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4

Because of the weak convergence, there is a probability space $(\Omega,\mathcal F,\Bbb P)$ and random variables $X_1,X_2,\ldots,X$ defined thereon, with values in $S$, such that (i) $P_n$ is the distribution of $X_n$ (i.e., $\Bbb P(X_n\in B)=P_n(B)$ for all Borel $B\subset S$), (ii) $P$ is the distribution of $X$, and (iii) $\lim_{n\to\infty} ... 4 If $$Y \mid X \sim \operatorname{Normal}(\mu = X, \sigma^2 = X^2),$$ then what is $$\operatorname{E}[Y \mid X]?$$ This is obviously simply$X$: Given the value of$X$,$Y$is normal with mean$\mu = X$, thus the expected value of$Y$given$X$is$X$. So $$\operatorname{E}[Y \mid X] = X.$$ Next, just take the expectation with respect to$X$: we have ... 4 Assuming the$X_n$are independent, then it follows from the second Borel-Cantelli lemma that $$\mathbb P\left(\limsup_{n\to\infty} \{X_n=1\}\right)=1.$$ (See for example here for a proof of the Borel-Canelli lemmas.) However, $$\mathbb P\left(\liminf_{n\to\infty}\{X_n=1\} \right) = \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty ... 3 Because t is fixed and will play no role, I'll drop it from the story. Thus we have an integrable random variable X such that E[\varphi(X)|\mathcal F]=\varphi(E[X|\mathcal F]) for all bounded and continuous \varphi. I will allow \varphi to be complex-valued. (Look at the real and imaginary parts of \varphi separately and then add.) Let's define ... 3 Fix \omega \in \Omega such that T(\omega)<\infty. Then there exists N \in \mathbb{N} such that T(\omega) \leq N. In particular,$$T(\omega) \wedge n = T(\omega) \qquad \text{for all } n \geq N.$$This implies$$X_{T \wedge n}(\omega) = X_{T}(\omega) \qquad \text{for all }n \geq N.$$Hence, obviously,$$\lim_{n \to \infty} X_{T \wedge ... 3 Just use Holder inequality with$p=q=2|F(x) - F(y)|=|\int_y^x f(t) dt|\le \sqrt{\int_y^x |f|^2 dt}\sqrt{\int_y^x 1 dt}\le C|x-y|^{\frac{1}{2}} $with$C=\sqrt{\int_\mathbb{R} |f|^2 dt}$3 I consider Terry Tao's Introduction to measure theory is a great book about measure theory and it covers the topics that you want. You can always give it a look and judge if it is too difficult for you or not. 3 Let$U$be a random variable which is uniform on$(0,1]$. Let$X=Y=U^{-1/2}$. Then$E[X]=E[Y]=2$but$E[XY]=+\infty$. You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly". On an infinite measure space you can also have the opposite phenomenon with tailing: ... 3 For the case$a+b\le x$where$0 \le x \le 1$the region is a triangle with base and height$x$which has the area$x^2/2.$To use an integral as you did, the limits on$a$should be from$0$to$x$rather than$0$to$1$as you have it. When$1 \le x \le 2$a good way is to subtract the area of the triangle above the line$a+b=x$from$1.$3 In the Kolmogorov formalism, you could call the entire sample space a "hidden variable", since it is usually not observed directly but only through actual random variables. The reason it doesn't conflict with Bell's inequality is that is is not local -- on the contrary it is the most global hidden variable conceivable, governing both which measurement the ... 3 Assume that$(X_i)_{i \in I} $are independent random variables. Let$Y_i := X_i \cdot 1_{|X_i|\leq C_i} $, where$C_i>0$is chosen so large that$Y_i \not\equiv c_i $for some constant$c_i $. This is possible since the$X_i $are non-constant. Then the$(Y_i - \Bbb{E}(Y_i))_i $form a family of independent and hence orthogonal random variables. Note ... 2 The Lebesgue integral and the proper Riemann integral always coincide provided the latter exists. Additionally, the Lebesgue integral and the improper Riemann integral also coincide provided they both exist. This follows from the monotone convergence theorem: given$f$satisfying the conditions and an increasing sequence of compact sets$K_n$whose union is ... 2 For a fixed$t$, $$\int_{-\infty}^\infty g(x)\mathrm{d}x=\int_{-\infty}^\infty g(x-t)\mathrm{d}x,$$ as can be verified with a variable substitution for$x-t$. This integral is a constant, so you are indeed justified in this step. 2 Notice that$u=(S_n+n)/2$is$Beta-Binomial(n,1,1)=U[0,n]$. Hence $$Y|S_n\sim Beta(1+\frac{n+S_n}2,1+\frac{n-S_n}2)=Beta(1+u,1+d)$$ with average$\hat y_n=\frac {1+u}{2+u+d}=\frac{1+u}{n+2}$and hence $$S_{n+1}-S_n\sim B(P((Y|S_n)>X_{n+1}))= B(\hat y_n)$$ where$B(p)$is Bernoulli on$\{-1,1\}$with$P(B(p)=1)=p$. 2 It is somewhat unusual to specify the mapping$X:\Omega\to\mathbb R^n$. Often what one knows about$X$is precisely the density or the cumulative distribution function. How does one specify what the "induced probability measure" is except by specifying the density or the c.d.f.? Suppose$X,Y$are independent random variables and$\Pr(X>t) = e^{-\alpha ...

2

Inclusion/Exclusion will work. It is easier to find the probability of the complement, that is, the probability that $X$ and $Y$ are both between $0$ and $2$. So our required probability is $$1-\int_0^2\int_0^2 \frac{1}{54}(x^2+y^2)\,dy\,dx.$$ Remark: Inclusion/Exclusion yields $$\int_{x=0}^3\int_{y=2}^3 k(x^2+y^2)\,dy\,dx+\int_{y=0}^3\int_{x=2}^3 ... 2 The probability that some number is drawn cannot be positive because the sum of the probabilities would be infinite. Probability 0 for an event not impossible is possible, if the number of events is uncountable. But for countably many events, P(X)=0 is equivalent to X is the impossible event. 2 Yeah, your proof is correct. Regarding your first question concerning integrability: If you prefer, you can restate the result as follows. The following statements are equivalent: \mathcal{F} and \mathcal{G} are independent. For all bounded X_F \in m \mathcal{F} and bounded X_G \in m \mathcal{G}, it holds that \mathbb{E}(X_G X_F ) = ... 2 As we have already discussed this depends on your definition of a "bounded process". If you mean that Y is (uniformly) bounded, i.e. there exists C>0 such that |Y_n| \leq C for all n \in \mathbb{N}, i.e.$$|Y_n(\omega)| \leq C \qquad \text{for all $\omega \in \Omega$ and $n \in \mathbb{N}$} \tag{1},$$then the claim does hold true. Indeed: Fix ... 2 Take a look only on one of these restaurants, hence the number of students that enters it distributed Binomial with n=200 and p=1/2. Denote it by Y. Then use the continuous correction and the Normal approximation to compute the probability of interest:$$ P(Y> 120) = P(Y\ge 120.5) \approx 1 - \phi \left(\frac{120.5 - np}{\sqrt{npq}} \right), $$... 2 Hint:$$E \left [ \sum_{i=1}^N X_i \right ] = \sum_{n=1}^\infty E \left [ \left. \sum_{i=1}^N X_i \right | N=n \right ] P(N=n) \\ = \sum_{n=1}^\infty \sum_{i=1}^n E[X_i|N=n] P(N=n).$$Can you compute the inner expectation? 2 The statement can be proved without invoking convolution concept. Clearly a_i \neq 0 for all i \in \{1, \ldots, m\}. Let P = \{i \in \{1, \ldots, m\}: a_i > 0\} and N = \{i \in \{1, \ldots, m\}: a_i < 0\}. If |P| = m (here |P| means the number of elements in P). Then the support of a_1 U_1 + \cdots + a_m U_m is [0, \sum_{i = 1}^m ... 2 For example you calculated$$P(R\text{ or }B|D_1)$$as 3/10. That is wrong. It should be$$P(R\text{ or }B|D_1) = \frac{1}{10}+\frac{3}{10} = \frac{4}{10}.$$There rest are probably similar mistakes. The very last line you gave is correct. 2 I was thinking say m = 2 and Y_1 = \infty, then \sup_{n \ge m} Y_n < \infty, but \sup_{n \ge 1} Y_n = \infty ? You almost answered you own question. You need further assumption for Y_n with n\geq 2 though. Say, Y_n\equiv 1 for each n. Then the Y_n's are independent. (Why?) Then you can check that the quoted statement is not true. 2$$\begin{align*} \operatorname{Var}[X \mid X \ge 10] &= \operatorname{E}[(X - \operatorname{E}[X])^2 \mid X \ge 10] \\ &= \operatorname{E}[X^2 \mid X \ge 10] - \operatorname{E}[X \mid X \ge 10]^2. \end{align*}$$To this end, we have$$\begin{align*} \operatorname{E}[X^k \mid X \ge 10] &= \frac{1}{\Pr[X \ge 10]} \int_{x=10}^\infty x^k f_X(x) \, ...

1

If I understand your question correctly, the easy answer is: no. As a counterexample, consider $Y_1,Y_2\sim U\left([0,1]\right)$ and let $X=Y_1+Y_2$. Your question is then: are $Y_2$ and $X$ independent? However, easy implications such as $$X\leq\frac{1}{2}\ \implies\ Y_2\leq\frac{1}{2}$$ $$Y_2\geq\frac{1}{2}\ \implies\ X\geq\frac{1}{2}$$ show that $Y_2$ and ...

1

This is quite a standard argument in measure theory. You don't need any additional assumptions (if I understand it correctly that you suppose the equality holds for all $D$ in the $\sigma$-algebra). You can easily reduce the problem to the setting $$\int_D X dP=0$$ for all $D\in \mathcal{D}$ by taking the difference of $X$ and $Y$ (both sides are finite ...

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

The density function is just the Radon derivative satisfying $$\int 1_A(x)dP_X(x)=P_X(A)=\int_A f(x)dx$$ Then when $g(x)$ is simple function we have $$\int g(x)dP_X(x)=\int g(x)f(x)dx$$ By monotone convergence and monotone convergence respectively, the above equation is also true when $g(x)$ is positive measurable function and bounded measurable function. ...

1

The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer.

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