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The pointwise convergence of the characteristic functions follows directly from the definition of weak convergence. Indeed, since $f(x) := e^{\imath \, x \xi}$ is for each $\xi \in \mathbb{R}$ continuous and bounded, we have $$\phi_n(\xi) := \int e^{\imath \, x \xi} \, d\mu_n(x) \to \int e^{\imath \, x \xi} \, d\mu(x) =: \phi(\xi).$$ The uniform ...


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Hint: Interchange integration and differentiation, i.e. use that $$\frac{d^k}{d^k t} \int e^{\imath \, t x} \, d\mu(x) = \int \frac{d^k}{d^k t} e^{\imath \, t x} \mu(dx).$$ (Here $\frac{d^k}{d^k t}$ denotes the $k$-th derivative with respect to $t$.) Remark: It follows from the theorem on differentiation of parametrized integrals and the integrability ...


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$$a_{n+2}=a_{n+1}+6a_n\\ a_{n+2}+2a_{n+1}=a_{n+1}+6a_n; \qquad \text{or}\qquad a_{n+2}-3a_{n+1}=-2a_{n+1}+6a_n$$ Put $u_n=a_n-a_{n-1}$ $$\begin{align} u_{n+2}&=3u_{n+1}; \qquad \text{or}\qquad &u_{n+2}&=-2u_{n+1}\\ u_n&=3u_{n-1} &u_n&=(-2)u_{n-1} \\ &=3^2u_{n-2} &&=(-2)^2u_{n-2}\\ &\vdots & &\vdots\\ ...


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Hint Since you seem to have found the roots of the characteristic equation, then you have, as usual, $$a_n=A(3-i)^n+B(3+i)^n$$ and the coefficients $A,B$ have to be determined from the first two terms. So, $$a_0=A+B=0$$ $$a_1=A(3-i)+B(3+i)=1$$ from which $A=\frac{i}{2}$ and $B=-\frac{i}{2}$. This makes $$a_n=\frac{1}{2} i \left((3-i)^n-(3+i)^n\right)$$ ...


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In terms of a generating function: Observe that we may use the recurrence relation to manipulate the formal power series $A(x):=\sum_{n=0}^\infty a_n x^n$ to give \begin{align} A(x)&=5+\sum_{n=0}^\infty a_{n+2} x^{n+2}\\ &=5+x\sum_{n=0}^\infty a_{n+1}x^{n+1}+x^2\sum_{n=0}^\infty a_{n}x^{n}\\ &=5+x(A(x)-5)+6x^2 A(x).\\ \end{align} From this we ...


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$x^2 - x - 6 = 0 \to x = 3, -2$, and the general solution is: $a_n = A\cdot 3^n + B\cdot (-2)^n$. For $a_0 = 5 \to A + B = 5$, and $a_1 = 0 \to 3A - 2B = 0 \to 3A - 2(5-A) = 0 \to A = 2, B = 3 \to a_n = 2\cdot 3^n + 3\cdot (-2)^n$.


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You do not follow the elementary indications given to you in comments hence it is a bit difficult to know what could help you (and the bounty is no substitute for that). Anyway, steps towards the solution could be as follows. The correct statement of the exercise involves $$Y_n = \frac{S_n/\tau_n - \lambda ...


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An approach which requires zero knowledge about modified Bessel functions $I_\nu$ and is valid for every nonnegative real number $u$ is to expand the function $$\phi:t\mapsto\exp(-4u\sin^2t),$$ using the identities $2-4\sin^2t=2\cos(2t)=\mathrm e^{2it}+\mathrm e^{-2it}$, as $$\phi(t)=\mathrm e^{-2u}\exp(u\mathrm e^{2it})\exp(u\mathrm e^{-2it})=\mathrm ...


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Hint: If $(X_k)_{1\leqslant k\leqslant n}$ is i.i.d. then $(Y_k)_{1\leqslant k\leqslant n}$ defined by $Y_k=X_{n+1-k}$ for every $1\leqslant k\leqslant n$ is distributed like $(X_k)_{1\leqslant k\leqslant n}$, in particular, $a_1X_1+a_2X_2+\cdots+a_nX_n$ and $a_1Y_1+a_2Y_2+\cdots+a_nY_n$ are identically distributed.


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Hints: Since $t \mapsto e^{-k |t|^{\beta}}$ is integrable, it follows from the Riemann-Lebesgue lemma that the distribution admits a continuous density (with respect to Lebesgue measure), i.e. there exists $p_{k,\beta} \geq 0$ such that $$\int e^{\imath \, t x} p_{k,\beta}(x)\, dx = e^{-k |t|^{\beta}}. \tag{1}$$ Show that $$p_{k,\beta}(x) = k^{-d/\beta} ...


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A compactly supported random variable has finite expectation. If $\Psi_{\beta}$ were the characteristic function of a compactly supported random variable $X$, then $\Psi_{\beta}$ would have to be differentiable at $t=0$ and $$E(X)=\frac{1}{i}\frac{d}{dt}\Psi_{\beta}(t) |_{t=0}\,. $$ But in case $0<\beta<1$, $\Psi_{\beta}$ is not differentiable at ...



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