import numpy as npimport pandas as pdimport matplotlib.pyplot as pltimport statsmodels.api as smfrom ISLP import load_data## Below are the new importsfrom scipy.stats import ( ttest_1samp, ttest_rel, ttest_ind, t as t_dbn)from statsmodels.stats.multicomp import pairwise_tukeyhsdfrom statsmodels.stats.multitest import multipletests as mult_test
1
one sample t test from the scipy.stats
2
two related samples t test functions from the scipy.stats
First we create 100 variables, each consisting of 10 observations. The first 50 variables have mean 0.5 and variance 1, while the others have mean 0 and variance 1.
To begin, we use ttest_1samp() from the scipy.stats module to test H_{0}: \mu_1=0, the null hypothesis that the first variable has mean zero.
result = ttest_1samp(X[:,0], 0)result.pvalue
0.9307442156164141
The p-value comes out to 0.931, which is not low enough to reject the null hypothesis at level \alpha=0.05. In this case, \mu_1=0.5, so the null hypothesis is false. Therefore, we have made a Type II error by failing to reject the null hypothesis when the null hypothesis is false.
We now test H_{0,j}: \mu_j=0 for j=1,\ldots,100. We compute the 100 p-values, and then construct a vector recording whether the jth p-value is less than or equal to 0.05, in which case we reject H_{0j}, or greater than 0.05, in which case we do not reject H_{0j}, for j=1,\ldots,100.
Therefore, at level \alpha=0.05, we reject 15 of the 50 false null hypotheses, and we incorrectly reject 5 of the true null hypotheses. Using the notation from Section~, we have V=5, S=15, U=45 and W=35. We have set \alpha=0.05, which means that we expect to reject around 5% of the true null hypotheses. This is in line with the 2 \times 2 table above, which indicates that we rejected V=5 of the 50 true null hypotheses.
In the simulation above, for the false null hypotheses, the ratio of the mean to the standard deviation was only 0.5/1 = 0.5. This amounts to quite a weak signal, and it resulted in a high number of Type II errors. Let’s instead simulate data with a stronger signal, so that the ratio of the mean to the standard deviation for the false null hypotheses equals 1. We make only 10 Type II errors.
Recall from that if the null hypothesis is true for each of m independent hypothesis tests, then the FWER is equal to 1-(1-\alpha)^m. We can use this expression to compute the FWER for m=1,\ldots, 500 and \alpha=0.05, 0.01, and 0.001. We plot the FWER for these values of \alpha in order to reproduce Figure~.
m = np.linspace(1, 501)fig, ax = plt.subplots()[ax.plot(m,1- (1- alpha)**m, label=r'$\alpha=%s$'%str(alpha))for alpha in [0.05, 0.01, 0.001]]ax.set_xscale('log')ax.set_xlabel('Number of Hypotheses')ax.set_ylabel('Family-Wise Error Rate')ax.legend()ax.axhline(0.05, c='k', ls='--');
As discussed previously, even for moderate values of m such as 50, the FWER exceeds 0.05 unless \alpha is set to a very low value, such as 0.001. Of course, the problem with setting \alpha to such a low value is that we are likely to make a number of Type II errors: in other words, our power is very low.
We now conduct a one-sample t-test for each of the first five managers in the Fund dataset, in order to test the null hypothesis that the jth fund manager’s mean return equals zero, H_{0,j}: \mu_j=0.
Fund = load_data('Fund')fund_mini = Fund.iloc[:,:5]fund_mini_pvals = np.empty(5)for i inrange(5): fund_mini_pvals[i] = ttest_1samp(fund_mini.iloc[:,i], 0).pvaluefund_mini_pvals
The p-values are low for Managers One and Three, and high for the other three managers. However, we cannot simply reject H_{0,1} and H_{0,3}, since this would fail to account for the multiple testing that we have performed. Instead, we will conduct Bonferroni’s method and Holm’s method to control the FWER.
To do this, we use the multipletests() function from the statsmodels module (abbreviated to mult_test()). Given the p-values, for methods like Holm and Bonferroni the function outputs adjusted p-values, which can be thought of as a new set of p-values that have been corrected for multiple testing. If the adjusted p-value for a given hypothesis is less than or equal to \alpha, then that hypothesis can be rejected while maintaining a FWER of no more than \alpha. In other words, for such methods, the adjusted p-values resulting from the multipletests() function can simply be compared to the desired FWER in order to determine whether or not to reject each hypothesis. We will later see that we can use the same function to control FDR as well.
The mult_test() function takes p-values and a method argument, as well as an optional alpha argument. It returns the decisions (reject below) as well as the adjusted p-values (bonf).
Is there evidence of a meaningful difference in performance between these two managers? We can check this by performing a paired t-test using the ttest_rel() function from scipy.stats:
The test results in a p-value of 0.038, suggesting a statistically significant difference.
However, we decided to perform this test only after examining the data and noting that Managers One and Two had the highest and lowest mean performances. In a sense, this means that we have implicitly performed {5 \choose 2} = 5(5-1)/2=10 hypothesis tests, rather than just one, as discussed in Section~. Hence, we use the pairwise_tukeyhsd() function from statsmodels.stats.multicomp to apply Tukey’s method in order to adjust for multiple testing. This function takes as input a fitted ANOVA regression model, which is essentially just a linear regression in which all of the predictors are qualitative. In this case, the response consists of the monthly excess returns achieved by each manager, and the predictor indicates the manager to which each return corresponds.
returns = np.hstack([fund_mini.iloc[:,i] for i inrange(5)])managers = np.hstack([[i+1]*50for i inrange(5)])tukey = pairwise_tukeyhsd(returns, managers)print(tukey.summary())
The pairwise_tukeyhsd() function provides confidence intervals for the difference between each pair of managers (lower and upper), as well as a p-value. All of these quantities have been adjusted for multiple testing. Notice that the p-value for the difference between Managers One and Two has increased from 0.038 to 0.186, so there is no longer clear evidence of a difference between the managers’ performances. We can plot the confidence intervals for the pairwise comparisons using the plot_simultaneous() method of tukey. Any pair of intervals that don’t overlap indicates a significant difference at the nominal level of 0.05. In this case, no differences are considered significant as reported in the table above.
Now we perform hypothesis tests for all 2,000 fund managers in the Fund dataset. We perform a one-sample t-test of H_{0,j}: \mu_j=0, which states that the jth fund manager’s mean return is zero.
fund_pvalues = np.empty(2000)for i, manager inenumerate(Fund.columns): fund_pvalues[i] = ttest_1samp(Fund[manager], 0).pvalue
There are far too many managers to consider trying to control the FWER. Instead, we focus on controlling the FDR: that is, the expected fraction of rejected null hypotheses that are actually false positives. The multipletests() function (abbreviated mult_test()) can be used to carry out the Benjamini–Hochberg procedure.
The q-values output by the Benjamini–Hochberg procedure can be interpreted as the smallest FDR threshold at which we would reject a particular null hypothesis. For instance, a q-value of 0.1 indicates that we can reject the corresponding null hypothesis at an FDR of 10% or greater, but that we cannot reject the null hypothesis at an FDR below 10%.
If we control the FDR at 10%, then for how many of the fund managers can we reject H_{0,j}: \mu_j=0?
(fund_qvalues <=0.1).sum()
146
We find that 146 of the 2,000 fund managers have a q-value below 0.1; therefore, we are able to conclude that 146 of the fund managers beat the market at an FDR of 10%. Only about 15 (10% of 146) of these fund managers are likely to be false discoveries.
By contrast, if we had instead used Bonferroni’s method to control the FWER at level \alpha=0.1, then we would have failed to reject any null hypotheses!
(fund_pvalues <=0.1/2000).sum()
0
Figure~ displays the ordered p-values, p_{(1)} \leq p_{(2)} \leq \cdots \leq p_{(2000)}, for the Fund dataset, as well as the threshold for rejection by the Benjamini–Hochberg procedure. Recall that the Benjamini–Hochberg procedure identifies the largest p-value such that p_{(j)}<qj/m, and rejects all hypotheses for which the p-value is less than or equal to p_{(j)}. In the code below, we implement the Benjamini–Hochberg procedure ourselves, in order to illustrate how it works. We first order the p-values. We then identify all p-values that satisfy p_{(j)}<qj/m (sorted_set_). Finally, selected_ is a boolean array indicating which p-values are less than or equal to the largest p-value in sorted_[sorted_set_]. Therefore, selected_ indexes the p-values rejected by the Benjamini–Hochberg procedure.
Here, we implement the re-sampling approach to hypothesis testing using the Khan dataset, which we investigated in Section~. First, we merge the training and testing data, which results in observations on 83 patients for 2,308 genes.
Khan = load_data('Khan') D = pd.concat([Khan['xtrain'], Khan['xtest']])D['Y'] = pd.concat([Khan['ytrain'], Khan['ytest']])D['Y'].value_counts()
Y
2 29
4 25
3 18
1 11
Name: count, dtype: int64
There are four classes of cancer. For each gene, we compare the mean expression in the second class (rhabdomyosarcoma) to the mean expression in the fourth class (Burkitt’s lymphoma). Performing a standard two-sample t-test
using ttest_ind() from scipy.stats on the 11th gene produces a test-statistic of -2.09 and an associated p-value of 0.0412, suggesting modest evidence of a difference in mean expression levels between the two cancer types.
However, this p-value relies on the assumption that under the null hypothesis of no difference between the two groups, the test statistic follows a t-distribution with 29+25-2=52 degrees of freedom. Instead of using this theoretical null distribution, we can randomly split the 54 patients into two groups of 29 and 25, and compute a new test statistic. Under the null hypothesis of no difference between the groups, this new test statistic should have the same distribution as our original one. Repeating this process 10,000 times allows us to approximate the null distribution of the test statistic. We compute the fraction of the time that our observed test statistic exceeds the test statistics obtained via re-sampling.
This fraction, 0.0398, is our re-sampling-based p-value. It is almost identical to the p-value of 0.0412 obtained using the theoretical null distribution. We can plot a histogram of the re-sampling-based test statistics in order to reproduce Figure~.
fig, ax = plt.subplots(figsize=(8,8))ax.hist(Tnull, bins=100, density=True, facecolor='y', label='Null')xval = np.linspace(-4.2, 4.2, 1001)ax.plot(xval, t_dbn.pdf(xval, D_.shape[0]-2), c='r')ax.axvline(observedT, c='b', label='Observed')ax.legend()ax.set_xlabel("Null Distribution of Test Statistic");
The re-sampling-based null distribution is almost identical to the theoretical null distribution, which is displayed in red.
Finally, we implement the plug-in re-sampling FDR approach outlined in Algorithm~. Depending on the speed of your computer, calculating the FDR for all 2,308 genes in the Khan dataset may take a while. Hence, we will illustrate the approach on a random subset of 100 genes. For each gene, we first compute the observed test statistic, and then produce 10,000 re-sampled test statistics. This may take a few minutes to run. If you are in a rush, then you could set B equal to a smaller value (e.g. B=500).
m, B =100, 10000idx = rng.choice(Khan['xtest'].columns, m, replace=False)T_vals = np.empty(m)Tnull_vals = np.empty((m, B))for j inrange(m): col = idx[j] T_vals[j] = ttest_ind(D2[col], D4[col], equal_var=True).statistic D_ = np.hstack([D2[col], D4[col]]) D_null = D_.copy()for b inrange(B): rng.shuffle(D_null) ttest_ = ttest_ind(D_null[:n_], D_null[n_:], equal_var=True) Tnull_vals[j,b] = ttest_.statistic
Next, we compute the number of rejected null hypotheses R, the estimated number of false positives \widehat{V}, and the estimated FDR, for a range of threshold values c in Algorithm~. The threshold values are chosen using the absolute values of the test statistics from the 100 genes.
cutoffs = np.sort(np.abs(T_vals))FDRs, Rs, Vs = np.empty((3, m))for j inrange(m): R = np.sum(np.abs(T_vals) >= cutoffs[j]) V = np.sum(np.abs(Tnull_vals) >= cutoffs[j]) / B Rs[j] = R Vs[j] = V FDRs[j] = V / R
Now, for any given FDR, we can find the genes that will be rejected. For example, with FDR controlled at 0.1, we reject 15 of the 100 null hypotheses. On average, we would expect about one or two of these genes (i.e. 10% of 15) to be false discoveries. At an FDR of 0.2, we can reject the null hypothesis for 28 genes, of which we expect around six to be false discoveries.
The variable idx stores which genes were included in our 100 randomly-selected genes. Let’s look at the genes whose estimated FDR is less than 0.1.
