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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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