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Note the Nate Eldredge comment on two different views of the unknown $\theta$. 1) If $\theta$ is a random variable then you can say $\{X_1, ..., X_{100}\}$ are conditionally i.i.d. given $\theta$, but they are not i.i.d. because they all depend on the common random variable $\theta$. This is the "Bayesian" view of $\theta$. 2) If $\theta$ is a fixed (...


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If two probability measures $P$ and $Q$ are equal on class of sets generating a sigma algebra you cannot immediately conclude that they are equal on the sigma algebra. This requires a proof unless the class you started with is an algebra. In this case sets of the form $X_1^{-1}(A_1)\cap ..\cap X_n^{-1}(A_n)$ do not form an algebra. This class of sets is ...


2

In general if random variables $Y$ and $W$ are independent, then (for any function $f$), $Y$ and $f(W)$ are independent as well. You should think about why this is true if you are not certain. If you are able to prove the above, then you can then prove the more general claim that if $Y$ and $W$ are conditionally independent given $X$, then $Y$ and $f(W)$ ...


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No, $Y$ does not have the no-memory property. Since $$Y_2 -Y_1 = \exp(-2) W_{\exp(2)} - \exp(-1/2) W_{\exp(1)}$$ we have \begin{align*} \mathbb{E}((Y_2-Y_1) Y_1) &= \exp(-5/2) \mathbb{E}(W_{\exp(2)} W_{\exp(1)}) - \exp(-1) \mathbb{E}(W_{\exp(1)}^2). \end{align*} Using that $$\mathbb{E}(W_s W_t) = \min\{s,t\}, \qquad s,t \geq 0,$$ we get $$\mathbb{E}((...


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Variables $X_1,X_2,...X_k$ are indeed mutually independent. Of course $X_1$ = $1$, lets say you received Lebrons picture. Imagine that $X_2$ = $1000$. So it means that you bought $999$ cards and for all of them you received Lebrons picture. Only on try $1000$ you received a new picture. Does it somehow affect the $X_3$? the fact that you had to buy $1001$...


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The converse is not true. Let $x=(1,0)$, $y=(2,0)$ and $z=(0,1)$. Then $x$ and $y$ are not linearly independent, but $x+z=(1,1)$ and $y+z=(2,1)$ are.


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Notice that $$\chi_A^{-1}(C) = \begin{cases} \Omega \quad \{0,1\} \subseteq C \\ A \quad 1 \in C, 0 \not \in C \\ A^c \quad 0 \in C, 1 \not \in C \\ \emptyset \quad \text{otherwise} \end{cases}. $$ Since $A \in \mathcal{F}$, each of these sets is in $\mathcal{F}$.


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We also know that $P(A\cup B) \le P(A\cup B\cup C)$, which leads to the inequality $2p-p^2 \le 3p-3p^2$; this is equivalent to $2p^2 \le p$, which implies that $p \le \frac12$.


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Because from $x_1\alpha_1+x_2\alpha_2+x_3\alpha_3=0$ we obtain: $$(x_1,x_1,0)+(x_2,0,x_2)+(0,x_3,x_3)=(0,0,0).$$ Can you end it now?


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I'm not sure if this answers your question but in order for a set of vectors to be a basis of a vector space it has to be linear independant and has to span the entire space. Now if you're given a set of $n$ linearly independant vectors in $\mathbb{R}^n$ you only have to show that it spans $\mathbb{R}^n$. But as you said yourself this is indeed the case ...


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The answer to your question is "no". Under A2 you will have $\ \mathbb{E}\left(Y_t\left\vert A_s\, \forall s\in \mathcal T\right.\right)= 3A_t +\mathbb{E}\left(B_t\left\vert A_s\, \forall s\in \mathcal T\right.\right)\ $, but $\ \mathbb{E}\left(Y_t\left\vert A_t\right.\right)= 3A_t +\mathbb{E}\left(B_t\right)\ $, and these would have to be equal for your ...


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If you want a bit of intuition, think like this: if you know $A$ happened you have no idea if $B$ happened or not. With probability half you got $TTT$ and then $B$ happened and with the same probability you got $HHH$ and then $B$ didn't happen. So you have no idea, the probability of $B$ happening is exactly the same as it was if you didn't know that $A$ ...


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You can still switch the summations as the summands are all non-negative. This is Tonelli's theorem. Here is a reference.


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Claim: If $ \Omega = \{ \omega_1,\ldots,\omega_n \}$ is a finite probability space with $\Pr(\omega_i) > 0$ for all $i$, and $E_1, \ldots, E_k$ are mutually independent events all different from $\Omega$ and $\emptyset$, then $k \leq \lfloor\log_2 n\rfloor$. Suppose $n < 2^{m+1}$ and suppose there are $m+1$ events $E_1, \ldots, E_{m+1}$. Then at ...


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In my interpretation of the question, $X$ and $Y$ are not independent. These random variables are functions of random variables: $$\eqalign{X=X(M,\vec{X},\vec{\epsilon})\\Y=Y(M,\vec{X},\vec{\epsilon})}$$ While the variation from $\vec{X}$ is not shared between $X$ and $Y$, the variations from $M$ and $\vec{\epsilon}$ are shared. As an example, suppose the ...


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We have $M(2t)=M(t)^3 M(-t)$ and $M(t)=M(-t)$, so $M(2t)=M(t)^4$. Hence we have a functional equation for $m(t)=t^{-2}\log M(t)$ (let $m(0)=\frac12s^2=\frac12$): $$m(2t)=m(t)\quad\text{for all }t\tag{1}$$ But we know $$ m(t)=\frac12+o(1)\text{ for small }t\tag{2} $$ from the expansion of $M(t)$. Now equations (1) and (2) together implies $m$ is constant $\...


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The OP's book's statement that events ${A_1, A_2, \ldots, A_n}$ are mutually independent if for any subset ${A_1, A_2, \ldots \cap A_m}$ (where $m \leq n$) of these events we have: $$P(A_1 \cap A_2 \cap \ldots \cap A_m) = P(A_1)P(A_2)...P(A_m)$$ is incorrect because the equality stated above must hold for every subset of ${A_1, A_2, ..A_n}$ (e.g. such ...


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