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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 ...


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

$$A^2 = \left( \begin{array}{ccc} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \\ \end{array} \right)$$ $$-4A = \left( \begin{array}{ccc} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \\ \end{array} \right)$$ $$A^2 - 4A = \left( \begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \\ ...


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.


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: ...


4

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.


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

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

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

$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$


4

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 ...


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 ...


3

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) = ...


3

$A=2J-I$, where $J$ is the all-one matrix with $J^2=3J$. Therefoer $$ A^2-4A-5I=(2J-I)^2-4(2J-I)-5I=4J^2-12J=0. $$


3

If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach. Let's take the equation $$ A^2 - 4A - 5I = 0.$$ And now forget about the matrices. What we have is simple quadratic equation in $x$ written as $$ x^2 - 4x - 5 = 0,$$ now just find the roots $$ (x-5)(x+1) = 0.$$ Let's get back to that matrix ...


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

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

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 ...


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

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

The characteristic function is $$\begin{align} \frac12 \int_{-\infty}^{\infty} dx\, e^{-|x|} e^{i t x} &= \frac12 \int_{-\infty}^0 dx \, e^{x+i t x} + \frac12 \int_0^{\infty} dx \, e^{-x + i t x}\\ &= \frac12 \int_0^{\infty} dx \, e^{-(1+i t) x} + \frac12 \int_0^{\infty} dx \, e^{-(1-i t) x} \\ &= \frac12 \frac{1}{1+i t} + \frac12 \frac{1}{1-i ...


2

Note that $(a+bi)+(a-bi)=2a$ so by splitting the integral at $0$ $$ \frac12\int_{-\infty}^\infty e^{itx}e^{-|x|}dx = \int_0^\infty \mathfrak{Re}(e^{itx}e^{-x})dx $$ $$ =\int_0^\infty \cos(tx)e^{-x}dx $$ $$ =e^{-x}\frac{t\sin(tx)-\cos(tx)}{t^2+1}\biggr|_{x=0}^{\infty} $$ $$ =\frac{1}{1+t^2}. $$ I will leave the calculation of the integral (by parts twice) to ...


2

Since $\varphi$ is the characteristic function of an infinitely divisible distribution we have that $$ \varphi(t)=\varphi_n(t)^n,\quad n\in\mathbb{N},\tag{1} $$ for a sequence of characteristic functions $\varphi_n$. Now we use that $|\varphi_n|^2$ is also a characteristic function for each $n$, and thus by $(1)$ we have that $|\varphi|^{2/n}$ is a ...


2

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 ...


2

Yes, thanks to Lévy's continuity theorem, the pointwise limit is a characteristic function as soon as it is continuous at $0$.


2

Hint: Split to two cases, $w \in A$ and $w \notin A$. In each case, think in which (if any) of the subsets $A_n$ you can expect to find $w$.


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

It's true because $\phi_n(-\omega)=\overline{\phi_n(\omega)}$ (as long as $\omega$ is real not complex). Thus, you get $$ |\phi_n(\omega)|^2 =\phi_n(\omega)\overline{\phi_n(\omega)} =\phi_n(\omega)\phi_n(-\omega). $$


2

Define $w=y-Cx$, then $$ E[\mathrm e^{\mathrm iu^Tw}|\mathcal F^x] =E[\mathrm e^{\mathrm iu^Ty}|\mathcal F^x]\,\mathrm e^{-\mathrm iu^TCx} = \mathrm e^{-u^T \tilde{Q}u/2} $$ is independent of $x$. Thus $w$ is independent of $x$ and centered normal with covariance $\tilde{Q}$.


2

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 ...



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