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8

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 ...


8

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

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

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

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


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

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

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 ...


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 $\prod\limits_{k=1}^\infty\varphi_k$ ...


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

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

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

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

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

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

Considering that $\phi(0)=1$ and that $\phi_k(0)=1$ for every $k$, one sees that the condition $$\sum_kp_k=1$$ is necessary. To prove that is is also sufficient and to avoid most of the technicalities, consider some random variables $N$ and $(X_k)$ defined on the same probability space, $N$ independent of $(X_k)$, each $X_k$ with characteristic function ...


3

Without loss of generality, assume that $X$ and $Y$ are Bernoulli random variables with the same parameter $p = P\{X=1\} = P\{Y=1\}$. Thus, $X-Y$ takes on values $-1, 0, 1$. Then, we have that $$\begin{align} P\{X=1\} = p &= P\{X=1, Y=1\} + P\{X=1, Y=0\}\tag{1}\\ P\{Y=1\} = p &= P\{X=1, Y=1\} + P\{X=0, Y=1\}\tag{2}. \end{align}$$ From $(1)$ and ...


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

$A=2J-I$, 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

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 ...



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