Synia
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Let $Y := Z^T \Sigma Z$. We have $\operatorname{var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2$ and $\mathbb{E}(Y) = \sum_i \sigma_{i, i}$. We thus need to compute $\mathbb{E}(Y^2)$. For this, ...

The (2) is just the independence of the Brownian motion $(\beta_t)_t$ with the clock'' $(H_T)_T$. This is a classical result due to Lévy (and more generally, this is the Dambis-Dubins-Schwartz ...

$\text{Dear James,}$ you can write your function as a(n infinite) sum of homogeneous symmetric functions \begin{equation*}%$h_\ell(x_1, \dots, x_k) := \sum_{1 \leqslant i_1 \leqslant \cdots \... View answer 2 votes$\textit{Lemma :}We have \begin{align*}% \frac{1}{d!} \sum_{\lambda\in \text{Hooks}(d)} (-1)^{\ell(\lambda)-1} \, \dim \lambda \prod_{\Box \in \lambda}(x + c(\Box) )(y + c(\Box) ) = [c_d]\...

Using my previous remark (namely the correction of the operator $\mathcal{O}$), an expression would just involve the "skeleton" of your continuous time random walk, namely the walk with initial law $... View answer 2 votes The answer is yes. In general, if a random variable$X$has a Lebesgue density$ f_X : \mathbb{R}^d \to \mathbb{R}_+ $, a Stein operator is given by$ g \mapsto \frac{\mathrm{div}(g f_X) }{ f_X } $, ... View answer Accepted answer 2 votes This is only a partial answer, as the computations can become quite involved. We have$$\mathbb{E}( X_t J_t ) = \mathbb{E}\left( X_t Q \int_{ \mathbb{R}_+ \times [0, t] } d\mu \right) = \mu_Q \... View answer Accepted answer 1 votes Concerning the finite sum, I am not sure you can get it (maybe for particular values of$ \nu $). Here is a possible "closed" expression for your sum. Start by writing$ \lambda := \mu + \alpha \$ with ...