We consider a generic situation that, for each gene i, I = 1,2,..., N, we have (relative) expression levels X_1i,..., X_mi from m microarrays under condition 1, and Y_1i,..., Y_mi from m arrays under condition 2. We need to assume that m is an even integer. A general statistical model is assumed for gene expression data:

X_ji = _(1),i + _ji,    Y_li = _(2),i + e_li,

where _(1),i and _(2),i are the mean expression levels for gene i under the two conditions respectively, and _ji and e_li are independent random errors with means and variances

for any j = 1,..., m, l = 1,..., m and i = 1,..., N. It is assumed that random errors _ji/_(1),i and are randomly taken respectively from one of two (not necessarily equal) distributions that are symmetric about their mean 0. Note that the above assumption on the distributions of random errors, not on that of gene expression levels (that is, X_ji and Y_li), is often reasonable, and similar assumptions are common in other statistical applications. In addition, we do not assume that the expression levels of all the genes have an equal variance, because some previous studies [3,4,6] have found that the variance ²_(c),i (for c = 1,2) of gene-expression levels may depend on the mean expression _(c),i. Also, we do not even need to assume that ²_(1),i = ²_(2),i unless _(1),i = _(2),i.

A goal is to detect all genes with _(1),i ≠ _(2),i. This can be accomplished through statistical hypothesis testing.

To test the null hypothesis H₀: _(1),i = _(2),i, we use a t-type test statistic or score

Note that the mean and variance of Z_i are

whereas the mean E(Z_i) = 0 under H₀. Hence, it can be seen that a large absolute value of Z_i, |Z_i|, gives evidence against H₀. As the number of arrays (that is, m) increases, the variance of the test statistic Z_i decreases. Hence, it is possible to reject H₀ (that is, detect differential expression for gene i) with any E(Z_i) ≠ 0 if m is large enough. In other words, if the Type I error rate and other parameters are fixed, then the statistical power of the test will increase as m increases. This is the key point that motivates the discussion on sample size calculations.

To determine the cut-off point for |Z_i| to reject H₀, we need to know or estimate the distribution of Z_i under H₀, the null distribution f₀. In a parametric approach, based on some full distributional assumptions for X_ji and Y_ji, one may derive the null distribution f₀, such as in a two-sample t-test. However, the validity of such a parametric method critically depends on the correctness of assumed distributions, which of course is not guaranteed. Here, we consider a nonparametric approach: a finite normal mixture model is used to estimate f₀ nonparametrically.

There may be various ways to estimate the null distribution f₀. For instance, using expression levels of some housekeeping genes that are known to have non-differential expression, one can construct their Z_i scores and then estimate f₀ using the obtained Z_i scores. In practice, however, there may be only a small number of or no housekeeping genes in a given experiment. Here, following the basic idea in a class of nonparametric methods [5,9,10], we construct a null score z_i for each gene and then use these null scores to estimate f₀ nonparametrically. The null score is constructed from the same observed gene expression data as used in Z_i:

Under the assumption that _ji and e_ji have symmetric distributions, then _ji and -_ji have the same distribution, and e_ji and -e_ji have the same distribution. Thus, by comparing the form of z_i with that of Z_i, we know that the distribution of z_i is exactly f₀, the null distribution for Z_i (under H₀). Note that under H₀, _(1),i = _(2),i, and hence _(1),i = _(2),I (since we assume that _(c),i only depends on _(c),i), then

Thus z_i and Z_i have the same distribution f₀ under H₀. We use all z_i values across all genes to estimate f₀.

In practice, _(c),i (for c = 1, 2) are unknown, and can be estimated using the sample standard deviations (SDs) s_(c),i. Although the sample SD s_(c),i is asymptotically unbiased, if m and n are small, s_(c),i may not be stable, and some modifications may be necessary. In any case, substituting _(c),i by any suitable estimates, we can calculate the scores z_i values and Z_i values, on the basis of which we can estimate f₀ and f respectively. By comparing f₀ and f, we can gain insight about genes with altered expression (that is, _(1),i ≠ _(2),i).

We assume that all the z_i values for i = 1,..., N are a random sample from f₀; thus we can use the observed z_i values to estimate f₀. Pan et al. [10] proposed estimating f₀ using a finite normal mixture model [19]. Specifically, it is assumed that

where (z; a_r, V_r) denotes the density function of a normal distribution N(a_r, V_r) with mean a_r and variance V_r, and _r values are mixing proportions. Ω_g0 represents all unknown parameters {_r, a_r, V_r) : r = 1,...g₀} in a g₀-component mixture model. Among others, a normal mixture is essentially nonparametric and flexible, and easy to use with stable tail probabilities.

A mixture model can be fitted by maximum likelihood using the expectation-maximization (EM) algorithm [19,20,21]. The number of components can be selected adaptively using the Akaike Information Criterion (AIC) [22] or the Bayesian Information Criterion (BIC) [23]. In using the AIC or BIC, one first fits a series of models with various values of g₀, then picks up the g₀ corresponding to the first local minimum of AIC or BIC [24]. Some empirical studies seem to favor the use of BIC [24].

Once we obtain an estimate of the null distribution f₀, we can determine the cut-off point of the rejection region for testing H₀. In general, as for a two-sample test, the rejection region can be selected in the tails of f₀ because, under the null hypothesis, Z_i should be close to the center of f₀, whereas if there is differential expression for gene i, Z_i is likely to be in one of the two tails of f₀. The specific choice may depend on the goal of the analysis. For example, if we are only interested in detecting upregulated genes, we can choose the rejection region at the right-tail of f₀. Our proposed method works for any specified way of determining the rejection region. As f₀ should be symmetric about its mean 0, and often we are interested in both up- and downregulated genes, we propose to take the rejection region at the two tails of f₀, {z : f₀(z) <C}, where the constant C > 0 is the cut-off point and depends on the specified (gene-specific) Type I error rate . As usual, C > 0 is chosen such that the rejection rate under H₀ is exactly :

where (.; a, V) is the corresponding cumulative distribution function for (.; a, V). Using a numerical algorithm, such as the bisection method [25], we can solve the above equation to obtain C for any given .

For microarray data, because we are testing H₀ for each gene, the multiple test problem arises and some control on it is necessary. Usually we can use Bonferroni's method. For instance, if we want to maintain the genome-wide Type I error rate at the usual 5% level, then the Bonferroni-adjusted gene-specific (that is, test-specific) Type I error rate is = 0.05/N, where N is the total number of genes to be tested.

Once C is determined, we can calculate the power as a function of d, the magnitude of the expression change targeted to be detected. Note that

is the difference of the coefficients of variation under the two conditions. If _(1),i = _(2),i, d can be interpreted as the change of the mean expression levels from condition 1 to condition 2. Otherwise, it can be regarded as the difference of (variation) standardized mean expression levels. Specifically, we have the power function

Unsurprisingly, we can see that (d, ) will increase as |d| increases. The effects of having more replicates will reduce the variability of f₀, leading to larger (d, ) for any given d.

Now we describe how to calculate replicate numbers based on some pilot data taken from earlier studies. We use z_m,i to explicitly denote the z_i scores in (2) with m replicates. Based on the data we can estimate the density function f_0,m (z;Ω_g0) of z_m,ivalues as a normal mixture

From now on, we treat f_0,m as known in Equation (5).

With estimated f_0,m, we want to estimate the density function f_0,mk for z_mk,i, the z_i scores based on mk replicates (with k > 1). If we can have an estimate of f_0,mk, then we can obtain the corresponding power function (d, ) for mk replicates in the same way as described earlier for m replicates. Of course, we assume that our pilot data are drawn from only m arrays under each of the two experimental conditions, and thus we do not observe any z_mk,i based on mk arrays. However, we show next that it is possible to generate z_mk,i values from z_m,i values. Note that we can draw random realizations of z_m,i from the estimated f_0,m (see Pan et al. [10] or the example below). Suppose z_m,i^(j) values (for j = 1,2,..., k) are k independent realizations of z_m,i, then it is easy to show that

have the distribution f_0,mk. Thus, the density function for z_mk,i values is

For example, if we triple the number of replicates, the resulting density function is

The number of components of f_0,mk may be too large. For example, if the number of components is g₀ = 3 for m = n = 2, the corresponding numbers of components for m = n = 4, m = n = 6 and m = n = 8 are, respectively, g₀² = 9, g₀³ = 27 and g₀⁴ = 81. In fact, some of these components maybe very similar or have a negligible role, hence the form of f_0,mk, may be simplified. In the extreme situation, as mk → ∞, by the Central Limit Theorem, the mixture model will reduce to a single-component normal distribution. Hence, we propose a simulation-based method to select a more parsimonious model for f_0,mk.

On the basis of the mixture model f_0,m in Equation (5), we can generate a random sample of z_m,i^(j) values [10], from which we can calculate z_mk,i values using Equation (6). Using z_mk,i values we can fit a normal mixture model for f_0,mk. As we shall show later, we find such a fitted mixture model often contains a smaller number of components than g^k₀, as dictated in Equation (7), leading to a simplified form of f_0,mk.

In summary, our proposed method of calculating the required replicate number works in the following steps.

Step 1. Suppose that we have pilot gene expression data X_ji and Y_ij from m arrays under each condition. Use formula (2) to calculate the scores z_i,m.

Step 2. Use z_i,m and the normal mixture model (5) to estimate f_0,m.

Step 3. For a specified Type I error rate , determine the cutoff point C for the rejection region using formula (3), in which f₀ is replaced with the estimated f_0,m.

Step 4. For any specified d, calculate the power function (d, ) using formula (4), in which f₀ is replaced with the estimated f_0,m.

Step 5. For any given k > 1, use formula (7) or (6) to estimate f_0,mk.

Step 6. For a specified Type I error rate , determine the cutoff point C for the rejection region using formula (3), in which f₀ is replaced with the estimated f_0,mk.

Step 7. For any specified d, calculate the power function (d, ) using formulae (4), in which f₀ is replaced with the estimated f_0,mk.

Step 8. Repeat Steps 5 to 7 until all k > 1 of interest have been tried.

After the power functions for many possible mk replicates have been obtained, we can determine an appropriate number of replicates by considering all the factors involved, the desired power and Type I error rate, the targeted expression changes and other experimental constraints.

To understand the pathogenesis of otitis media, a study was conducted to identify genes involved in response to pneumococcal middle-ear infection and to study their roles in otitis media. Radioactively labeled DNA microarrays were applied to the mRNA analysis of 1,176 genes in middle-ear mucosa of rats with and without subacute pneumococcal middle-ear infection [26]. The data are available for the control group and for the pneumococcal middle-ear infection group. A more detailed description of how the data were collected and their public availability was provided in Pan et al. [26]. For the purpose of sample size calculations and to mimic many practical situations with only a small number of replicates, we only use m = n = 2 arrays from each group. We first take a natural logarithm transformation for all the observed gene-expression levels (that is, radioactive intensities) so that the resulting distributions are less skewed (which will reduce the number of components of a fitted mixture model). Then, for each microarray, we standardize the transformed gene-expression levels by subtracting their median.

Because of the small m = 2, the sample SDs may not be stable. One way is to add a small constant as suggested by Efron et al. [5]. Here we follow the idea of Lin et al. [27] and use a loess smoother [28] to nonparametrically model the sample SDs in terms of the mean expression levels (Figure 1). Then we plug in the smoothed SD to calculate z_2,i. Note that an alternative use of SD or its modification in calculating z_2,i values will not change the basic idea and the following steps in sample size calculations.

We fitted three mixture models for f_0,2 with g₀ ranging from 1 to 3. Table 1 summarizes the model-fitting results. g₀ = 1 was selected as both AIC and BIC achieve their minima there. So the fitted f₀ is a normal distribution, N(-0.0013, 0.1278). However, for the purposes of general illustration, we choose g₀ = 2 as the fitted model:

f_0,2(z) = 0.76 (z;-0.0415, 1.3117) + 0.24 (z;0.0700, 2.6970).

Figure 2a presents the histogram of z_i values and the fitted f₀ with g₀ = 1 and 2. There is not much difference between the two fitted f_0,2, both of which fit the data well. In particular, f_0,2 does not look like a t-distribution with small degrees of freedom, as predicted from the t-test.

A realization of z_2,i can be simulated in the following two steps. First, we draw a random number p_i from {1, 2} with probability 0.76 and 0.24 respectively. Second, if the drawn p_i = 1, z_i is randomly drawn from a normal distribution (z; -0.0415, 1.3117); otherwise, it is drawn from (z; 0.0700, 2.6970). From the generated z_2,i values, following expression (6) we generated three simulated data sets: z_2k,i values, I = 1,..., 1,176 for k = 2, 3 and 4. Then a normal mixture model was fitted to each data set. From Table 1, it can be seen that a single-component normal distribution was selected in each case. In Figure 2, each of the fitted normal distributions, N(-0.0494, 0.8226), N(-0.0644, 0.5383) and N(-0.0438, 0.4206), is compared with its theoretically derived mixture model in Equation (7); they are all very close. Here we see that using simulated data to fit a mixture model results in a much-simplified model. For example, for k = 4, it is a fitted single-component model versus a 2⁴ = 16-component model in Equation (7). Note that, as predicted, all the means of the fitted models are all essentially 0, and their variances decrease as k increases.

If we want to have only one expected false-positive result from testing each of 1,176 non-differentially expressed genes, the gene-specific (or test-specific) Type I error rate is = 1/1176 = 0.09%. Using formula (3) and fitted-mixture model f_0,2k, the cut-off points C are determined. Then the power functions (d, ) are drawn in Figure 3, which may help make a decision on the required number of replicates. For instance, if we want to detect an expression change d = 3 with probability at least 80% and with = 0.09%, then six replicates are needed. Also, with just two replicates, the power to detect a change as high as 4 is very low, smaller than 30%. Note that the choice of d may depend on some prior knowledge. For instance, based on the pilot data, we can estimate the d values for some selected genes (with the sample means and sample SDs substituting the true means and SDs in the formula for d), from which one can determine a range of d values of interest.

Figures 4,5,6 give the results for testing N = 1,000, 5,000 and 10,000 genes, respectively, while controlling the genome-wide Type I error rate at the usual 5% level. It can be seen that as N increases, we also need a larger number of arrays to maintain the power of the statistical test when other parameters are fixed. For instance, for N = 10,000 (Figure 6), even eight replicates cannot detect a change as large as d = 3 with 80% power, but six replicates can detect a change d = 4 with 80% power.