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9

Define $X_n$ to be such that $X_n$ is $0$ with probability $1-\frac{1}{n}$ and $n^2$ with probability $\frac{1}{n}$. It is the case that $E[X_n]=n \to \infty$. But for any positive $k$ we have $\mathbb{P}(X_n > k) = \frac{1}{n} \to 0$. Thus showing a counter example.


6

From the definition of a (probability) measure, you know that $$P\left(\bigcup_{_{n\in\mathbb{N}}}E_n\right) = \sum_{n\in\mathbb{N}}P(E_n)\qquad (1)$$ if the $E_i$ are pairwise dijoint ($E_i\cap E_j =\emptyset,\ i\neq j$). Now, define $$E_1 = A_1,\ E_2 = A_2\backslash A_1,\ E_3=A_3\backslash (A_1\cup A_2), .. $$ or, more formally, $$ E_n = ...


5

If you can argue that $\Pr(A \cap B) \le \Pr(A)$, then you can also argue that $\Pr(A \cap B) \le \Pr(B)$ by symmetry, giving the desired result. If you need a formal argument to show that $\Pr(A \cup B) \ge \Pr(A)$, consider writing $A \cup B$ as a union of the disjoint events $A$ and $B \setminus A$.


5

Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure $\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$ we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ...


4

Let $X_L$ be $L$ times a Bernoulli random variable (that is, it takes values $0$ and $L$ with equal probability). Its expectation is $L/2$ which tends to infinity with $L$. However, $P(X_L>0)=1/2$ for all $L$.


4

Since $$\begin{align*} \left\| \sum_{k=1}^n (X_k-\mathbb{E}(X_k)) - \sum_{k=1}^m (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 &= \left\| \sum_{k=m+1}^n (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 \\ &= \sum_{k=m+1}^n \sum_{\ell=m+1}^n \text{cov}(X_k,X_{\ell}) \end{align*}$$ for all $n \geq m$, we find by the Cauchy Schwarz inequality $$\begin{align*} \left\| ...


3

By the Borel-Cantelli lemma, if the series $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$ converges for each $\varepsilon>0$, $Y_n\to0$ almost surely as $n\to\infty$. Using Chebyshev's inequality, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$ since $\operatorname EY_n=0$. By ...


3

Call three heads and three tails when tossing six coins a success. Then the probability of success is $\frac{\binom{6}{3}}{2^6}$, which simplifies to $\frac{5}{16}$. Let $X$ be the number of trials until the first success. Then $X$ has geometric distribution with parameter $p=\frac{5}{16}$. It is a standard result that if $X$ has geometric distribution ...


3

First, let us observe that the distribution is symmetric since \begin{align*} P(-X_1\le x) &=P(X_1\ge-x)\\ &=c\int_{-x}^\infty(1+t^2)^{-1}(\log(1+t^2))^{-1}dt\\ &=-c\int_{-x}^\infty(1+(-s)^2)^{-1}(\log(1+(-s)^2))^{-1}ds\\ &=c\int_{-\infty}^x(1+s^2)^{-1}(\log(1+s^2))^{-1}ds\\ &=P(X_1\le x). \end{align*} Since the distribution is symmetric, ...


3

(From my comment above.) This follows from Fatou's lemma.


2

The way you worded that is a little strange. I have an alternative approach. Let $I$ be the event that a person is infected, $D$ be the event that a person develops the disease, and $\bar I, \bar D$, not those events. Then $$P(I\bar D)= P(\bar D|I)P(I) = (1-P(D|I))P(I) = (1-.30)(.5) = 0.35 = 35\%$$ where in the second step I used the product rule.


2

To answer your question about infinity specifically, one fixes an infinite hypernatural $H$, and works with the collection $$\{1,2,\ldots,n-1,n,n+1,\ldots,H-1, H\}.$$ As you suggested, one can assign probability $\frac{1}{H}$ to the occurrence of each individual number in this collection. This is the basic idea behind using infinitesimals in probability. ...


2

Consider an $1>\epsilon>0$. For all $n$, $|X_n-0|=0$ with probability $1-1/n$, and $|X_n-0|=1$ with probability $1/n$. So $$Pr(|X_n-0|>\epsilon)=\frac{1}{n}$$ since $|X_n-0|>\epsilon$ only when $X_n=1$. $1/n$ converges to 0 as $n \to \infty$. If $\epsilon \geq 1$, then $Pr(|X_n-0|>\epsilon)=0$ for all $n$.


2

Hint: Use the optional sampling theorem for the submartingale $|M_n|$. Direct proof: First, note that it suffices to show $\mathbb{E}(|M_S|) \leq \mathbb{E}(|M_n|)$ for any stopping time $S$ such that $S \leq n$. Since $(M_n)_{n \in \mathbb{N}}$ is a martingale, we know from Jensen's inequaliy that $(|M_n|)_{n \in \mathbb{N}}$ is a submartingale, i.e. ...


2

Let the sample space be $[0,1]$ and let each element of the filtration be the borel $\sigma-algebra$. Let $I$ be some non-measurable subset and $\tau_x=\mathbb{1}(x), x\in [0,1]$. Then $\{\tau_x<t\}$ is the complement of a singleton for $x\le 1$ and $[0,1]$ otherwise so it belongs to each $F_t$, but $\{sup_{x\in I}\tau_x<t\}=\cap_{x\in I}\{\tau_x ...


2

Recall that the arrival times in a Poisson process with rate $\lambda$ follow an Exponential distribution with mean $1/\lambda$. Since inter-arrival times are independent, and you have two Poisson process, then it sounds like you are asking for $$P(T<S)$$ where $T = X_1+X_2$ is the sum of two waiting times for one process, and $S = Y_1+Y_2$ is the sum ...


2

For the first one, for any $t>0$, $$P(X\geq Y)=P(tX\geq tY)=P(t(X-Y)\geq0)=P(e^{t(X-Y)}\geq1)\leq E(e^{tX})E(e^{-tY})$$ by Markov inequality and independence. Noticing that $E(e^{tX})$ is the moment generating function of $X$ (and noticing the same about $Y$), and substituting the forms of the moment general function, we get a bound that is equal to ...


2

For the first part, (i): $$ \mathbb{P}\{X \geq Y\}\} = \mathbb{E}[\mathbb{1}_{\{X \geq Y\}}] = \mathbb{E}[\mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] ]. $$ Dealing with the inner conditional expectation first, $$ \mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] = e^{-\lambda} \sum_{k=0}^\infty \mathbb{1}_{\{k \geq Y\}} \frac{\lambda^k}{k!} = e^{-\lambda} ...


2

Let $B_{n}=\cup_{i=1}^{n} E_{i}$. Notice, that $\lim\limits_{n \to \infty} B_{n} =E$ (The details are left to you!, but as a hint, $B_{n}$ is ncreasing!). $$\mu (E)=\mu(\lim B_{n})= \lim \mu(B_{n})=\lim \mu(\cup_{i=1}^{n} E_{i})= \lim \sum\limits_{i=1}^{n} \mu (E_{i}) =\sum\limits_{i=1 }^{\infty}\mu(E_{i}).$$ The seccond lime is because $B_{n}$ is increasing ...


2

I believe you are overthinking this assignment, and perhaps you feel the need to do some calculations before giving any kind of answer. You should not approach this task in that manner, as I hope my answers to your questions will illustrate. (i) is correct as long as you have "seniors" in your answer. They are the people contacted for responses to the ...


2

This probability is $0$. You can find it by conditioning on $g_1,\dots,g_{N-1}$. $span(g_1,\dots,g_{N-1})$ is an $N-1$ dimensional subspace. Let $S$ be the subspace orthogonal to this and pick a line $L$ in $S$. Observe that $L$ always exists since $N\leq M$. Due to rotational invariance of normal distribution, the projection of $g_N$ onto $L$ is a standard ...


2

Let $W_n=n^{-1/2}Y_n$. Then $$n^{-1/2}E[Y_n\mid Y_n>0]=E[W_n\mid W_n>0]\le \frac{\sqrt{E|W_n|^2}}{P\{W_n>0\}}=\frac{\sigma}{P\{W_n>0\}},$$ and $P\{W_n>0\}\to 1/2$ (by the CLT), i.e. for any $\epsilon>0$, $P\{W_n>0\}\ge \frac{1}{2}-\epsilon$ for $n$ large enough.


2

Note that the $X_n$ are bounded by $1$, so the absolute convergence follows from the convergence of the sum $\sum \limits_{n = 1}^\infty \frac{2}{3^n}$.


2

In thise case, the sequence of random variables is given by $X_n = \frac{X}{n}$. You just need to apply the definition of convergence in probability resp. almost sure convergence to solve this exercise.


2

Using that $P(A\cup B) = P(A)+P(B)-P(A\cap B)$, you get that $P(A\cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ (simply because $P(A \cup B)\geq P(A)$). So you can say that $P(A\cap B) \leq \operatorname{min}\{P(A),P(B)\}$. Now, if $A=B$ then you get the equality, so that is the largest possible value.


2

$A \cap B\subseteq A$; therefore $\Pr(A\cap B)\le\Pr(A)$. In the same way, prove $\Pr(A\cap B)\le\Pr(B)$.


2

Hint: use the Borel—Cantelli lemma. In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$ In even more detail: (place your ...


2

Fix $\varepsilon$. First us tightness to find a compact subset $K=K(\varepsilon)$ of $C[0,+\infty)$ such that $\mathbb P_n(K)\gt 1-\varepsilon$. Use the uniform convergence of $(f_n)_{n\geqslant 1}$ to $f$ in order to handle the integral of $f_n$ over $K$. Use the uniform bound to handle the integral of $f_n$ over the complement of $K$ (which has a measure ...


2

This is not a correct MGF, because $M_X(0)$ always exists and is equal to $1$ but here $$M_X(0)=-0(1-e^0)=0\neq1$$


2

Which one is true, a high probability or $0.5$? If indeed every new and discrete event is independent of the results of the previous independent events, then the probability is indeed $0.5$. So the only question remaining is - are those events really "physically" independent of each other? I don't think that science has a decisive answer for this ...



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