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The function $x \mapsto t^{x^2}$ decreases on $\mathbb{R}_+$, so for every $n \in \mathbb{N}$, $$t^{(n+1)^2} \le \int_n^{n+1}t^{x^2}dx \le t^{n^2}$$ By summation over $n$, $$\sum_{n=1}^{\infty}t^{n^2} …

I assume that $(a_n)_{n \ge 1}$ are random variables taking values on a finite subset $B$, and that $\nu_l(b) \le P[a_n = b|a_1,\ldots,a_{n-1}] \le \nu_u(B)$ almost surely for every $n \ge 1$ and $b \ …

I call $f$ the density and $F$ the cumulative distribution function of $\mathcal{N}(0,1)$. Since $X_{(n)} \ge X_{(n-1)}$, $$X_{(n)}-X_{(n-1)} = \int_\mathbb{R} 1_{[X_{(n-1)} \le x < X_{(n)}]}dx.$$ Tak …

The answer is negative. I set $a_n = 1_B(n)$, where $B$ is the union of all intervals of the form $[(2k)!,(2k+1)![$. Let $N_s$ be a random variable with geometric distribution on $\{0,1,\ldots\}$ with …

I prove the convergence of the series. For $n \ge 1$, let $$S_n = \sum_{k=1}^n k^{-i\beta-\alpha} \text{ and } S'_n = \sum_{k=1}^n (-1)^ {k-1} k^{-i\beta-\alpha}$$ Then $$S_{2n}-S'_{2n} = 2 \sum_{k=1} …

Once we have convergence $F_n \to F$ in measure (this is the case in your situation), the natural condition for convergence in $L^1$ is uniform integrability. This is equivalent to convergence $||F_n| …

The notations are contradictory. Once $p$ is fixed, and then it varies. Do you set $\mu_{n}:=\frac{1}{n}\sum_{i=0}^{n-1}T_{\ast}^{i}\nu$ for ALL $n \ge 1$ and assume that $T_{\ast}^{p} \nu = \nu$ for …