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18

I will give two answers: Do it without complex numbers, notice that $$ \begin{eqnarray} \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 ...


13

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

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 ...


6

Finally, the semi-explicit (but quite useable) answer for $K \neq 0$, which I jot down quickly before coming back with clean derivations and proofs. $$\psi[\alpha, \beta,\sigma,K]=\sigma \frac{\Gamma \left(\frac{\alpha -1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan ...


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 ...


6

$$T=ZX+(1-Z)Y\implies E(\mathrm e^{\mathrm itT})=pE(\mathrm e^{\mathrm itX})+(1-p)E(\mathrm e^{\mathrm itY})$$


5

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) = ...


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

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 ...


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

Suppose from the start that I have three vector bundles $E_1, E_2, E_3$ and an open cover of the base space on which all three of these trivialize. Then if the transition functions for $E_i$ are denoted $\phi_{i,U}: U \to GL_n$, then the tensor product $E_1 \otimes E_2$ is the vector bundle given by transition functions $\phi_1 \otimes \phi_2$, where I mean ...


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

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

As user75064 already pointed out, the answer is "no". But there is the following result: Let $X,Y \in L^1$ $\mathbb{R}$-valued random variables. Then the following statements are equivalent. $X,Y$ are independent $\forall \eta,\xi \in \mathbb{R}: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot ...


4

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 ...


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 of any random variable is continuous. Say $X$ is a random variable and $t_n\to t$. Then $$\Bbb E[e^{it_n X}]\to\Bbb E[e^{it X}]$$by the Dominated Convergence Theorem.


4

It is not true for the characteristic polynomial. For example take the zero matrix of order 2 ,and take the 2×2 matrix with zero's everywhere except the top right corner. They have the same characteristic polynomial X^2 and therefore the condition holds, but the first is diagonizable and the second isnt.


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

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 ...


4

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 ...


4

No. For example this won't work for $P(Z=a)=P(Z=b)=1/2$, $a\not= b$. Since the convolution can only make measures less singular, $X_1$ must have purely discrete distribution also, but then it's easy to see that $\sum w_j X_j$ simply takes too many values (with positive probability).



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