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## Hot answers tagged characteristic-functions

12

Theorem (Kac's theorem) Let $X,Y \in L^1$ $\mathbb{R}^d$-valued random variables. Then the following statements are equivalent. $X,Y$ are independent $\forall \eta,\xi \in \mathbb{R}^d: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot \mathbb{E}e^{\imath \, Y \cdot \eta}$ Proof: $(1) \Rightarrow ... 10 Denote by$\Phi(t) = \mathbb{E}e^{\imath \, t \cdot X}$the characteristic function of$X. We have $$X = \frac{1}{2} \big((X+Y)+(X-Y) \big).$$ Thus, \begin{align*} \Phi(t) &= \mathbb{E}e^{\imath \, \frac{t}{2} (X+Y)} \cdot \mathbb{E}e^{\imath \, \frac{t}{2} (X-Y)}= \left( \mathbb{E}e^{\imath \, \frac{t}{2} X} \right)^2 \cdot \mathbb{E}e^{\imath \, ... 7 If \phi is a characteristic function, then, for every real values of s and t, K(t,s)\geqslant0 where K(t,s) is the determinant K(t,s)=\det\begin{pmatrix}\phi(0) & \phi(t) & \phi(t+s) \\ \phi(-t) & \phi(0) & \phi(s) \\ \phi(-t-s) & \phi(-s) & \phi(0)\end{pmatrix}. $$Using \phi_\alpha(t)=\mathrm e^{-c|t|^\alpha} for ... 6$$T=ZX+(1-Z)Y\implies E(\mathrm e^{\mathrm itT})=pE(\mathrm e^{\mathrm itX})+(1-p)E(\mathrm e^{\mathrm itY})$$6 To prove these are characteristic functions, let us use random variables. This yields simpler, and more intuitive, proofs. In the first case, assume that \phi_1(t)=\mathrm E(\mathrm e^{itX_1}) and \phi_2(t)=\mathrm E(\mathrm e^{itX_2}) for some random variables X_1 and X_2 defined on the same probability space and introduce a Bernoulli random ... 5 I should choose for induction: E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]. By induction: E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]. Further we have: ... 5 Below, there is a discrete example because I misread the word continuous. However, if X has the density f given by$$f(x)=\begin{cases}0&\mbox{ if }|x|\leqslant 2,\\ \frac C{x^2\log |x|}&\mbox{ if }|x|>2,\end{cases}$$then X is not integrable and the characteristic function is given by ... 5 Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ... 5 It's probably better to use the cumulative distribution function, since it behaves better with respect to the maximum of an independent sequence. We have$$\mu\{n(1-M_n)\leqslant t\}=\mu\{1-t/n\leqslant M_n\}=1-(1-t/n)^n$$and we recognize a well-known limit. 5 If X is a random variable with values in \mathbb{N}, then$$ {\rm E}[f(X)]=\sum_{k=0}^\infty f(k)P(X=k) $$for any 'nice' function f. This is the law of the unconscious statistician for discrete random variables. 4$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$4 If X_n\to 0 in distribution, then any \varepsilon does the job. The converse is harder. Here it's the proof of Levy's continuity theorem which will be used. Denoting by \varphi_n the characteristic function of X_n and \mu_n its distribution, we indeed have the equality ... 4 Hint: The Cayley-Hamilton theorem implies that A^n is a linear combination of I,A,A^2,\ldots,A^{n-1}. 4 The characteristic function is an expectation:$$ \varphi_S(t) = \mathbb{E}\left(\exp(i S t)\right) = \mathbb{E}\left(\exp\left(i \left(U_1 + U_2 + \cdots + U_n \right) t\right)\right) $$Now, if U_i is independent, the expectation factors into product of expectations, because :$$ \varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot ... 4 The probability density function is non-negative. Hence,|f(x)|=f(x)$and $$\int |f(x)|\mathrm dx=\int f(x)\mathrm dx=1.$$ It follows that$f\in L^1$. 4 This is a consequence of Levy's Inversion Formula (aka Fourier Inversion Theorem). If$\varphi$is the CF of$X$and$\int_{\mathbb{R}}|\varphi(\theta)|d\theta < \infty$then$X$is absolutely continuous with density $$f(x)= \frac{1}{2\pi}\int_\mathbb{R}e^{-i\theta x}\varphi(\theta)d\theta.$$ (Here we are using the definition$\varphi(\theta) = ...

4

No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that $|\phi(t)|\leq 1$ and the inequality $1-\cos(2t)\leq 4(1-\cos(t))$ which allows you to conclude $1-\Re \phi(2t)\leq 4[1-\Re \phi(t)]$ which is essentially a statement that says the behavior of $\phi(t)$ ...

4

Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite. Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line. Factoid 3: The function $t\mapsto|\cos(t)|$ is ...

4

This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine: Define $\quad U = X+Y, \quad V=(X-Y)^2$ and characteristic functions $$\phi(t) = E e^{itX}\\ ... 4 Since S_n=\sum\limits_{k=1}^nX_k converges almost surely to S=\sum\limits_{k=1}^\infty X_k, \varphi_{S_n}\to\varphi_S pointwise by dominated convergence. By independence, \varphi_{S_n}=\prod\limits_{k=1}^n\varphi_k. Thus, \prod\limits_{k=1}^n\varphi_k converges pointwise when n\to\infty and its limit is both \prod\limits_{k=1}^\infty\varphi_k ... 4 The letters a,b and c are just letters. I think your problem is with the quadratic formula itself. Think like this:$$\color{red}{\rm stuff}\,x^2 + \color{blue}{\rm stuff}\,x + \color{green}{\rm stuff} = 0 \implies x = \frac{-\color{blue}{\rm stuff} \pm \sqrt{(\color{blue}{\rm stuff})^2 - 4\,\color{red}{\rm stuff}\,\color{green}{\rm ...

3

Just to iron out the details in my comment, since others have also posted complete answers, $\phi(t)$ is infinitely differentiable at $0$; in fact, $\left.\frac{d^n \phi}{dt^n} \right|_{t = 0} = 0$ and hence if $\phi(t)$ is the cf of some random variable $X$ it must be that $E[X^n] = 0$ for all $n$. In particular, $\mbox{Var}(X) = 0$ and $E[X] = 0$ so that ...

3

By Bochner's theorem, a function $\phi : \mathbb{R} \to \mathbb{C}$ is the characteristic function of a probability measure if and only if $\phi$ is positive definite, $\phi(0) = 1$, and $\phi$ is continuous at the origin. Since these properties are conserved under convex combination, your second statement is true whenever $\alpha_i$ are non-negative. ...

3

In the approach with characteristic functions, we have to use the fact that $\varphi_{S_n}(t)=\prod_{j=1}^n\varphi_{X_j}(t)$. here is an other approach: we have, noting by $[n]$ the set $\{1,\dots,n\}$, $$\mathbb E\left(\sum_{i=1}^nX_i\right)^3=\sum_{(i,j,k)\in [n]^3}\mathbb E(X_iX_jX_k).$$ The set $[n]^3$ can be divided into the $(i,j,k)$ such that: ...

3

Let $A$ be independent of $X$ with $P(A=1)=P(A=0)=\frac{1}{2}$. Then $$E\left[e^{it\{AX+(1-A)(-X)\}}\right]=\frac{1}{2}E\left[e^{itX}\right]+\frac{1}{2}E\left[e^{it(-X)}\right]=\frac{\phi(t)+\phi(-t)}{2},$$ but using that $\cos$ is even and $\sin$ is odd, we obtain $$\phi(-t)=E\left[e^{i(-t)X}\right]=E[\cos(-tX)]+iE[\sin(-tX)]=E[\cos(tX)]-iE[\sin(tX)]$$ ...

3

Let $P$ denote the probability distribution of $X$ and $\mu$ the Lebesgue measure on $\mathbb{R}$. Then \begin{align*} \mathsf{E}[g(X)] &= \int_{\mathbb{R}} g(x) \, d{P(x)} \\ &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} G(t) e^{i t x} \, d{\mu(t)} \right] d{P(x)} \\ &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} G(t) e^{i t ...

3

Let me just state the theorem I linked to in my comment, so that this question does not go unanswered. Bochner's theorem If $\varphi:\mathbb{R}^d\to \mathbb C$ is a complex-valued function with $\varphi(0)=1$, continuous at $0$ and nonnegative-definite in the sense that for $n\geq 1$ we have that $$... 3 We have$$\begin{align} \frac{e^{-ita}-e^{-itb}}{it}e^{itx} &= \frac{e^{it(x-a)}-e^{it(x-b)}}{it}\\ &= \frac{\cos\left( t(x-a)\right) - \cos \left(t(x-b)\right)}{it} + \frac{\sin \left(t(x-a)\right) - \sin \left(t(x-b)\right)}{t} \end{align} by the addition theorem for the exponential function and Euler's formula $e^{iz} = \cos z + i\sin z$. The ...

3

Hint: Let $G$ be a group and $H \leq K$ be two subgroups of $G.$ If $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G.$ EDIT: Let $g \in G$ and consider the automorphism $\phi_g:G \to G, x \mapsto gxg^{-1}.$ Since $K$ is normal, $gKg^{-1} = K.$ So $\phi_g|_K$ is an automorphism of $K.$ Since $H$ is a characteristic subgroup of ...

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