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3

To find the p.d.f of the ratio $\frac{Y}{X+Y}$, let us first write its c.d.f. Since $X$ and $Y$ are always positive, their ratio is also positive and, therefore, for $0\leq t\lt1$ we can write: $P\left(\frac{Y}{X+Y}\leq t\right)=P\left(Y\leq \frac{t}{1-t}X\right)=\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx$ as ...

2

For $x>1$, we have $$\mathbb P(\xi_n\leqslant x) = \mathbb P\left(\bigcap_{i=1}^n \left\{\eta_i\leqslant x\right\}\right)=\prod_{i=1}^n\mathbb P\left(\eta_i\leqslant x \right) = \left(1 - x^{-\alpha}\right)^n.$$ Hence \begin{align} \mathbb P(\zeta_n\leqslant x) &= \mathbb P\left(\xi_n n^{-\frac1\alpha}\leqslant x\right)\\ &= \mathbb ...

2

For any $p \geq 1$, we have $$|x+y|^p \leq 2^p (|x|^p+|y|^p),$$ and therefore \begin{align*} \mathbb{E}(|X_n-X|^p \mid \mathcal{F}) &\leq 2^p \mathbb{E}(|X_n|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}) \\ &\leq 2^p \mathbb{E}(|X|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}). \end{align*} This shows that ...

2

Hint: $X^2< \frac12\iff |X|<\frac1{\sqrt2}$. If $X$ and $Y$ are independent, then $|X|$ and $|Y|$ are also independent. Complete the sentence: if $A$ and $B$ are independent, then $P(A, B) = ...$

2

Since $X$ and $Y$ are independent, we have $$\Pr[(X^2 \le 1/2) \cap (|Y| \le 1/2)] = \Pr[X^2 \le 1/2]\Pr[Y \le 1/2].$$ Since they are uniform on $[-1,1]$, we then have $$\Pr[X^2 \le 1/2] = \Pr[-1/\sqrt{2} \le X \le 1/\sqrt{2}] = \frac{1/\sqrt{2} - (-1/\sqrt{2})}{1 - (-1)} = \frac{1}{\sqrt{2}},$$ and $$\Pr[|Y| \le 1/2] = \Pr[-1/2 \le Y \le 1/2] = ... 1 Notice that because of independence: $$P\left(X^2 < \frac{1}{2}, |Y| < \frac{1}{2} \right) = P\left(X^2 < \frac{1}{2}\right) P\left(|Y| < \frac{1}{2} \right)$$ Analyzing the X term: P\left(X^2 < \frac{1}{2}\right) = P \left( -\frac{1}{\sqrt{2}} < X < \frac{1}{\sqrt{2}} \right) = ... 1 The answer to your first question is yes: if c \ne 0 and c \in \mathbb R, then cX \sim \operatorname{Normal}(c\mu, |c|\sigma) if X is normal with mean \mu and standard deviation \sigma. This is because the normal distribution belongs to a location-scale family: its PDF is$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad ...

1

The probability of extinction is the smallest positive root of $$G_O(z)=z$$ Where $O$ denotes the offspring distribution, and $G_O(z)$ its generating function at $z$. It is easily seen that $G_O(0)$ is the probability of extinction in the first generation. Second, if you know about generating functions, then you know that the sum: , where $X$ is ...

1

For any positive integer $n$, we have $T_n-S_n = g(T_n,S_n)$ where $g:\mathbb R^2\to \mathbb R$ is the map $(x,y)\mapsto x-y$. Given $t\in\mathbb R$, it is clear that $$g^{-1}((-\infty,t]) = \{(x,y):x-y\leqslant t\}$$ is a Lebesgue-measurable set in $\mathbb R^2$, and so $g$ is a measurable function. Since $\sigma(g(T_n,S_n))\subset \sigma(T_n,S_n)$, ...

1

Let $X, Y$ be random variables in $(\Omega, \mathcal B, \mathbb P)$. If $A \in \mathcal B$, then $1_A$ and $1_{A^C}$ are random variables. Note that $$Z = X1_A + Y1_{A^C}$$ Since sums or products of random variables in $(\Omega, \mathcal B, \mathbb P)$ are random variables in $(\Omega, \mathcal B, \mathbb P)$, $Z$ is a random variable in $(\Omega, ... 1 'if' Let$A_n^c := \{X_n > M\}$. By BCL1, we have $$P(\limsup A_n^C) = 0$$ $$\to P(\liminf A_n) = 1$$ $$\to \lim P(A_n) = 1$$ $$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$ $$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$ $$\to \forall n \in \mathbb N, P(A_n) = 1$$ $$\to \forall n \in \mathbb N, P\{X_n \le M\} = 1$$ $$\to P( \sup_{n \ge 1} (X_n) ... 1 No, there is nothing to be said about the Cesàro averages of the whole sequence. They can be as bad as the sequence itself. Indeed, given any weakly convergent sequence \{f_n\}, we can consider another sequence \{g_n\} defined as$$f_1,f_1, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_3, \dots $$where the term f_n appears ... 1$$E[(a-\bar{a})(a+\bar{a})]\leq E[|a-\bar{a}|(a+\bar{a})]\leq 2E[|a-\bar{a}|],$$where the second inequality follows from the fact that (a+\bar{a})\leq 2. 1 As 0 \leq a, \bar{a} \leq 1, we have$$(a-\bar{a})(a + \bar{a}) \leq |(a-\bar{a})(a+\bar{a})| = |a-\bar{a}| \cdot \underbrace{|a+\bar{a}|}_{= a+\bar{a} \leq 2}.$$Taking expectation on both sides, proves the inequality. 1 Assuming that$(\sup\limits_n Y_n)(x)=\sup\limits_{n}\{Y_n(x)\}$then indeed, there is nothing guaranteeing that the first random variable$Y_1$should be finite a.s and therefore there is no reason that your claim should be true in general. The example you posited disproves it, but you need to fix values for all the later$Y_n$as well (i.e. with$n\geq2$). ... 1 I was thinking say$m = 2$and$Y_1 = \infty$, then$\sup_{n \ge m} Y_n < \infty$, but$\sup_{n \ge 1} Y_n = \infty ?$You almost answered you own question. You need further assumption for$Y_n$with$n\geq 2$though. Say,$Y_n\equiv 1$for each$n$. Then the$Y_n\$'s are independent. (Why?) Then you can check that the quoted statement is not true.

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