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4

Write $$\frac{S_n - n\mu}{\sqrt{S_n}} = \frac{S_n - n\mu}{\sqrt{n}\sigma}\cdot \frac{\sqrt{n}\sigma}{\sqrt{S_n}}.$$ where $\sigma^2 = \text{Var}(X_1)$. By CLT, $\frac{S_n - n\mu}{\sqrt{n}\sigma} \Rightarrow N(0, 1)$. By WLLN, $\frac{S_n}{n} \to \mu$ in probability. The result follows from the Slutsky's theorem. Clearly, $b^2 = \frac{\sigma^2}{\mu}$.

3

Let $X$ be your random variable on $\mathbb R$. If the (non central) 4th moment is zero, we have $$\mathbb E X^4 = \mathbb E |X|^4 = 0.$$ That is, $X$ must be $0$ almost surely. Quite boring :)

2

If $X$ and $Y$ are independent and $Y$ takes values in $\{-1,1\}$ while the distribution of $X$ is symmetric about $0$, then $XY$ has the same distribution as $X$. Hint: condition on $Y$.

2

If $X,Y,Z$ are independent then: $$\mathbb{P}[X \in A, Y \in B Z \in C] =\mathbb{P}[X \in A] \mathbb{P}[Y \in B] \mathbb{P}[Z \in C]$$ consider now $s^1(x) = \sum_{i =1}^{k_1} 1_{A_i}(x) \alpha_i$ $s^2(y) = \sum_{j =1}^{k_2} 1_{B_j}(y) \beta_j$ $s^3(z) = \sum_{l =1}^{k_3} 1_{C_l}(z) \gamma_j$ Now compute $$\mathbb{E}[s^1(X) s^2(Y) s^3(Z)] = ... 2 First write,$$ \begin{align} \lim_{n \to \infty} \prod_{i=1}^{n} X_i &= \exp \left ( \lim_{n \to \infty} n \frac{1}{n} \sum_{i=1}^{n} \log(X_i) \right ) . \end{align} Now by the SLLN we have n^{-1} \sum_{i=1}^{n} \log(X_i) \stackrel{\text{a.s.}}{\to} \text{E}[\log(X_1)]. But Jensen's inequality (along with strict concavity of the log function ... 2 \begin{align} \operatorname{E}(e^{itY}\mid X=x) & = \operatorname{E}(e^{it\sqrt{x} \, Z})\quad\text{where }Z\sim N(0,1), \\[10pt] & = \varphi_Z(t\sqrt x) = \exp \left( \frac{-1}2 t^2 x \right). \end{align} \begin{align} \operatorname{E}(e^{itY}) & = \operatorname{E} \left( \operatorname{E}(e^{itY}\mid X) \right) = \operatorname{E}\left( ... 2 Z=(X,Y)' is not necessary bivariate normal (it works if X and Y are independent). There are lots of counterexamples (e.g. link). Another example: \psi_1 and \psi_2 are independent standard normal r.v.s and(X, Y)=(\psi_1,|\psi_2|)1\{\psi_1\ge 0\}+(\psi_1,-|\psi_2|)1\{\psi_1< 0\}$$which is not bivariate normal. However, marginals X and Y ... 2 Let \bar X_n=n^{-1}S_n. By the CLT you have$$\sqrt{n}(\bar X_n-\mu)\rightarrow N(0,\sigma^2)$$Now, apply the mean value theorem to get$$g(\bar X_n)-g(\mu)=g'(\bar X_n')(\bar X_n-\mu)$$where \bar X_n' lies between \bar X_n and \mu so that g'(\bar X_n')\rightarrow^p g'(\mu) (by the WLLN and the continuous mapping theorem) and ... 1$$G_{S_n}(z)=\left[G_{X_{n1}} (z)\right]^n=\left[\mathbb{E}z^{X_{n1}}\right]^n=\left[\frac{z}{n}+\left(\frac{z}{n}\right)^2+\left(1-\frac{1}{n}-\frac{1}{n^2}\right)\right]^n\lim_{n\rightarrow \infty}G_{S_n}(z)=e^{z-1}$$which corresponds to the probability generating function of \text{Poisson(1)}. Note: For large n ... 1 If you have \lambda\rightarrow \infty then$$\varphi_{Y/\sqrt{\lambda}}(t)=\varphi_{Y}\left(\frac{t}{\sqrt{\lambda}}\right)=e^{-\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}\rightarrow e^{-\frac{t^2}{2}}$$because$$\left(1-\frac{\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}{\lambda}\right)^\lambda=e^{-\frac{t^2}{2}}$$1 First, k=1,\dots,n (otherwise, the total prob. does not sum up to 1). Then$$\varphi_{X_n}(t)=\mathbb{E}[e^{\mathrm{i}tX_n}]= \sum_{k=1}^{n}e^{\mathrm{i}t\frac{k}{n}}\cdot\frac{1} {n}= \frac{\exp\left(\mathrm{i}t\cdot\frac{n+1}{n}\right)-\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)}{\left(\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)-1\right)\cdot n} ...

1

If every $X_n^1$ has standard normal distribution and $X_n^2=-X_n^1$ then: $$X_n=(X_n^1,X_n^2)\stackrel{d}{\rightarrow}(U,V)$$ where $(U,V)$ has a bivariate normal distribution such that $U$ and $V$ both have standard normal distribution and $U+V=0$. So we have: $$g(X_n^1,X_n^2)=0=g(U,V)$$ for each $n$.

1

Suppose: $A_i$ and $B_i$ for each $i = 1, 2, \ldots$ represent times for A and B to complete their $i^{\rm th}$ race, respectively. Once the $i^{\rm th}$ race is complete, the $(i+1)^{\rm th}$ race is begun immediately, without regard to the status of the other racer (i.e., the two processes are independent). Let $N_A(t)$ and $N_B(t)$ represent the total ...

1

Denote by $F$ the event $B>A$ (B wins before A ) since the events are independent and denote $S$ the event $A>B$ $$\Bbb{P}(B>A, n \text{ times}) = \Bbb{P} (FF \ldots FS) = \bigg(\frac{\alpha}{\alpha + \beta}\bigg)^n \frac{\beta}{\alpha + \beta}$$ since $\Bbb{P}(A>B) = P(A \geq B) = 1 - P(B>A)$

1

This seems to fall under eigenvalue perturbation problem. The following assumes that eigenvectors $x_{0j}$ of the unperturbed matrix are chosen such that $||x_{0j}||^2 = x_{0j}^{*} x_{0j} = 1$. An eigenvalue of perturbed matrix is $\lambda_j = \lambda_{0j} + \mathbf{x}_{0j}^{*} \delta A \mathbf{x}_{0j} + o(||\delta A||)$, where $\lambda_{0j} = -2 - 2 \cos ... 1 Since$X$has a symmetic distribution about$0$, you have$P(X-0 \le x) = P(0-X \le x)$, i.e.$P(X \le x) = P(-X \le x)$. Now consider the cumulative distribution function$P(Z \le x)$. This is$P(XY \le x)=P(XY \le x|Y=1)P(Y=1)+P(XY \le x|Y=-1)P(Y=-1)$which is$P(X \le x|Y=1)P(Y=1)+P(-X \le x|Y=-1)P(Y=-1)$which by the symmetry and independence is ... 1 If$e_j\in\{-1,+1\}$for$0\leq j\leq n$, and if we set$e_{-1}=1$, then $$\mathbb{P}(\cap_{j=0}^n [X_j=e_j]) =\mathbb{P}(\cap_{j=0}^n [Y_j=e_j/e_{j-1}])=(1/2)^{n+1}.$$ 1 First off, since you are using that the$\sigma$-algebra is$2^{\Omega}$(or a subset of it), we have to assume that$\Omega$is finite or countable. I think the main confusion here is the relationship between a probability measure and a random variable. A probability measure$\mathbb P$is a function$\mathbb P:\mathcal F \rightarrow [0, 1]$for some ... 1 well if exponentiation preserves a property then it will continue to hold. For example,$f$holder cts implies that$e^f$is so yes that property goes over. 1 Yes, otherwise how do you plan on taking intersection of sets which don't live in the same space? You should think of random variable as just measurable functions, which then would be kinda odd if they didn't share the same domain. Perhaps a better question is how do we know that, given a sequence of random variable, that there is a probability space which ... 1 Given that$N_m=k$, where$k\ge 1$, and$0\lt x\lt 1$, we have $$\Pr(M_m\le x)=x^k.$$ Now we need to decide what$\Pr(M_m\le x)$is if$N_m=0$. There is no obvious definition. With not much conviction, we call this probability$1$. Then $$\Pr(M_m\le x)=\sum_{k=0}^\infty e^{-m} \frac{m^k}{k!}x^k.$$ We recognize the sum as$e^{-m}e^{mx}$. 1 You have a minor issue with the expression for$g_Y\$: $$g_Y(t)=\frac{q(qt-1)}{p^2+qt-1}$$ Then $$P\{Y=k\}=\frac{g_Y^{(k)}(0)}{k!}=(-1)^{k+1}\frac{(q-1)^2}{(q-2)^{k+1}} \text{, }k=0,1,\dots$$

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