# Integral of a sum of complex exponentials

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.

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.

• Thanks :) I continued your calculation. I put it in a new answer. Is it correct? – Bo25 Aug 1 '14 at 17:13

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?

• Yes it is correct. Carry on. – Mhenni Benghorbal Aug 2 '14 at 9:05
• 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 4 '14 at 11:21