# Tag Info

## New answers tagged characteristic-functions

1

If $p( \lambda )$ is the characteristic polynomial of $A$ then $p(A)=0$ for the Hamilton Cayley theorem

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Note that $-x^3+6x^2+9x-14=-(x-1)(x+2)(x-7)$, so we may assume that $M$ is the diagonal matrix with entries $1,-2,7$ on the diagonal. Then it's easy to see that the characteristic polynomial of $M^{-3}$ is given by $(x-1)(x+(1/2)^3)(x-(1/7)^3)$.

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Unfortunately, your result for the convolution of the first two terms is not correct. We have \begin{align*} \chi_{[-a,a]}(x-y) \chi_{[-a,a]}(y) &= \chi_{[-a,a]}(y-x) \chi_{[-a,a]}(y) \\ &= \chi_{[-a+x,a+x]}(y) \chi_{[-a,a]}(y). \end{align*} This product equals $1$ if, and only if, $$-a + x \leq y \leq a+x \qquad \text{and} \qquad -a \leq y ... 0 The for language is much simpler and is able to express all primitive recursive functions : You can have as many variables (a,b,c,...) as you want that contains a natural integer. The first variables are usually used as input (all other variable are initialised to 0) and the first variable can be used also as the output. Operations : incr a : to ... 3 Let Z_n=a_n(X_n-X). We are told that \frac{1}{a_n}\to 0 so, in particular, \frac{1}{a_n}\overset{P}{\to}0. So by Slutsky's Theorem, X_n-X=\frac{1}{a_n}\times Z_n\overset{d}{\to}0\times Z=0. But convergence in distribution to constant is equivalent to convergence in probability to that same constant, so we conclude that X_n-X\overset{P}{\to}0. 0 a_n(X_n-X)\stackrel{\mathcal{D}}{\to}Z implies that \mathbb{P}\left(|Z|<\infty\right)=1 - Z should be a proper random variable. If the claim would not hold for some \varepsilon, then:$$ \exists\varepsilon>0: \lim_{n\to\infty}\mathbb{P}\left(|X_n-X|>\varepsilon\right)\neq 0\Rightarrow ...

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The probability distribution $p(x)$ can be obtained via inverse Fourier-transformation $$p(x) = \int \frac{dt}{2\pi} e^{-i t x} \frac{e^{it}}{1-it}.$$ Performing the integral, we obtain the distribution $$p(x) = e^{1-x}$$ on $x\geq 1$.

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Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book): Anyway, this answer might be weird or very wrong or stupid but... Consider iid $X_i$ ~ $Poi(\lambda/n)$. Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$ Then (\frac{1}{n} \sum_{i=1}^{n} X_i - ...

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This follows directly from the weak law of large numbers. Finite variance is implied by bounded variance.

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Hint: First show that if $\phi$ has even order derivative then all moments exist upto that order of derivative. Here you have infinitely differentiable $\phi$ so obviously all even order derivatives exist and hence all moments exist. By the way, the first part is a standard inclusion in probability textbooks.

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You're just missing one tiny piece of complex analysis to finish your proof, namely Morera's Theorem.

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Pointwise convergence $f_n \to 0$ is per definitionem the following: $f_n(x) \to 0$ for all $x \in \mathbf R$. So, let $x \in \mathbf R$. Choose $N \in \mathbf N$ with $N > x$ (such $N$ exist), then for all $n \ge N$ we have that $x \not\in [n,\infty)$, hence $f_n(x) = 0$. So $f_n(x) = 0$ for all $n \ge N$. This implies $f_n(x) \to 0$.

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$ $\dfrac{dy}{dt}=x+y=t+y$ , we have $y=y_0e^t-t-1=y_0e^x-x-1$ $\dfrac{du}{dt}=1$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=t+f(y_0)=x+f((x+y+1)e^{-x})$ $u(x,-x)=0$ : $f(e^{-x})=-x$ $f(x)=\ln x$ \$\therefore ...

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