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We have $\Omega$ = outcome space. $\omega$ = a particular outcome (that is, $\omega \in \Omega$). If $Z$ is an event then it is a subset of $\Omega$ (that is, $Z \subseteq \Omega)$. (*See footnote for an additional detail.) Indeed the random variable $X$ is a function $X:\Omega \rightarrow \mathbb{R}$. Suppose $A$ is some given subset of real numbers. ...


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$M=\underset{1\le i\le n}{\max} X_i.$ This means $M<T\,\,$ iff $\,\, X_i< T\,\,\forall \,\,i.$ Hence $\mathbb{P}(M<T)=\left[\mathbb{P}(X_i<T)\right]^n=\left[1-(1-p)^T\right]^n.$ $$\text{You need }\left[1-(1-p)^T\right]^n=1-\delta$$ $$\iff 1-(1-\delta)^{1/n}=(1-p)^T\iff \,\,T=\dfrac{\log\left(1-(1-\delta)^{1/n}\right)}{\log(1-p)}.$$


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Define $$\mu_{i,j}=\sigma((X_{i,k})_{1 \leq k \leq j}).$$ Define $$\mu_i=\bigcup_{j \geq 1}{\mu_{i,j}}.$$ $\mu_i$ is stable under finite intersection so the monotone class it generates is $$\sigma_i=\sigma((X_{i,j})_{j \geq 1}).$$ Note that all $X_{i,j}$ and X_i are $\sigma_i$-measurable. We prove that all $\sigma_i$ are independent. Define $$\mathscr{...


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Your solution to (b) looks fine. You can solve (a) in a similar fashion, but it may be easier to first note that $X_n$ is a function of $S_0, \dots, S_n$ (write $Y_k = S_k - S_{k-1}$). Then to show it is a martingale, it is equivalent to show that $\mathbb{E}[X_{n+1} - X_n \mid S_0, \dots, S_n] = 0$, which saves you from having to work with the sum. Each ...


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Some observations on i.i.d. vectors $\{X_i\}_{i=1}^{\infty}$ through a function $f:\mathbb{R}^k\rightarrow\mathbb{R}$. 1) As in my above comments: Lipschitz-like property: If there is a real-valued constant $L>0$ such that: $$ ||f(X_i+\delta_i)-f(X_i)|| \leq L||\delta_i||\quad \forall i \in \{1, 2, 3, ...\}$$ then the desired probability 1 ...


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Your $a$ is the golden ratio, $\phi$. We don't need to know this, but the golden ratio is worth noticing. You have already shown that $$ \int_1^x a^y \ln a \,\mathrm{d}y = a^x - a^1 \text{,} $$ so why not use it? $\phi^{3/2} - \phi$ is just some number near $0.4$.


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Hint on b) $\overline{Y}^{4}=n^{-4}\left(\sum_{i=1}^{n}Y_{i}\right)^{4}=n^{-4}\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{n}\sum_{i_{3}=1}^{n}\sum_{i_{4}=1}^{n}Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}$ so that $\mathbb{E}\overline{Y}^{4}=n^{-4}\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{n}\sum_{i_{3}=1}^{n}\sum_{i_{4}=1}^{n}\mathbb{E}\left[Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\...


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No you are not supposed to assume SLLN, but to prove it in this case, and I assume you mean $\mathbb{E}X_i=\mu_i\to\mu$ instead of $\mathbb{E}(\lvert X_i\rvert)\to\mu$. (a) Use $\mathbb{E}X_n^4<B$ to bound $\mathbb{E}X_n^2$ and hence $\operatorname{Var}X_n$. Similarly expand $\mathbb{E}Y_n^4$ and use e.g., Cauchy-Schwarz. (b) Expand $\overline{Y_n}^4$ ...


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The hierarchical model is: $$N \mid M \sim \operatorname{Poisson}(\lambda M), \quad M \sim \operatorname{NegBinomial}(r,p)$$ where $\lambda, r, p$ are fixed parameters, and $M$ is parametrized such that $M \in \{0, 1, 2, \ldots \}$. Then the PMF of the unconditional distribution of $N$ is $$\Pr[N = n] = \sum_{m=0}^\infty \Pr[N = n \mid M = m]\Pr[M = m] = \...


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Both are wrong . For the first one $X_1=1,X_2=2$ and $\alpha =2$ gives a counterexample. For the second one you need continuity of $F_{X_2}$ at the point $\frac {\alpha} 2$. In general you should take the left hand limit of $F_{X_2}$ at the point $\frac {\alpha} 2$ on RHS.


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Note that $B_T = B_0e^{rT}$, not $e^{-rT}$! What I think your professor has done is the following (I will set $X = \mathbb{I}_{S_T\geq K}$ for simplicity): \begin{align} e^{-rT}\text{E}_Q\left[S_T X\right] &= S_0 e^{-rT}\text{E}_Q\left[\frac{1}{S_0}S_T X\right]\\ &= S_0\text{E}_Q\left[\frac{S_T}{S_0} e^{-rT} X\right]\\ \end{align} Now, using the ...


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If $\{X_i\}$ is an i.i.d sequence of random variables, then so is $\{|X_i|\}$. Now you can apply the weak law of large numbers to the sequence $\{|X_i|\}$.


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