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6

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 ...

5

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 : $$\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot ... 5 To prove these are characteristic functions, let us use random variables. This yields simpler, and more intuitive, proofs. In the first case, assume that \phi_1(t)=\mathrm E(\mathrm e^{itX_1}) and \phi_2(t)=\mathrm E(\mathrm e^{itX_2}) for some random variables X_1 and X_2 defined on the same probability space and introduce a Bernoulli random ... 4 No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that |\phi(t)|\leq 1 and the inequality 1-\cos(2t)\leq 4(1-\cos(t)) which allows you to conclude 1-\Re \phi(2t)\leq 4[1-\Re \phi(t)] which is essentially a statement that says the behavior of \phi(t) ... 3 Let A be independent of X with P(A=1)=P(A=0)=\frac{1}{2}. Then$$ 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)] $$... 3 Any random variable Z:(\Omega,\mathcal F)\to(\mathbb R^n,\mathcal B(\mathbb R^n)) such that \mathrm E(\mathrm e^{\mathrm i\langle u,Z\rangle})=\mathrm e^{-\kappa\|u\|^2} for every u in \mathbb R^n and for some positive \kappa, is centered normal with variance-covariance matrix 2\kappa I. A quick way to see this is to note that the function ... 3 Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite. Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line. Factoid 3: The function t\mapsto|\cos(t)| is ... 3 Let me just state the theorem I linked to in my comment, so that this question does not go unanswered. Bochner's theorem If \varphi:\mathbb{R}^d\to \mathbb C is a complex-valued function with \varphi(0)=1, continuous at 0 and nonnegative-definite in the sense that for n\geq 1 we have that$$ ...

2

W. Feller, An Introduction to Probability Theory and Applications, Volume I, XIX.4, Theorem 1. A continuous function $\phi$ with period $2\pi$ is a characteristic function iff its Fourier coefficients (4.2) satisfy $\phi_k \ge 0$ and $\phi(0) = 1$. $$\phi_k = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i k \zeta} \phi(\zeta)\,d\zeta \tag{4.2}$$

2

No. The property in your post is called subindependence, and it is strictly weaker than independence. (Note that some people use the term "subindependent" as a synonym for "uncorrelated".) In addition to the references given in Wikipedia, you can find an example in this short note. Unfortunately it's behind a paywall. The example consists of two random ...

2

Choosing the $z$ axis along $k$ and denoting $|k|$ by $q$, we have \begin{align} \int_{\mathbb R^3}\chi_{|x|\lt r}(x)\exp(-\mathrm ikx)\,\mathrm dx &=\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm d\theta\int_0^{2\pi}\mathrm d\phi\exp(-\mathrm iqR\cos\theta) \\ &=2\pi\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm ... 2 Recall the definition of a derivative i.e.f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}h$$Hence, we get that$$f'(0) = \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \dfrac{2h^2 \sin(1/h)-0}h = \lim_{h \to 0} 2h \sin(1/h)$$Now recall that \vert \sin(y) \vert \leq 1. Hence, we have that$$\left \vert 2h \sin(1/h) \right \vert \leq \left \vert 2h ...

2

If $X$ is a random variable with density $f$ (with respect to Lebesgue measure on the real line) and $G$ is a well-behaved function, then $$E[G(X)]=\int_{\Bbb R}G(x)f(x)d\lambda(x).$$ In the particular case of the exponential law, this gives $$\phi(t)=\int_0^{+\infty}e^{itx}e^{-\lambda x}\lambda dx.$$ If $X$ is a random variable with values in the set of ...

2

One way to do this is by a simple induction on the number of terms after proving it works when $n=2$. $$E((X_1+X_2)^3) = E(X_1^3)+3E(X_1^2X_2)+3E(X_1X_2^2)+E(X_2^3).$$ Because of independence this becomes $$E(X_1^3)+3E(X_1^2)E(X_2)+3E(X_1)E(X_2^2)+E(X_2^3).$$ Then the middle two terms are $0$ because each has a factor that is $0$. (But this doesn't ...

2

Yes, for example when $\mu_n$ is normally distributed with mean $0$ and standard deviation $n^2$: the sequence of characteristic functions converges pointwise to the map $\varphi$ such that $\varphi(t)=0$ if $t\neq 0$ and $\varphi(0)=1$. But such a sequence of measure cannot converge weakly (it's not tight). However, if the sequence of characteristic ...

2

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 ...

2

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. ...

2

Simply because the characteristic function of a random variable $Y$ is defined as $$\phi(t) = E(e^{itY})$$ And so with $Y= X^2$ and using the density of $X \sim \mathcal{N}(0,1)$ we have $$\phi(t) = E(e^{itY}) = E(e^{itX^2}) = \int_{-\infty}^\infty e^{itx^2} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$$ Remember: When a random variable $X$ has a ...

1

Assume only that $X$ is square-integrable with $\mathrm E(X)=0$ and $\mathrm E(X^2)=m_2$. Since $|\phi''(t)|\leqslant m_2$ for every $t$ and $\phi'(0)=0$, the mean value theorem for vector-valued functions shows that $|\phi'(t)|\leqslant m_2|t|$ for every $t$. Since $\phi(0)=1$, a second application of the mean value theorem yields ...

1

Yes, that expansion is all you need. Let $\ln[\phi(\omega)]=c(\omega)$, then $$\kappa_n = i^{-n}c^{(n)}(\omega)|_{\omega=0}$$ where $c^{(n)}$ is the $n$-th derivative of the cumulant generating function with respect to $\omega$. Also, it is possible to get the cumulants from the moments using the recursions in this link ...

1

If you have $a t^2 F^{(2)}(t) + b t F^{(1)}(t) + c F(t) = 0$ Then suppose $F(t) = t^x$ is the form of a solution Then $F^{(1)}(t) = x t^{x-1} = x \frac{f}{t}$ and $F^{(2)}(t) = x(x-1) t^{x-2} = x(x-1) \frac{f}{t^2}$ Substituting back into the first equation gets: $a x(x-1) f + b x f + c f = 0$ So $a x(x-1) + b x + c = 0$ The characteristic equation is ...

1

This is a Cauchy–Euler equation: $$x^2y''+\frac{2\mu}{\sigma^2}xy'-\frac{2r}{\sigma^2}y=2\frac{-Ax+b}{\sigma^2}$$ (I denoted $y=F(p_t)$, $x=p_t$ and multiplied by $\frac{2}{\sigma^2}$) The characteristic equation comes from the substitution $y=x^p$ to the homogeneous equation $$x^2y''+\frac{2\mu}{\sigma^2}xy'-\frac{2r}{\sigma^2}y=0$$ We have ...

1

Hints: $$N^2=0\iff \;\text{the characteristic polynomial of}\;N\;\;p_N(x)=x^2\implies\;\text{the minimal polynomial is either}\;\;x^2\;\;\text{or}\;\;x\ldots$$ Added: Trying to avoid working with the minimal polynomial: if $\,N^2=0\,$ then the only eigenvalue of $\,N\,$ is zero, so we have two possibilities: either there are two linearly independent ...

1

Hints: prove the following: 1) If $N^2=0$ then $\operatorname{Im}(N)\subseteq\ker(N)$. 2) Moreover, if $N\neq0$ and $N$ is $2\times2$, then $\operatorname{Im}(N)=\ker(N)$. 3) Take any $0\neq v\in\operatorname{Im}(N)=\ker(N)$. There exists $w\neq 0$ such that $Nw=v$ and $B=(w,v)$ form a basis to $\mathbb{C}^2$. 4) Denote $P=[I]^B_e$ - the identity matrix from ...

1

Recall that $\varphi_X(t) = E[e^{itX}]$. Hence, $\varphi_{X-a}(t) = E[e^{it(X-a)}] = e^{-ita}\varphi_X(t)$. You can compute the moments using your formula for $\varphi_{X-a}(t)$. Regarding your second question, I guess it corresponds to: "When is a distribution completely specified by it's moments?". An answer is here. PS: Any analytic function is ...

1

If $E(\vert Z\vert)<\infty$ which is the case here since the mean and variance assumed to be well-defined, then the first relation is derived by just linearity of expectation $$E(c_1Z) = E(c_1 (X+ j Y)) = E(c_1X + jc_1 Y) = c_1 E(X) + j c_1 E(Y) = c_1 E(Z)$$ Note that $E(\vert X\vert)< E(\vert Z\vert)<\infty$ and the similar relation is also true ...

1

Let us use this as PDF for the inverse chi-square distribution $$f(x; \nu) = \frac{2^{-\nu/2}}{\Gamma(\nu/2)}\,x^{-\nu/2-1} e^{-1/(2 x)}$$ The characteristic funcion is $$\phi_X(t)=\int_0^{+\infty} f(x; \nu) e^{itx}\,\mathrm{d}x$$ $$=\int_0^{+\infty} \frac{2^{-\nu/2}}{\Gamma(\nu/2)}\,x^{-\nu/2-1} e^{-1/(2 x)} e^{itx}\,\mathrm{d}x$$ Not an easy ...

1

There is no reason why $\varphi_Z(t)=\mathbb E(\mathrm e^{\mathrm itZ})$ with $Z=\max(Y,0)$ should be a simple function of $\varphi_Y(t)=\mathbb E(\mathrm e^{\mathrm itY})$, in general. In the specific case when $Y$ is standard normal, the tedious residue computations one imagines yield $\varphi_Y(t)=\mathrm e^{-t^2/2}$ and  \varphi_Z(t)=\tfrac12+\mathrm ...

1

Using the power series of exponential function and the fact that for $|z|<R/2$, the sequence $(M_n\frac{z^n}{n!},n\geqslant 1)$ is bounded, one can define a power series (hence an holomorphic function) on $B(0,R/2)$ which extends $\phi$. This proves that the values of $\phi$ are determined on a non degenerate interval. Now the last thing to check is ...

1

I think I've figured this out. Let $F$ be a possible distribution function of $X$. We know that the moment-generating function $M(t) = \mathbb{E_F}e^{Xt}$ is uniquely defined and analytic on a ball with radius $R$ around zero. Then characteristic function $\phi_F(z) = \mathbb{E_F}e^{iXz}$ then exists on a stripe $S$ with width $R$ around the real axis, ...

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