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## Hot answers tagged st.statistics

5

Let $A={\rm diag}[a_1,\dots,a_n]$ and $B={\rm diag}[b_1,\dots,b_n]$. Let $\Delta(a)$ be the Vandermonde product in the $a_j$, and similarly $\Delta(b)$ be the Vandermonde product in the $b_k$. Suppose ${\rm min} \, a_j \ge {\rm max} \, b_k$. Let ${\rm d}U$ denote the normalised Haar measure on the unitary group $U(n)$. Then by Eq. (3.21) in Gross and ...

3

If $\Sigma=I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ is the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\vec{\mu}^\top\vec{\mu}$. If $\Sigma\ne I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},... 2 We have $$EX_{r-j:r}=\frac{r!}{(r-j-1)!\,j!} \int_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j \\ =r\binom{r-1}j \int_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j$$ and $$(1-u)^j=\sum_{i=0}^j(-1)^i \binom ji u^i.$$ So, $$\lambda_r=\int_0^1 du\,x(u)\,p_r(u),$$ where ... 2 If we assume$A$constant $$\frac{d}{dt}\mathbb{E}(X_t )=A \mathbb{E}(X_t )$$so$\mathbb{E}(X_t)=e^{tA}X_0$and$\mathbb{E}Y_t = Y_0 + \int_0^t H_s e^{sA}X_0ds$. For the variance, we can assume$X_0=0$and$Y_0=0$. And we have $$\frac{d}{dt}\mathbb{E}(X_tX_t^T )=A\mathbb{E}(X_tX_t^T )+\mathbb{E}(X_tX_t^T )A^T+C_tC_t^T$$so$\$\mathbb{E}(X_tX_t^T )=\int_0^t e^...

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