Tag Info

New answers tagged

0

Your elementary divisors are primary, so powers of irreducible polynomials. Clearly the irreducible (over $\Bbb Q$) polynomials $x$, $x^2-2$ and $x^2+4$ are the only ones relevant here, and they can be considered independently. For each the power in $p_A$ gives the product of the elementary divisors for this irreducible, and the power in $m_A$ gives the ...


0

As commenters said, the answer is negative: e.g., $f(x) =|x|^{-1} e^{-|x|}$ is a nonintegrable function such that $|f(x)\sin x|$ is globally integrable. More generally, the following are equivalent: $f$ is integrable whenever $fg$ is; $\operatorname{ess\,inf}|g|>0$ Indeed, if 2) fails then the sets $A_n = \{x: 2^{-n}\le |g(x)|<2^{1-n}\}$ have ...


0

$\frac{\sin tu-\sin t(u-x)}{t}$ has a removable singularity at $t=0$, that is $ \lim_{t\to 0} \frac{\sin tu-\sin t(u-x)}{t} $ exists and is finite. So no principle value integral is required. It is a double integral of a bounded function over two finite measure spaces, and so Fubini applies.


1

If $X$ is a Poisson type random variable i.e. for $k\in \mathbb{N}_0$ and $\lambda>0$ $$P\{X=x_0+k\cdot h\}=\frac{\lambda^k e^{-\lambda}}{k!}$$ its characteristic function is given by $$\varphi_X(t)=\sum_{k=0}^{\infty}e^{it(x_0+kh)}\frac{\lambda^k e^{-\lambda}}{k!}=e^{itx_0}\sum_{k=0}^{\infty}e^{it(kh)}\frac{\lambda^k ...


1

For complex numbers, by definition $\log z = \log |z| + i \arg z$. Consequently, since $I_n(t) \to \log f(t) = \log |f(t)| + i \arg f(t)$, where only the first term is real, $\Re I_n(t) \to \log |f(t)|$.


1

The default topology on $\{0,1\}$ is the subspace topology induced from $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$, i.e. the discrete topology. If a different topology is considered, that is (usually, at least) explicitly mentioned. Typically, characteristic functions are considered in contexts where other real- or complex-valued functions occur too, ...


1

Once you've shown that $n^{-1} S_n$ has the same distribution as $X_1$, then it's pretty much immediate that $n^{-\gamma} S_n$ does not convergence in distribution for $\gamma < 1$. If $\gamma > 1$ the characteristic function argument shows that $n^{- \gamma} S_n \stackrel{\text{d}}{\to} 0$. So $n^{-\gamma} S_n$ converges in distribution for $\gamma ...


0

For each $u$ consider the random variable $X_u: \Omega_u\to \Bbb{R} $ here $\Omega_u= \Bbb{R}$ but we consider it endowed with the probability $\mu_u$ associated with the characteristic function $f(u,\cdot)$. The probability space is $(\Bbb{R}, \mathbb{B}, \mu_u)$ Extend this to $(\mathbb{R}^{(-\infty,\infty)}, \mathcal{B}(\mathbb{R})^{\otimes ...


0

Partial answer: According to a theorem (Gil-Pelaez), if $0$ is a point of continuity of $F_X$, then $$ F_X(0) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt, $$ so, since a random variable $X$ is a.s. non-negative, then $F_X(0) = 0$, $\int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt = \pi / 2$ is the condition for a.s. ...


0

You're saying the pair $(X,Y)$ has the same distribution as the pair $(\bar X,\bar Y)$ and $\bar X,\bar Y$ are independpendent and you want to prove $X,Y$ are independent. \begin{align} & \Pr(X\in A\ \&\ Y\in B) \\[10pt] = {} & \Pr((X,Y)\in A\times B) \\[10pt] = {} & \Pr((\bar X,\bar Y)\in A\times B) & & \text{(since the joint ...


0

Denote by $\mathbb{P}_X$ the distribution of a random variable $X$ and by "$\stackrel{d}{=}$" equality in distribution. Since $\tilde{X}$ and $\tilde{Y}$ are independent, the distribution of $\tilde{Z}$ equals $$\mathbb{P}_{\bar{Z}} = \mathbb{P}_{\tilde{X}} \otimes \mathbb{P}_{\tilde{Y}}.$$ Moreover, $\tilde{X} \stackrel{d}{=}X$ and $\tilde{Y} ...



Top 50 recent answers are included