, when the results we are observing are simply due to chance.

Challenge 5

1. If we run 5000 tests, even if there are no true differences, how many of them (on average) will be statistically significant at a threshold of $p < 0.05$?
2. Why would we want to be conservative in labelling features as significantly different? By conservative, we mean to err towards labelling true differences as “not significant” rather than vice versa.
3. How could we account for a varying number of tests to ensure “significant” changes are truly different?

Solution

1. By default we expect $5000 \times 0.05 = 250$ features to be statistically significant under the null hypothesis, because p-values should always be uniformly distributed under the null hypothesis.
2. Features that we label as “significantly different” will often be reported in manuscripts. We may also spend time and money investigating them further, computationally or in the lab. Therefore, spurious results have a real cost for ourselves and for others.
3. One approach to controlling for the number of tests is to divide our significance threshold by the number of tests performed. This is termed “Bonferroni correction” and we’ll discuss this further now.

Adjusting for multiple tests

When performing many statistical tests to categorise features, we are effectively classifying features as “non-significant” or “significant”, that latter meaning those for which we reject the null hypothesis. We also generally hope that there is a subset of features for which the null hypothesis is truly false, as well as many for which the null truly does hold. We hope that for all features for which the null hypothesis is true, we accept it, and for all features for which the null hypothesis is not true, we reject it. As we showed in the example with permuted age, with a large number of tests it is inevitable that we will get some of these wrong.

We can think of these features as being “truly different” or “not truly different”¹. Using this idea, we can see that each categorisation we make falls into four categories:

If the null hypothesis was true for every feature, then as we perform more and more tests we’d tend to correctly categorise most results as negative. However, since p-values are uniformly distributed under the null, at a significance level of 5%, 5% of all results will be “significant” even though we would expect to see these results, given the null hypothesis is true, simply by chance. These would fall under the label “false positives” in the table above, and are also termed “false discoveries.”

There are two common ways of controlling these false discoveries. The first is to say, when we’re doing $n$ tests, that we want to have the same certainty of making one false discovery with $n$ tests as we have if we’re only doing one test. This is “Bonferroni” correction,² which divides the significance level by the number of tests performed, $n$. Equivalently, we can use the non-transformed p-value threshold but multiply our p-values by the number of tests. This is often very conservative, especially with a lot of features!

p_raw <- toptab_age$P.Value
p_fwer <- p.adjust(p_raw, method = "bonferroni")
plot(p_raw, p_fwer, pch = 16, log="xy")
abline(0:1, lty = "dashed")
abline(v = 0.05, lty = "dashed", col = "red")
abline(h = 0.05, lty = "dashed", col = "red")

Bonferroni correction often produces very large p-values, especially with low sample sizes.

You can see that the p-values are exactly one for the vast majority of tests we performed! This is not ideal sometimes, because unfortunately we usually don’t have very large sample sizes in health sciences.

The second main way of controlling for multiple tests is to control the false discovery rate (FDR).³ This is the proportion of false positives, or false discoveries, we’d expect to get each time if we repeated the experiment over and over.

1. Rank the p-values
2. Assign each a rank (1 is smallest)
3. Calculate the critical value \(q = \left(\frac{i}{m}\right)Q\), where $i$ is rank, $m$ is the number of tests, and $Q$ is the false discovery rate we want to target.⁴
4. Find the largest p-value less than the critical value. All smaller than this are significant.

Challenge 6

1. At a significance level of 0.05, with 100 tests performed, what is the Bonferroni significance threshold?
2. In a gene expression experiment, after FDR correction with an FDR-adjusted p-value threshold of 0.05, we observe 500 significant genes. What proportion of these genes are truly different?
3. Try running FDR correction on the p_raw vector. Hint: check help("p.adjust") to see what the method is called.
   Compare these values to the raw p-values and the Bonferroni p-values.

Solution

1. The Bonferroni threshold for this significance threshold is \(\frac{0.05}{100} = 0.0005\)

2. We can’t say what proportion of these genes are truly different. However, if we repeated this experiment and statistical test over and over, on average 5% of the results from each run would be false discoveries.

3. The following code runs FDR correction and compares it to non-corrected values and to Bonferroni:

   p_fdr <- p.adjust(p_raw, method = "BH")
   plot(p_raw, p_fdr, pch = 16, log="xy")
   abline(0:1, lty = "dashed")
   abline(v = 0.05, lty = "dashed", col = "red")
   abline(h = 0.05, lty = "dashed", col = "red")
   

   

   Benjamini-Hochberg correction is less conservative than Bonferroni

   plot(p_fwer, p_fdr, pch = 16, log="xy")
   abline(0:1, lty = "dashed")
   abline(v = 0.05, lty = "dashed", col = "red")
   abline(h = 0.05, lty = "dashed", col = "red")
   

   

   plot of chunk plot-fdr-fwer

Feature selection

In this episode, we have focussed on regression in a setting where there are more features than observations. This approach is relevant if we are interested in the association of each feature with some outcome or if we want to screen for features that have a strong association with an outcome. If, however, we are interested in predicting an outcome or if we want to know which features explain the variation in the outcome, we may want to restrict ourselves to a subset of relevant features. One way of doing this is called regularisation, and this is the topic of the next episode. An alternative is called feature selection. This is covered in the subsequent (optional) episode.

Further reading

• limma tutorial by Kasper D. Hansen
• limma user manual.
• The VariancePartition package has similar functionality to limma but allows the inclusion of random effects.

Footnotes

Key Points

• Performing linear regression in a high-dimensional setting requires us to perform hypothesis testing in a way that low-dimensional regression may not.

• Sharing information between features can increase power and reduce false positives.

• When running a lot of null hypothesis tests for high-dimensional data, multiple testing correction allows retain power and avoid making costly false discoveries.

• Multiple testing methods can be more conservative or more liberal, depending on our goals.