A stationary bootstrap test about two mean vectors comparison with somewhat dense differences and fewer sample size than dimension

Abstract

Two sample mean vectors comparison hypothesis testing problems often emerge in modern biostatistics. Many tests are proposed for detecting relatively dense signals with somewhat dense nonzero components in mean vectors differences. One kind of these tests is based on some quadratic forms about two sample mean vectors differences. Another kind of these tests is based on some quadratic forms about studentized version of two sample mean vectors differences. In this article, we propose a bootstrap test by adopting stationary bootstrap scheme to calculate p value of a typical test which is based on a quadratic form about studentized version of two sample mean vectors differences. Extensive simulations are conducted to compare performances of the bootstrap test with other existing typical tests. We also apply the bootstrap test to a real genetic data analysis about breast cancer.

Appendix

1.1 Proof of Theorem 1

Proof

For any p-dimension real vector $\mathbf{l }=(l_1,l_2,\ldots ,l_p)'_{p\times 1}$ with $\vert \vert \mathbf{l }\vert \vert ^2_2=\sum \nolimits _{s=1}^p l_s^2<M_p$, Denote $f_X(t)$ character function of one-dimension random variable $\mathbf{l }' \mathbf{X } =\sum \nolimits ^p_{q=1}l_qX_q$, and $f_Y(t)$ character function of one-dimension random variable $\mathbf{l }' \mathbf{Y } =\sum \nolimits ^p_{q=1}l_qY_q$. For any positive integer p, there exists a positive real number $\alpha$, conditions $\frac{\mathbf{l }'\Sigma _1\mathbf{l }}{p^\alpha }<C$ and $\frac{\mathbf{l }'\Sigma _2\mathbf{l }}{p^\alpha }<C$ are satisfied, where C is a finite positive constant for any p. Then ,we can get

$$\begin{aligned} f_X(t) = 1 + \frac{\mathbf{l }' \mu _1}{p^{\alpha /2}}(it) - \frac{\mathbf{l }'\Sigma _1\mathbf{l }}{2p^{\alpha }}t^2 + o(t^2), \end{aligned}$$

and

$$\begin{aligned} f_Y(t) = 1 + \frac{\mathbf{l }' \mu _2}{p^{\alpha /2}}(it) - \frac{\mathbf{l }'\Sigma _2\mathbf{l }}{2p^{\alpha }}t^2 + o(t^2). \end{aligned}$$

Characteristic function of $\left( \frac{(m+n)^{1/2}({{\bar{X}}}_1 -\bar{Y}_1)}{p^{\alpha /2}}+\cdots +\frac{(m+n)^{1/2}({{\bar{X}}}_p -\bar{Y}_p)}{p^{\alpha /2}}\right)$ is

$$\begin{aligned} g_1(t;l_1,l_2,\ldots ,l_p)= & {} \left[ f_X\left( \frac{(m+n)^{1/2}}{m}t\right) \right] ^m \left[ f_Y\left( -\frac{(m+n)^{1/2}}{n}t\right) \right] ^n\\= & {} \left[ 1 + \frac{\mathbf{l }' \mu _1}{p^{\alpha /2}}\frac{(m+n)^{1/2}}{m}(it) - \frac{\mathbf{l }'\Sigma _1\mathbf{l }}{2p^{\alpha }}\frac{(m+n)}{m^2}t^2 + o(\frac{(m+n)}{m^2}t^2)\right] ^m\\&\left[ 1 - \frac{\mathbf{l }' \mu _2}{p^{\alpha /2}}\frac{(m+n)^{1/2}}{n}(it) - \frac{\mathbf{l }'\Sigma _2\mathbf{l }}{2p^{\alpha }}\frac{(m+n)}{n^2} t^2 + o\left( \frac{(m+n)}{n^2}t^2\right) \right] ^n . \end{aligned}$$

Under null hypothesis,as $m,n \rightarrow \infty$,we have

$$\begin{aligned}&\log [g_1(l_1,l_2,\ldots ,l_p)] \\&\quad = m \log \left[ 1 + \frac{\mathbf{l }' \mu _1}{p^{\alpha /2}}\frac{(m+n)^{1/2}}{m}(it) - \frac{\mathbf{l }'\Sigma _1\mathbf{l }}{2p^{\alpha }}\frac{(m+n)}{m^2}t^2 + o\left( \frac{(m+n)}{m^2}t^2\right) \right] \\&\qquad + n \log \left[ 1 - \frac{\mathbf{l }' \mu _2}{p^{\alpha /2}}\frac{(m+n)^{1/2}}{n}(it) - \frac{\mathbf{l }'\Sigma _2\mathbf{l }}{2p^{\alpha }} \frac{(m+n)}{n^2}t^2 + o\left( \frac{(m+n)}{n^2}t^2\right) \right] \\&\quad =- \frac{\mathbf{l }'\Sigma _1\mathbf{l }}{2p^{\alpha }}\frac{(m+n)}{m}t^2 - \frac{\mathbf{l }'\Sigma _2\mathbf{l }}{2p^{\alpha }} \frac{(m+n)}{n}t^2 + o\left( \frac{(m+n)}{m}t^2\right) + o\left( \frac{(m+n)}{n}t^2\right) \\&\quad =-\frac{\mathbf{l }'(\frac{\Sigma _1}{\theta }+\frac{\Sigma _2}{1-\theta })\mathbf{l }}{2p^{\alpha }}t^2 + o(t^2). \end{aligned}$$

Namely, $\log [g_1(t;l_1,l_2,\ldots ,l_p)] = -\frac{\mathbf{l }'(\frac{\Sigma _1}{\theta }+\frac{\Sigma _2}{1-\theta })\mathbf{l }}{2p^{\alpha }}t^2 + o(t^2),$ let $t=1$, we can get

$$\begin{aligned} \log [h_1(l_1,l_2,\ldots ,l_p)] = \log [g_1(1;l_1,l_2,\ldots ,l_p)] = -\frac{\mathbf{l }'(\frac{\Sigma _1}{\theta }+\frac{\Sigma _2}{1-\theta })\mathbf{l }}{2p^{\alpha }} + {o(1)}, \end{aligned}$$

as $m,n \rightarrow \infty$. This completes proof of Theorem 1. $\square$

1.2 Proof of $T_{m,n,p}= \sum \nolimits _{j=1}^{p}\left( {{\bar{X}}}_{j}-{{\bar{Y}}}_{j}\right) ^{2} -\left( \frac{1}{m}{\text {tr}} \widehat{\varvec{\Sigma }}_1+\frac{1}{n}{\text {tr}} \widehat{\varvec{\Sigma }}_2\right)$ under null hypothesis with $\varvec{\mu }_1=\varvec{\mu }_2$

Proof

We can obtain that,

$$\begin{aligned}&T_{m,n,p}-\left\| \varvec{\mu }_{1}-\varvec{\mu }_{2}\right\| ^{2}\\&\quad = \frac{\sum \nolimits _{(i \ne j)=1}^{m} ({\varvec{X}}^{(i)})^{T} {\varvec{X}}^{(j)}}{m\left( m-1\right) } +\frac{\sum \nolimits _{(i \ne j)=1}^{n} ({\varvec{Y}}^{(i)})^{T} {\varvec{Y}}^{(j)}}{n\left( n-1\right) }-2 \frac{\sum \nolimits _{i=1}^{n} \sum \nolimits _{j=1}^{m} ({\varvec{X}}^{(i)})^{T} {\varvec{Y}}^{(j)}}{n m}\\&\quad = \frac{\sum \nolimits _{i ,j=1}^{m} ({\varvec{X}}^{(i)})^{T} {\varvec{X}}^{(j)}-\sum \nolimits _{i =1}^{m} ({\varvec{X}}^{(i)})^{T} {\varvec{X}}^{(i)}}{m\left( m-1\right) } + \frac{\sum \nolimits _{i ,j=1}^{n} ({\varvec{Y}}^{(i)})^{T} {\varvec{Y}}^{(j)}-\sum \nolimits _{i =1}^{n} ({\varvec{Y}}^{(i)})^{T} {\varvec{Y}}^{(i)}}{n\left( n-1\right) }\\&\qquad -2 \frac{\sum \nolimits _{i=1}^{n} \sum \nolimits _{j=1}^{m} ({\varvec{X}}^{(i)})^{T} {\varvec{Y}}^{(j)}}{n m} \\&\quad = \frac{m^2 {\bar{{\varvec{X}}}} ^T {\bar{{\varvec{X}}}} -(m-1)tr {\widehat{\Sigma }}_1-m {\bar{{\varvec{X}}}}^T {\bar{{\varvec{X}}}}}{m\left( m-1\right) } + \frac{m^2 {\bar{{\varvec{Y}}}} ^T {\bar{{\varvec{Y}}}} -(m-1)tr {\widehat{\Sigma }}_1-m {\bar{{\varvec{Y}}}}^T {\bar{{\varvec{Y}}}}}{n\left( n-1\right) } \\&\qquad -2 \frac{mn {\bar{{\varvec{X}}}}^{{\varvec{T}}}{\bar{{\varvec{Y}}}}}{n m}\\&\quad = ({\bar{{\varvec{X}}}}-{\bar{{\varvec{Y}}}})^T({\bar{{\varvec{X}}}}-{\bar{{\varvec{Y}}}}) -\left( \frac{1}{m}{\text {tr}} \widehat{\varvec{\Sigma }}_1+\frac{1}{n}{\text {tr}} \widehat{\varvec{\Sigma }}_2\right) \\&\quad = \sum \limits _{j=1}^{p}\left( {{\bar{X}}}_{j}-\bar{Y}_{j}\right) ^{2} -\left( \frac{1}{m}{\text {tr}} \widehat{\varvec{\Sigma }}_1+\frac{1}{n}{\text {tr}} \widehat{\varvec{\Sigma }}_2\right) . \end{aligned}$$

Under null hypothesis with $\varvec{\mu }_1=\varvec{\mu }_2$, we can obtain the above conclusion. $\square$