Showing that a characteristic function is “positive semi-definite”

I found this exercise while preparing for an exam.

Let $X$ a real random variable, let $\varphi(t):=E[e^{itX}]$ its characteristic function. Show that $$\forall t_1,\cdots,t_n\in\mathbb{R},\,\forall z_1,\cdots,z_n\in\mathbb C,\, \sum_{k,h=1}^n\varphi(t_h-t_k)z_h\overline{z_k}\ge 0$$

I've spent some time on it, but aside from some small observations like

• if $X$ satisfies the theorem, then $\forall\lambda\in\mathbb R,\ \lambda X$ satisfies it as well

• the case $z_k=e^{i\theta_k},\ \theta_k\in[0,2\pi)$ is sufficient

I'm quite far from the crucial idea that might allow me to show that $\forall (t_k)_1^n$ the hermitian matrix $A_{h,k}:=E[e^{i(t_h-t_k)X}]$ is positive semi-definite.

Since it should be a rather easy exercise (and solving it is in my best interest), a hint will suffice. Thank you in advance.

• $$\sum_{k,h}\varphi(t_h-t_k)z_h\overline{z_k}=E\sum_{k,h}z_{h}e^{it_hX}\ \overline{z_{k}e^{it_kX}}=E\left(\left|\sum_{k}z_ke^{it_kX}\right|^2\right)$$ – Did Jun 19 '15 at 7:59
• I really need to go on vacation, if don't see that coming. Thank you, @Did – Gae. S. Jun 19 '15 at 8:09
• Not sure about the vacation diagnostic. From my experience one can forget this trick, then be remembered of it, then reforget it, and so on. In particular I was convinced at a time that $e^{it_hX}$ and $e^{it_kX}$ being not independent was a problem... You see? – Did Jun 19 '15 at 8:14
• The property is called positive semi-definite and not positive semi-defined. – saz Jun 19 '15 at 8:30