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8

The problem is the claim that all $x^2$ values between $0$ and $4$ are equally likely. That's just not true, except in the sense that each one of them has probability zero. But in any more reasonable sense, it fails. In particular, if all were equally likely, then the probability that the square was between 0 and 2 would be the same as the probability that ...


8

Write $A \setminus B$ as $A \cap B^{c}$.


5

"$\Leftarrow$" is obvious. For "$\Rightarrow$" we set $\sigma_k^2 = \mathbb{E}(X_k^2)$, $$M_n := \sum_{k=1}^n X_k \qquad \quad A_n := \sum_{k=1}^n \sigma_k^2.$$ Using the independence of the random variables, it is not difficult to see that $(M_n^2-A_n)_{n \geq 0}$ is a martingale (with respect to the canonical filtration). For $r>0$ we define a stopping ...


4

You are on a good start. We already know that $\mathbb{P} (A) \in \{0, 1\}$, but how do we decide which one? First remark: a sequence $(X_n)$ is in $A^c$ if and only if its sum is bounded from above, i.e., if and only if there exists $N \in \mathbb{N}$ such that $S_n \leq N$ for all $n$. Now, let's use the symmetry of the random walk. The map $(X_n)_{n \in ...


4

If $X$ has density $f$ with respect to $\mu$, i.e. $$X(P)(A) = \int_A f \, d\mu, \qquad A \in \mathbb{E},\tag{1}$$ then this implies $$\int g \, dX(P) = \int g \cdot f \, d\mu \tag{2}$$ for any measurable $g: X \to [0,\infty)$. Proof: Let $g= 1_A$ be an indicator function. Then, by $(1)$, $$\int 1_A \, dX(P) \stackrel{\text{def}}{=} X(P)(A) ...


4

A probability space is a triplet $(\Omega,\mathcal F,P)$ where $\Omega$ is a set, $\mathcal F$ a sigma-algebra on $\Omega$, and $P$ a probability measure on $(\Omega,\mathcal F)$. Thus, a probability is the third element in a triplet defining a probability space. Example: To model the throw of a die, one could use $\Omega=\{1,2,3,4,5,6\}$, $\mathcal ...


4

They don't mean you literally can't win the lottery, they just mean that the expected value of buying a ticket is negative, or that winning is so unlikely that buying a ticket is a bad investment. However, you should read How Not to be Wrong by Jordan Ellenberg, which (among other things) tells the story of some students from MIT who noticed that some ...


4

The random variables $X_n := 1_{(0,1/n)}$ are not independent. This follows from the fact that $$\lambda(X_n=1,X_m=1) = \lambda(X_m=1) = \frac{1}{m} \neq \frac{1}{n} \frac{1}{m} = \lambda(X_n=1) \lambda(X_m=1)$$ for any $m \leq n$. (Here $\lambda$ denotes the Lebesgue measure restricted to $(0,1)$.) Consequence: The assumptions of the Borel-Cantelli Lema ...


3

Because the line $x_i=x_j$ has Lebesgue measure zero in the $(x_i,x_j)$-plane hence the integral of every function on it is zero.


3

Recall the following statement from probability theory: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent random variables. Then $Y:=\limsup_{n \to \infty} Y_n$ is constant almost surely. Fix $t \geq 0$ and a sequence $(t_n)_{n \in \mathbb{N}}$, $t_n \neq t$, such that $t_n \to t$. By the continuity of the process $(X_s)_{s \geq 0}$, ...


3

$B \cup A=B$ means that $A$ is a subset of $B$. Now we will "disjointify": Let $C=B - A$. Then $A$ and $C$ are disjoint, and $A \cup C=B$. So, $P(B)=P(A \cup C)=P(A)+P(C) \geq P(A)$ as $P(C) \geq 0$


3

Assume that $\sum\limits_{x\in I}\delta_x$ is finite, then $n\sum\limits_{x\in I}\delta_x\geqslant\sum\limits_{x\in I}n\delta_x\mathbf 1_{n\delta_x\geqslant1}\geqslant\#\{x\in I\mid n\delta_x\geqslant1\}$. Hence, every set $\{x\in I\mid n\delta_x\geqslant1\}$ is finite, which implies that their union $\{x\in I\mid \delta_x\ne0\}$ is at most countable.


3

Hint I: $\quad\mathsf P(X\!=\!-1)+\mathsf P(X\!=\!1)+\mathsf P(X\!=\!3) = 1$ Hint II: $\quad\mathsf P(X+Y\!=\!2) = \mathsf P(X\!=\!1, Y\!=\!1) +\mathsf P (X\!=\!-1, Y\!=\!3)$ Hint III: Independence means what? $$\;\\[2ex] \begin{bmatrix} ~ & -1 & 1 & 3 \\ 1 & 0 & 2 & 4 \\ 2 & 1 & 3 & 5 \\ 3 & 2 & 4 & 6 ...


3

If you develop the product $(0.2+0.5+0.3)\times(0.3+0.4+0.3)$, you will get nine terms corresponding to the probabilities of occurrence of all pairs of values from $X$ and $Y$, each pair giving a certain sum. Grouping the terms with equal sums and adding their probabilities, you will get the requested distribution. Hint: there are $7$ distinct sums.


3

Notice that in the theorem, Exercise 1.3.3 from Klenke's 2nd edition, you need to show that $$\mu_n(A)\to\mu(A)\ \ \ \ \mathbf{\text{ for all }}\ \ \ A\in\mathcal{A}$$ But we have only seen $\mu_n(A)$ converging for $A=(a,b]$. $$\mu_n((a,b])\to\begin{cases}0&\text{ if }a>0\\1&\text{ if }a=0\end{cases}$$ When checking the convergence for other ...


3

We need to show that $\left\{Y_n\right\}$ is uniformly integrable, given that $\left\{X_n\right\}$ and $\left\{Z_n\right\}$ are each uniformly integrable. To do this, there are two defining properties that need to be checked. In what follows, I assume that the random variables are all defined on the same probability space. Boundedness: Firstly, we show that ...


3

You're a beginning in probability? In that case, you shouldn't be learning about probability spaces, I think. You should be learning about Kolmogorov axioms, counting techniques, distribution functions, Bayes' Theorem, etc. Anyway, a probability space is often denoted $(\Omega, \mathfrak{F}, P)$, a triple of three things: $\Omega$ is the set of possible ...


3

I am at a loss for an inequality that ends up being free of $n$. However, in my attempts I came up with what appears to be a counterexample to what you are trying to prove. Assuming that $0\leq X_i\leq 4n^2$ is not a typo: Let $\mu_n=E[S],p_n:=P(S\geq n^2/2), \;U:=E[S|S\geq n^2/2,\mu_n>n^2],\; L:=E[S|S<n^2/2,\mu_n>n^2]$ We know that: ...


3

Sure. Take $g = \dfrac 1 \lambda f_X(x), h = \lambda f_Y(y)$, for some constant $\lambda > 0$ (other than $\lambda = 1$, of course).


3

If $D=1$, which has probability $\frac 1k$, the chance a ticket is picked is $\frac 12$. If $D=2$, also probability $\frac 1k$, the chance a ticket is picked is $\frac 1{2^2}$ The total probability a ticket is picked is then $$\frac 1k\sum_{i=1}^k \frac 1{2^i}=\frac 1k\left(1-\frac 1{2^k}\right)$$


3

The Borel $\sigma$-algebra on the reals is the smallest $\sigma $-algebra that contains all the open sets. It's a very large class of subsets of reals and it is very difficult to understand in its entirety. Luckily, you don't need to understand each and every Borel set. It suffices to be able to reason about certain sets that interest you. So, the interval ...


3

$A_n \uparrow A$ means $A_m \subseteq A_n$ for $m \leq n$ and $A = \bigcup_{n \in \mathbb{N}}A_n$. Similarly, $A_n \downarrow A$ means $A_m \supseteq A_n$ for $m \leq n$ and $A = \bigcap_{n \in \mathbb{N}} A_n$.


3

The two uparrows $\uparrow$ have different meanings. $A_n \uparrow A$ means that $A_n$ is an increasing sequence of sets, i.e. $A_n \subseteq A_{n+1}$, and the countable union $\cup_{n \in \mathbb{N}} A_n = A$. $P(A_n) \uparrow P(A)$ means that $P(A_n)$ is an increasing sequence of real numbers, i.e. $P(A_n) \leq P(A_{n+1})$. Since $P(E) \leq 1$, such a ...


3

I would say "Yes" to all your questions. The so-called weak* topology is an example of a weak topology. Weak convergence of measures can be considered a weak* topology. To clarify a new example of convergence of this kind, you should master that $\sigma (\mathcal{F}, X)$ notation. (I think it goes back to Mackey, and was popularized by Bourbaki.)


3

Hint: Applying the inequality $$\sum_{n \geq 1} \mathbb{P}(|Y| \geq n) \leq \mathbb{E}(|Y|)$$ for $Y := \epsilon^{-1} X_1^2$ yields $$\sum_{n \geq 1} \mathbb{P}\left(|X_1|^2 \geq n \epsilon \right)<\infty$$ for any $\epsilon>0$. Now use the Borel-Cantelli theorem and the fact that $$\mathbb{P}(|X_1|^2 \geq n \epsilon) = \mathbb{P}\left(|X_n|^2 \geq n ...


3

Hint: $$\mathbb{P}(|X_n-X|^2 \geq \epsilon) = \mathbb{P}(|X_n-X| \geq \sqrt{\epsilon}).$$


3

$f(0^+)$ or less common $f(0+)$ refers to $$\lim_{\epsilon\searrow 0} f(\epsilon) = \lim_{\epsilon\to0, \epsilon>0} f(\epsilon)$$ i.e. the right-sided limit of $f$ at $0$.


2

Its a common misunderstanding to equate $P(Z)=0$ with the statement " Z cannot happen". If $|\Omega|<\infty$, then $P(Z)=0$ does mean that $Z$ cannot happen. However, if $|\Omega|=\infty$ then we need to use the transfinite numbers to determine the "size" of $\Omega$. In these cases, all we know about events with $0$ probability is that $|\omega \in ...


2

$A_p=\{p,2p,\dots,mp\}$ where $m$ is a nonnegative integer and $mp\leq n<(m+1)p$. This leads to $|A_p|=m=\lfloor np^{-1}\rfloor$ so that: $$P(A_p)=n^{-1}\times\lfloor np^{-1}\rfloor$$ This is valid for any fixed positive integer $p$. It does not have to be a prime.


2

You need to be more careful: The support of $X$ is $(0,1)$; consequently, $-\infty < \log X < 0$, and $\infty > -2\theta \log X > 0$. So the transformation is order-reversing (larger values of $X$ give smaller values of $Y$), and you need to take this into account: $$\begin{align*} \Pr[Y \le y] &= \Pr[0 < Y \le y] = \Pr[0 < -2\theta ...



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