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Let $$\hat{\varphi_n}(t)=\frac{1}{n}\sum_{j=1}^n{exp(i{t}Y_j)}\quad(t\in\mathbb{R})$$ denote the empirical characteristic function of the residuals $Y_j\,=\,S_n^{-\frac{1}{2}}(X_j-\bar{X}_n),\quad j=1,\dots,n,$ where $\bar{X}_n=\frac{1}{n}\sum_{j=1}^n{X_j}$ ist the sample mean and $S_n^{-\frac{1}{2}}$ is (almost sure) the positiv square root of the inverse of the sample variance $ S_n=\frac{1}{n}\sum_{j=1}^n{(X_j-\bar{X}_n)^2}$.

I have to solve the integral of the test statistic $$ T_n= n \int_{\mathbb{R}}{\left|\hat{\varphi_n}(t)-exp\left(-\frac{1}{2}t^2\right)\right|^2}\psi(t)dt, $$ where $\psi(t)=(2\pi)^{-1/2}exp(-\frac{1}{2}t^2)$.

I know carrying out the integration leads us to

$$ T_n=\frac{1}{n}\sum_{j,k=1}^n{exp(-\frac{1}{2}R_{jk})}-2^{1-\frac{1}{2}}\sum_{j=1}^n{exp(-\frac{1}{4}R_{j}^2)}+n3^{-\frac{1}{2}}, $$ where $\quad R_{jk}=(X_j-X_k)^2S_n^{-1}=\left|Y_j-Y_k\right|^2\quad$ and $\quad R_{j}^2=(X_j-\bar{X}_n)^2S_n^{-1}=\left|Y_j\right|^2$.

I dont have any idea how to proceed. Can someone help me?

The test statistic can be found in "A Consistent Test for Multivariate Normality Based on the Empirical Characteristic Function" by L. Baringhaus and N. Henze.

Bo25
  • 111

2 Answers2

1

Here is a start

$$ T_n= n \int_{\mathbb{R}}{\left|\hat{\varphi_n}(t)-e^\left(-\frac{1}{2}t^2\right)\right|^2}\psi(t)dt $$

$$= n \int_{\mathbb{R}}{\left(\hat{\varphi_n}(t)-e^\left(-\frac{1}{2}t^2\right)\right)\overline{ \left(\hat{\varphi_n}(t)-e^\left(-\frac{1}{2}t^2\right)\right) } }\psi(t)dt $$

$$ = n \int_{\mathbb{R}}{\left(\hat{\varphi_n}(t)-e^\left(-\frac{1}{2}t^2\right)\right){ \left(\overline{\hat{\varphi_n}(t)}-e^\left(-\frac{1}{2}t^2\right)\right) } }\psi(t)dt. $$

We used the the complex conjugate in the above manipulations. You should be able to advance now. See related techniques.

Community
  • 1
Mhenni Benghorbal
  • 48,099
0

I got

$$ T_n= n \int_{\mathbb{R}}{\left(\hat{\varphi_n}(t)-e^\left(-\frac{1}{2}t^2\right)\right){ \left(\overline{\hat{\varphi_n}(t)}-e^\left(-\frac{1}{2}t^2\right)\right) } }\psi(t)dt $$ $$ = n \int_{\mathbb{R}}{\left[\left(\hat{\varphi_n}(t)\overline{\hat{\varphi_n}(t)}\right)-\left(\hat{\varphi_n}(t)e^\left(-\frac{1}{2}t^2\right)\right)-\left(\overline{\hat{\varphi_n}(t)}e^\left(-\frac{1}{2}t^2\right)\right)+\left(e^{-t^2}\right)\right] }\psi(t)dt $$ $$ =n\int_{\mathbb{R}}{\left[\left(\frac{1}{n^2}\sum_{j,k=1}^n{e^{it(Y_j-Y_k)}}\right)-\left(\frac{1}{n}\sum_{j=1}^n{e^{it(Y_j)}}e^{-\frac{1}{2}t^2}\right)-\left(\frac{1}{n}\sum_{j=1}^n{e^{-it(Y_j)}}e^{-\frac{1}{2}t^2}\right)+\left(e^{-t^2}\right)\right]\psi(t)dt} $$ $$ =\frac{1}{n}\sum_{j,k=1}^n{\int_{\mathbb{R}}{e^{it(Y_j-Y_k)}}}\psi(t)dt-\sum_{j=1}^n{\int_{\mathbb{R}}{e^{it(Y_j)}}e^{-\frac{1}{2}t^2}\psi(t)dt}-\sum_{j=1}^n{\int_{\mathbb{R}}{e^{-it(Y_j)}}e^{-\frac{1}{2}t^2}\psi(t)dt}+n\int_{\mathbb{R}}{e^{-t^2}\psi(t)dt} $$ And know I have to solve the integrals. Is that correct?

Bo25
  • 111
  • Yes it is correct. Carry on. – Mhenni Benghorbal Aug 02 '14 at 09:05
  • After carrying out the integration i got $$ =\frac{1}{n}\sum_{j,k=1}^n{-\frac{ie^{-\frac{1}{2}t(t-2iY_j+2iY_k)}}{\sqrt{2\pi}t}}+c_1+\sum_{j=1}^n{\frac{ie^{-t(t-iY_j)}}{\sqrt{2\pi}t}}+c_2-\sum_{j=1}^n{\frac{ie^{-t(t+iY_j)}}{\sqrt{2\pi}t}}+c_3+n\frac{erf\left(\sqrt{\frac{3}{2}t}\right)}{2\sqrt{3}}+c_4 $$ where erf(x) ist the error function. But I still do not know yet, how do I reach the form $$ T_n=\frac{1}{n}\sum_{j,k=1}^n{exp(-\frac{1}{2}R_{jk})}-2^{1-\frac{1}{2}}\sum_{j=1}^n{exp(-\frac{1}{4}R_{j}^2)}+n3^{-\frac{1}{2}} $$ – Bo25 Aug 04 '14 at 11:21