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12

You want to prove the statement: $$\lim_{n\to\infty}\sum_{i=1}^{n}c=c \implies c=0$$ Instead, you can prove the equivalent statement: $$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c \neq c$$ And this is rather simple, as you can use the exact trick that you were trying to avoid: $$c\neq0 \implies ...


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.


8

In the context of probability theory, you can write $$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$ Then if you subtract $\Pr(\varnothing)$ from both sides, you get $$ 0 = \Pr(\varnothing). $$ Thus you don't need infinite series. That $\infty\cdot0=0$ does not make sense in certain broad contexts, since for ...


6

"$\Rightarrow$": Use $$|\mathbb{E}(X_n 1_A)-\mathbb{E}(X 1_A)| \leq \mathbb{E}(|X_n-X|).$$ "$\Leftarrow$": Show $$\mathbb{E}(|X_n-X|) = \mathbb{E}[(X_n-X) 1_{\{X_n-X \geq 0\}})]+ \mathbb{E}[(X-X_n) 1_{\{X_n-X<0\}}],$$ and conclude that $$\mathbb{E}(|X_n-X|) \leq 2 \sup_{A \in \mathcal{F}} |\mathbb{E}[(X_n-X) 1_A]|.$$


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 = ...


6

Yes, it's true. By the very definition of $\liminf$, there exists a subsequence such that $$\liminf_{n \to \infty} \mathbb{E}(X_n ) = \lim_{k \to \infty} \mathbb{E}(X_{n_k}). \tag{1}$$ Since, by assumption, $X_{n} \to X$ in probability (hence in particular $X_{n_k} \to X$ in probability), we can choose a further subsequence of $(X_{n_k})_{k \in ...


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 $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ...


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

Your reasoning is basically right but your notation is objectionable. What does "$X=x$ is a constant" mean? If $x$ is a constant, then it would came outside the outer expectation - and the law of total expectation would not apply. You should simply write $$E (X\, Y) =E (E(X \, Y \mid X))=E (X \, E(Y\mid X))$$ This is indeed true for any $X,Y$. And, by the ...


5

Note that $E[X+Y]=E[X]+E[Y]$ holds in full generality, even if $X$ and $Y$ are not mutually independent. Proofs of linearity of expectation do not assume independence of $X$ and $Y$. Here's one for example. In other words you do not need to impose that restriction.


4

To simplify the subscripts, I'm going to consider just the first few letters the monkey types. Add a variable $k$ to all my subscripts if you want to apply the reasoning at an arbitrary point in the string. It is true that $E(X_1) = 26^{-5}$ and also that $E(X_2) = 26^{-5}$. We just have to cycle through the $26^6$ equally-likely possibilities for the first ...


4

Community wiki answer so the question can be marked as answered: As Alex remarked: yes, this is correct.


4

Note that $$\hat\theta_n\sim \frac 1 {1+\sum_{i=1}^{n-1}U_i} $$ for i.i.d. standard uniforms $U_i$. Now see this and this


4

Hint: Consider each of the $10$ possible values for $A$ For example $P(A=2)=P(X=1 \text{ and } Y=1)=0.1^2$ while $P(A=3)=P(X=1 \text{ and } Y=2)+P(X=2 \text{ and } Y=1)=0.1\times 0.4+0.4 \times 0.1$


4

The complement of the middle-thirds Cantor set is a countable union of (open) intervals. However, the Cantor set itself has uncountably many elements, and any two of them are separated by a point not in the Cantor set. So no countable union of intervals can produce it. Thus $B$ is not closed under complement and therefore it is not a $\sigma$-algebra.


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

If you require a sigma algebra to be closed under uncountable union, then the only measure on the real numbers would be the zero measure, because every subset of $\mathbb{R}$ would have to be in the sigma algebra (at least assuming that the intervals were.)


4

The variable $X$ is the sample mean and the variable $Y$ is the sample variance times $(n-1)$. So Basu's theorem implies that they are independent. The distribution of $Y$ is $\chi^2_{n-1}$ as the sum of the squares of the $n$ iid normal random variables $X_i-X$, (where $X$ is used and so there are $n-1$ degrees of freedom instead of $n$).


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

Write $f(X) = f(X)^+ - f(X)^-$. Then, observe that the bound $f \geq -c$ implies $f(X)^- \leq c$ a.s., so that $$ \mathbb{E}[f(X)] = \mathbb{E}[f(X)^+ - f(X)^-] = \mathbb{E}[f(X)^+] - \mathbb{E}[f(X)^-] \geq \mathbb{E}[f(X)^+] - c $$ and reorganizing the terms give what you want: $\mathbb{E}[f(X)^+] \leq \mathbb{E}[f(X)] + c$. Note that this does not ...


3

Without any additional assumptions on the Lévy process $(X_t)_{t \geq 0}$, a central limit theorem does not hold true. Let $(X_t)_{t \geq 0}$ be a (one-dimensional) Lévy process with Lévy triplet $(b,\sigma^2,\nu)$. Define $$T(x) := \nu((x,\infty)) + \nu((-\infty,-x))$$ and $$U(x) := \sigma^2+2 \int_0^x y T(y) \, dy$$ for $x>0$. There is the ...


3

Because the normal curve has a bell shape, there is more area under it near the average. It appears that we are dealing with a standard normal distribution, so you want to solve for little $x$ Using the CDF, \begin{align*} .95 &= P(X\leq x) -P(X\leq -x) \\ &=P(X\leq x)-[1-P(X\leq x)]\tag{1}\\ &=2P(X\leq x)-1\\ &=2\Phi(x)-1 \end{align*} where ...


3

Observe that $$\max_{1\leq m \leq n} X_m \geq \lambda \Leftrightarrow \max_{1\leq m \leq n} X_m + c \geq \lambda + c \implies \max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c$$ $$\max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c \Leftrightarrow \max_{1\leq m \leq n} \lvert X_m + c \rvert^2 \geq (\lambda + c)^2$$ The second relation ...


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

If the distribution is discrete, there may be a problem. For example, imagine tossing a fair coin that has a $0$ on one side and a $1$ on the other. Then all possibilities $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$ are equally likely. When we order, the result $(0,1)$ is twice as likely as either $(0,0)$ or $(1,1)$. Remark: If the underlying distribution is ...


3

I think that almost everything you might want to know about this space ($\mathrm{MALG}_\lambda$, called the Lebesgue measure algebra) is in exercises (17.42)-(17.46) in Kechris' book. This metric $d$ turns $\mathrm{MALG}_\lambda$ into a Polish space, and the Boolean operations ($\cup$, $\cap$, and relative complement in $[0,1]$) are well defined and ...


3

If you've already proved the base case $P(A_1\cup A_2) \le P(A_1)+P(A_2)$, then you can do this: \begin{align} & P(A_1 \cup \cdots \cup A_n \cup A_{n+1}) \\[10pt] = {} & P( \Big( A_1 \cup \cdots \cup A_n \Big) \cup A_{n+1}) \\[10pt] \le {} & P(A_1 \cup \cdots \cup A_n) + P(A_{n+1}) & & \text{by the base case applied to these two events,} ...


3

Through the Spitzer identity, it is possible to find some kind of transform of the distribution of $M_n$. Well, not exactly. The Spitzer identity involves the expressions $M^+_n = \max_{0\le k\le n} S_k$, where $S_0 = 0$, $S_k = X_1 + \dots + X_k$, $k\ge 1$. So this translates to the positive part of expression you are interested in. But it is possible to ...


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

Note that your limit is $\lim_{n\to\infty} n \, c$. If $c \ne 0$, the sequence diverges, so the limit cannot be $c$.



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