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Consider a real random vector $\vec X =(X_1, X_2)$ with characteristic function $\phi(\vec t) \equiv \mathbb{E} \big[ e^{i \vec t \cdot \vec X} \big]$ (where $\vec{t}=(t_1, t_2) \in \mathbb{R}^2$) given by $$ \phi(\vec t) =\exp \bigg\{ - \int_{S^1} \mu(d \vec{s}) \, | \vec{t} \cdot \vec{s}| \Big( 1 + \frac{2i}{\pi} \, \text{sign} ( \vec{t} \cdot \vec{s}) \log | \vec{t} \cdot \vec{s}| \Big) \bigg\} \, , $$ where $S^1$ is the 1-sphere $S^1 = \{ \vec{s} \in \mathbb{R}^2 \mathrel: \vec{s}^2 = 1 \}$ and $\mu(d \vec{s})$ is a measure on $S^1$ ($\mu \geq 0$). I know that $\phi$ is a characteristic function of some probability distribution*. Therefore, according to Bochner's theorem, $\phi$ is a positive-definite function, i.e. it satisfies

$$ 0 \leq \sum_{k,l=1}^N a_k^* a_l \, \phi(\vec{t}_l - \vec{t}_k) \, , \qquad \forall a_1, ..., a_N \in \mathbb{C} \, , \, \, \forall \, \vec{t}_1, ..., \vec{t}_N \in \mathbb{R}^2 \, , \, \text{ and } \, \forall N \in \mathbb{N} \, . $$

My question is: how do I check this condition for the particular $\phi$ given above?


*In particular, $\phi$ is the characteristic function of a bivariate Cauchy distribution, as parametrized by Samorodnitsky and Taqqu - Stable non-Gaussian random processes, but this is not relevant to the question.

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  • $\begingroup$ I removed the positive-characteristic tag which is about something completely different $\endgroup$
    – Yemon Choi
    Commented May 4, 2021 at 21:37
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    $\begingroup$ I find this question very unclear, just as essentially every question "how does one prove X without using Y". Perhaps if you explain your motivation for finding an alternative argument, your question will attract more attention. $\endgroup$ Commented May 5, 2021 at 12:16

1 Answer 1

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$\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\s}{\vec s}\newcommand{\ttt}{\vec t}$For any real $c>0$, the function $\vpi_c$ given by the formula \begin{equation*} \vpi_c(t):=\exp\{-c|t|(1+\tfrac{2i}\pi\,\text{sign}(t)\ln|t|)\} \end{equation*} for real $t$ is the characteristic function (c.f.) of the (univariate) stable distribution with parameters $\alpha=1$, $\beta=1$, $c$, and $\mu=\frac2\pi\,c\ln c$, so that \begin{equation*} \vpi_c(t)=Ee^{itY_c} \end{equation*} for some real-valued random variable $Y_c$ and all real $t$.

So, for each real $c>0$, each $\s\in\R^2$, and all $\ttt\in\R^2$, \begin{equation*} f_{c,\s}(\ttt):=\vpi_c(\s\cdot\ttt)=Ee^{i(Y_c\s)\cdot\ttt}, \end{equation*} so that $f_{c,\s}$ is a c.f.; specifically, $f_{c,\s}$ is the c.f. of the random vector $Y_c\s$ in $\R^2$.

Approximating now the integral in the definition of the function $\phi$ in the OP by corresponding integral sums, we see that $\phi$ is the pointwise limit of functions of the form \begin{equation*} \prod_{j=1}^n f_{c_j,\s_j} \tag{1} \end{equation*} for some natural $n$, some real positive $c_j$'s, and some $\s_j$'s in $\R^2$. All functions of the form (1) are c.f.'s (of bivariate distributions), because the product of c.f.'s of distributions is a c.f. (namely, the c.f. of the convolution of the distributions). So, obviously all functions of the form (1) are positive definite. Also, obviously the pointwise limit of positive-definite functions is positive definite.

Thus, $\phi$ is indeed positive definite.

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  • $\begingroup$ OK but I am asking for a way to check that $\phi$ is positive-definite, not that it is a c.f. In the last sentence you use Bochner's theorem, so that showing it is a c.f. implies that it is positive-definite, but I don't want to use Bochner's theorem. I know $\phi$ is a c.f. and I want to check that it is positive-definite. $\endgroup$
    – MBolin
    Commented May 5, 2021 at 10:46
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    $\begingroup$ @MBolin : The nontrivial part of Bochner's theorem is that any continuous positive-definite function is a c.f. The converse part is trivial, and I do not need the nontrivial part to show that $\phi$ is positive definite. Do you want to avoid even trivialities in a proof you desire? How can such a proof be at all possible? $\endgroup$ Commented May 5, 2021 at 11:53
  • $\begingroup$ OK, then probably I haven't explained very well what I am looking for. Maybe I just accept your answer and post a new question. $\endgroup$
    – MBolin
    Commented May 5, 2021 at 12:34
  • $\begingroup$ @MBolin : As stated above, I believe that the fact that any c.f. is positive definite is trivial; anyway, it is much more trivial than the fact that $\varphi_c$ is a c.f. So, I interpreted your sentence "I know that $\phi$ is a characteristic function" as something like this: "It is stated in a source that I consider reliable that $\phi$ is a characteristic function [but I do not know how to prove that]". $\endgroup$ Commented May 5, 2021 at 14:08
  • $\begingroup$ OK, I understand. But can you prove that your $\varphi_c$ is positive-definite (maybe it's trivial and I don't see it)? I think I am not asking you to "prove X without using Y", as is said in a comment. I am just asking you to provide the full proof. If, in order to prove that $\varphi_c$ is positive-definite you first need to prove that it is a c.f., fine, I would accept it as a valid answer, but I wonder whether there is a simpler way. $\endgroup$
    – MBolin
    Commented May 7, 2021 at 17:37

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