Question about proving positive definiteness

Question

Good afternoon, I'm currently studying stochastic processes and would like to understand a step in a proof of the semi definiteness of the covariance function of the fractional Brownian Motion.

A lemma we use: For $\alpha \in (0,2]$, $t \mapsto \exp(-|t|^{\alpha})$ is a characteristic function of a probability measure on $\mathbb{R}$.

Claim: Let $K_{B^H}(t,s) := \frac{1}{2} (|t|^{2H}+|s|^{2H}+|t-s|^{2H})$. Then for $\lambda_1, \dots ,\lambda _d \in \mathbb{C}, t_1, \dots , t_d \in \mathbb{R}$ we get: $\sum _{i=1}^d\sum_{j=1}^d K(t_i, t_j) \lambda _i \bar{\lambda}_j \geq 0$

Main steps:

• W. l. o. g., let $\lambda_1, \dots ,\lambda _d \in \mathbb{R}$
• Define $t_0 := 0$, $\lambda _0 := -\sum_{i=1} ^d \lambda_i$
• (By some algebraic transformations) $\sum _{i=1}^d\sum_{j=1}^d (|t_i|^{2H}+|t_j|^{2H}+|t_i - t_j|^{2H})\lambda_i \lambda_j = \dots = -\sum _{i=0}^d\sum_{j=0}^d |t_i - t_j|^{2H} \lambda_i \lambda_j$ (keep attention to the zeroes as the start)
• Now, for each $\varepsilon$ we have (1): $\sum _{i=0}^d\sum_{j=0}^d \exp(-\varepsilon |t_i - t_j|^{2H})\lambda_i \lambda_j \overset{(2)}{=} \sum _{i=0}^d\sum_{j=0}^d [\exp(-\varepsilon |t_i - t_j|^{2H}) - 1]\lambda_i \lambda_j \overset{(3)}{=} -\varepsilon \sum _{i=0}^d\sum_{j=0}^d |t_i - t_j|^{2H}\lambda_i \lambda_j + o(\varepsilon)$
• From the last conclusion by the lemma should follow, that $K$ is positive definite.

My questions:

• Why can we assume w. l. o. g. $\lambda_1, \dots ,\lambda _d \in \mathbb{R}$ for $\lambda_1, \dots ,\lambda _d \in \mathbb{C}$?
• Why can we state (1)?
• Why do (2) and (3) hold?
• Why does the conclusion hold?

I'd be very grateful for an answer

Best regards

Answer 1

1. Having real numbers instead of complex doesn't play a role in the proof so I don't know why they make this assumption

2. (2) comes from $\sum_{i=0}^d \lambda_i =0$. It implies $\sum_{i=0}^d \sum_{j=0}^d \lambda_i \lambda_j = \left(\sum_{i=0}^d \lambda_i \right)\left(\sum_{j=0}^d \lambda_j \right)$

3. (3) comes from the limited development of $\exp$, for $\epsilon \rightarrow 0$

4.Use the lemma and say that $\phi(\epsilon^{1/2H}(t_i-t_j))=\exp(−\epsilon |t_i−t_j|^{2H})$ is the characteristic function of $X$ : $\phi(u)= E[e^{iuX}]$ Then (pay attention to the fact that the $i$ in $\exp$ is not the index of the sum): \begin{align} \sum _{i=0}^d\sum_{j=0}^d \exp(-\varepsilon |t_i - t_j|^{2H})\lambda_i \bar{\lambda_j} &= E[\sum _{i=0}^d\sum_{j=0}^d \exp(i\varepsilon^{1/2H}(t_i - t_j)X)\lambda_i \bar{\lambda_j} ]\\ &=E[\left(\sum_{i=0}^d \exp(i\varepsilon^{1/2H}t_iX)\lambda_i \right) \left(\sum_{j=0}^d \exp(-i\varepsilon^{1/2H}t_jX)\bar{\lambda_j} \right) ]\\ &=E[\left|\sum_{i=0}^d \exp(i\varepsilon^{1/2H}t_iX)\lambda_i \right|^2] \ge 0 \end{align}

I think you can try to conclude with this