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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, $$\lim_{n \to \infty} X_{T \wedge ... 3 Assuming the X_n are independent, then it follows from the second Borel-Cantelli lemma that$$\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 ...

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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

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 ... 1 If you meant $$P\left(\bigcap_{n=1}^{\infty}A_n\right)$$ then that is the probability that given$\omega \in \Omega$, $$\omega \in A_n \ \forall n \ge 1$$ or $$\omega \in A_1, A_2, A_3, ...$$ That is, almost surely, all the events$A_1, A_2, A_3, ...$occur. Sometimes this arises in the case of independent events where we have ... 1 $$P(-b<X<b) = P(X<b)-P(X<-b)$$ $$= 1 - P(X>b) -P(X<-b)$$ $$= 1 - P(X<-b) -P(X<-b) \ \text{Why?}$$ $$= 1 - 2P(X<-b)$$ Thus, $$1 - 2P(X<-b) = 0.9$$ $$\to 0.1/2 = P(X<-b)$$ $$\to 0.05 = P(X<-b) = F_X(-b)$$ Let$c := - b$. Then$0.05 = F_X(c)$Use a computer or your t table to find such a c. Then find b. Note: ... 1 Suppose$\{X_n\}$is uniformly bounded, i.e. there exists$M>0$such that$\sup_n|X_n|\leqslant M$a.s., and that$X_n\stackrel{\mathrm{a.s.}}\longrightarrow X$. Then we may choose$N$such that$n\geqslant N$implies$|X_n-X|<1$a.s. Hence with probability$1$, $$|X| = |X-X_N+X_N| \leqslant |X-X_N|+|X_N|\leqslant M+1,$$ so that$X$is bounded. 1 Since the stochastic integral process $$Y(t)=\int_{0}^{t}X_{s}\,\text{d}W_{s}\ , \quad t \geq 0$$ is a martingale and$\tau_{b}$is an (almost-surely) bounded stopping time (adapted to the same filtration as process$Y$) then by the Optional Stopping Theorem the expected value of the random variable$Y(\tau_{b})$is equal to that of the random variable ... 1 Hint/outline: Consider the$i$th string. Its length$X_i$follows a _____ distribution? Do you have just one string? No, you have 60. Assume they are all independent. Are you interested in their sum? No, you want the average, call it$\bar X$. Thus you want$P(20<\bar X< 38)$Since you have 60 strings, then$\bar X$is approximately normal by the ... 1 The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer. 1 Thanks Andre, I found it. For any event A, let IA be the indicator random variable of A, i.e. IA equals 1 if A occurs and 0 otherwise. Then 1 Here is the problem stated in full context: Let$(\Omega, \mathcal H,\mathbb P)$be a probability space. Let$H\in\mathcal H$and let$\mathcal F:=\sigma(H) = \{\varnothing, H, H^c, \Omega\}$. Show that $$\mathbb E[X\mid \mathcal F](\omega) = \mathbb E[X\mid H]$$ for all$\omega\in H$. The conditional expectation of$X$given the event$H\$ is ...

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Note that when you say proportion of people who have the disease are correctly tested, I would assume that you are referring to "how many of the people who actually have the disease were correctly tested." In which case, the answer is 100% - 2% = 98%. If you mean "how many of the people who were diagnosed as having the disease, who actually do have the ...

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