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0

$$ \langle e^{\mathrm{i}t N}\rangle=\sum_{n=1}^\infty (1-r)^2 n r^{n-1}e^{\mathrm{i}t n}=-\mathrm{i}\frac{(1-r)^2}{r}\frac{d}{dt}\sum_{n=1}^\infty (r e^{\mathrm{i}t})^n\ , $$ and then apply the geometric series $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ (and noting that the term corresponding to $n=0$ does not contribute thanks to the derivative) to get $$ ...


1

$$ \phi_W(t)=\langle e^{\mathrm{i}tx}\rangle=\int_0^\infty dx\ x e^{-x}e^{\mathrm{i}tx}=-\mathrm{i}\frac{d}{dt}\int_0^\infty dx\ e^{-(1-\mathrm{i}t)x}=-\mathrm{i}\frac{d}{dt}\frac{1}{1-\mathrm{i}t}=\frac{1}{(1-\mathrm{i}t)^2}\ . $$


1

The characteristic polynomial of a linear map is defined without any additional structure beyond that of a finite dimensional vector space while the notion of unitarity or normality requires the structure of an inner product so you should be suspicious if someone tells you that the characteristic polynomial tells you something about unitarity or normality. ...


4

User John Brevik gives as a hint: What is the characteristic polynomial of $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ How could you have come up with this hint yourself? Well, as you say, the characteristic polynomial only cares about eigenvalues, so you need to find two matrices which are not similar but which have the same eigenvalues. ...


2

This is not a correct MGF, because $M_X(0)$ always exists and is equal to $1$ but here $$M_X(0)=-0(1-e^0)=0\neq1$$


1

No distribution has such a moment generating function, for the trivial reason that $M_X(0) = 0$, which would imply that $$\operatorname{E}[e^{0X}] = \operatorname{E}[1] = 0,$$ a contradiction.


0

The first form that you have written is way much better! Exactly as you say, we should "avoid strange notation like that". Most of the time we only use them to shorten some specific localized notation that otherwise would be awkward to write in some other manner, say for example inside a specific integral. Personally I have always avoided them, on purpose.


1

I will write the answer for $X$ one dimensional. Your approach of bringing the derivative inside the integral is right. You need to estimate: $ \frac {e^{iux}-1}{u}.$ Writing down this function you find: $$\frac {\cos (ux) -1}{u} + i \frac {\sin (ux)}{u} = x \sin (\xi_1 x) - ix \cos (\xi_2 x)$$ by the mean value theorem. So in norm this can be ...


1

Denote by $$M(t) := \int e^{tx} \, \mathbb{P}(dx)$$ the moment generating function of the measure $\mathbb{P}$. Suppose that there exist $t_0 \in \mathbb{R}$ and $\epsilon>0$ such that $M(t)<\infty$ for all $t \in [t_0-\epsilon,t_0+\epsilon]$. Then If $t_0>0$ and $\int_{(-\infty,0)} |x| \, \mathbb{P}(dx)<\infty$, then $M$ is ...


1

Let us do the case $a=1$. Define for $x\neq 0$, $$I(c):=\int_{-\infty}^{\infty} \cfrac{\cos(xt) }{\pi t^2} \mathrm dt.$$ Then using the expansion of $\cos(a\pm b)$, we have $$ \int_{-\infty}^{+\infty} \left [ \cfrac{\cos(xt) \cdot (1 - \cos t)}{\pi t^2} \right ]\mathrm dt =I(x)-\frac{I(x+1)+I(x-1)}2. $$ Now, to compute $I(x)$, use the substitution $s=xt$ ...



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