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10

If we divide every roll by $n$, rolling the die and dividing by $n$ approximates the uniform distribution on $[0,1]$ for arbitrarily large $n$. We then are looking for the expected number of samples from a uniform distribution required to get a sum above $1$. For any integer $k \in \mathbb{N}$, Let $X_1, \ldots , X_k, \ldots$ be the random variables in ...


8

Here is a proof of the first question using Zorn's lemma, and a counterexample to the second question using an ultrafilter. (So both cases used some form of the axiom of choice!) Theorem (Sierpinski): For a non-atomic probability space $(\Omega, \mathcal{F}, \mu)$, $\mu$ is surjective onto $[0,1]$. Proof: Let $x \in [0,1]$, and let $$ \mathcal{G} = ...


6

Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega_1,F_1,P_1)$ and let $(\Omega_2,F_2,P_2)$ be an arbitrary probability space. Define a new probability space $(\Omega,F,P)$ by $$\Omega := \Omega_1 \times \Omega_2 \qquad F := F_1 \otimes F_2 \qquad P := P_1 \otimes P_2.$$ If we set $$\tilde{B}_t(\omega_1,\omega_2) := B_t(\omega_1) ...


6

If I toss a coin an infinite amount of times, can I be sure to get an infinite amount of heads? According to the Borel-Cantelli lemma, since each coin toss is an event of probability $\frac12$ and a sum of $\frac12$ diverges, the probability of $\limsup_{n\to\infty}\{\text{heads at $n$-th flip}\}$ is 1. But the $\limsup$ is precisely the event of ...


5

The output of a function is uniquely determined by its input. However, for a measurable function as a random variable, we don't know the input: only after a realization can we know $x$.


4

The second statement is false. Suppose you flip a fair coin, $X$ is the indicator of "heads" and $Y$ is the indicator of "tails". Then they are both identically distributed but if you let $K$ be the event "heads", their integrals over $K$ are clearly not equal. In fact, the equality for all $K$ holds iff $X=Y$ almost surely, which is a far stronger ...


4

The philosophical setup is that $\Omega$ is a sample space, $\mathscr{F}$ a collection of events that could happen. A measurable $X$ is called a random variable because we are pretending that we do not know which $\omega$ will occur. I am going to flip a coin and let $X$ be the indicator of heads. I model it as $X(\omega_1) = 1$ and $X(\omega_2) = 0$ (i.e. ...


4

The CDF $F$ of a random variable $X$ is unique, since it is defined as $$F(x) = P(X \le x).$$


4

Let $x\leq x^\prime$ be two continuity points of $G$. If there were a subsequence with $a_{n(k)}\leq 0$ for all $k$, then $a_{n(k)}x^\prime +b_{n(k)}\leq a_{n(k)}x +b_{n(k)}$ and hence $F_{n(k)}(a_{n(k)}x^\prime +b_{n(k)}) \leq F_{n(k)}(a_{n(k)}x +b_{n(k)}).$ Letting $k\to\infty$ gives $G(x^\prime)\leq G(x)$ so that $G$ is a non-increasing function, ...


4

Set $$\begin{align*} A &:= \left\{ \limsup_{n \to \infty} \frac{S_n}{n} = \infty \right\} \\ B &:= \left\{ \liminf_{n \to \infty} \frac{S_n}{n} = - \infty \right\}. \end{align*}$$ Hints: By symmetry, we have $\mathbb{P}(A) = \mathbb{P}(B)$. By Hewitt-Savage's 0-1-law, $\mathbb{P}(A)=\mathbb{P}(B) \in \{0,1\}$. Deduce from the first two steps and ...


4

The two versions are equivalent. By the law of total probability, $P(A) = P(A,B) + P(A,B^c) = P(A|B)P(B) + P(A|B^c)P(B^c)$. The second version is more compact (and arguably gives a better intuition as to what the theorem means), but when you go to evaluate $P(A)$ you often have to decompose it as in the first version, so its application is more direct.


4

The field of probability can be made mathematically rigorous. Introductory textbooks on probability tend to use terminology like 'sample space', 'outcome', 'event', and outcomes are given names like 'Heads' or 'HTT' or 'King of Spades', all in an attempt to keep things informal and intuitive. These texts often refrain from defining such concepts precisely, ...


4

EDIT: New question is a question about conditional probability. The trick is to apply Bayes' theorem. That is $P(A \mid B) = P(A)P(B\mid A)/P(B)$. So say event $A$ is picking the two tails coin. And say event $B$ is seeing a tails. Then $P(A) = \frac12$ and $P(B) = \frac34$ and $P(B\mid A) = 1$, because we're guaranteed to see tails on the two tails coin. So ...


3

It doesn't. A measure $\mu$ is absolutely continuous with respect to another measure $\mu$ if $\mu(A) = 0$ implies $\nu(A) = 0$. However, the Radon-Nikodym theorem doesn't hold if you remove the sigma-finiteness of the two measures [cf. here]. Note that there are generalizations that weaken the assumption of sigma-finiteness, but you can't simply remove ...


3

The full equation reads: $$P(\text{rain}) = P(T>25)\cdot P(\text{rain if } T>25) + P(T\le 25)\cdot P(\text{rain if } T\le 25).\tag 1$$ The problem statement explicitly contains: The temperature at noon is equally likely to be above 25°C, or at/below 25°C. That is, $P(T>25) = P(T\le 25)$. Since $P(T>25) + P(T\le 25) = 1$, it immediately ...


3

Hints: Fix $x \in [a,b]$. Instead of $\mathbb{E}_{X_0=x}$ I will use the notation $\mathbb{E}^x$. Suppose that $f \in C_b^2$ solves the differential equation $$x f'(x) + \frac{\sigma^2}{2} f''(x)=0. \tag{1}$$ Show, using Itô's formula, that $(f(X_t))_{t \geq 0}$ is a martingale. Define $$f(x) := \Phi \left( \frac{2}{\sigma} x \right)$$ where $$\Phi(x) := ...


3

This is not true if the $X_i$ can be negative : take $n=1$, $x=-2$ and $X_1 = -1$. Then $$x+ X_{1}\Bbb 1_{X_{1}>x}= -2+(-1) = -3 < -2 = X_{(1)}$$ It is also not true if $x$ can be negative : take $n=1$, $x=-1$ and $X_1 = 1$ Then $$x+ X_{1}\Bbb 1_{X_{1}>x}= -1+1 = 0 < 1 = X_{(1)}$$ Now, if all the $X_i$ are positive, and that $x\geq 0$, ...


3

For $x \in \mathbb{R}_+$ $$\max_{1\le i\le n}\{|X_i|\}=\max_{1\le i\le n}\{|X_i|1\{|X_i|\le x\}+|X_i|1\{|X_i|> x\}\}$$ $$\le \max_{1\le i\le n}\{x+|X_i|1\{|X_i|> x\}\}=x+\max_{1\le i\le n}\{|X_i|1\{|X_i|> x\}\}$$ $$\le x+\sum_{i=1}^n |X_i|1\{|X_i|> x\}$$ If $X_i\ge 0, i=1,\dots,n$ then $X_i=|X_i|$ and $$\max_{1\le i\le n}\{X_i\}\le ...


3

This is an interesting fact about exponential random variables: if $X$ and $Y$ are independent exponential random variables, say both with mean $\lambda$, the the mean of $\min \{ X,Y \}$ is $\frac{\lambda}{2}$. More generally, if they have means $\lambda_1$ and $\lambda_2$, then the mean of $\min \{ X,Y \}$ is $\frac{1}{\frac{1}{\lambda_1}+\frac{1}{\lambda_ ...


3

Consider the following example: $P\{Y=1\}=1$ and $P\{X=0\}=1/3, P\{X=2\}=2/3$. So, $$P\{X\ge Y\}=\frac{2}{3}$$ However, $$\frac{1}{3}=P\left(X\le\frac{1}{2}\right)\ge P\left(Y\le\frac{1}{2}\right)=0$$


3

No, the converse is, in general, not true. Just consider $((0,1),\mathcal{B}(0,1))$ endowed with Lebesgue measure and the sequence of random variables $$X_n(x) := 2n \cdot 1_{(0,1/n)}(x).$$ Then $$\mathbb{E}(|X_n|) = 2$$ for all $n \in \mathbb{N}$, but $$\limsup_{M \to \infty} \sup_{n \in \mathbb{N}} \int_{|X_n|>M} |X_n| \, d\mathbb{P} \geq ...


3

The bound 2 is arbitrary in the sense that it can be replaced by any real number larger than 1 - the real requirement is that you set some bound on the summed conditional variances of the array. Note that you already assume that $V_{n,nt} \xrightarrow{P} t$ - what does this condition entail on the variances? One reason that this is allowed in general is that ...


3

The assumption underlying Benford's law is that since the physical units are arbitrary, the distribution of physical constants should be invariant under multiplication by constant, which corresponds to a change of units (though no such distribution can really exist). Another way of stating this is that the distribution of the logarithm of a physical ...


3

Clearly $F'(x)=f(x)$ (wherever CDF is continuous). Substitute $F(x)=t$. Then $dt=f(x)dx$. $$\int_{-\infty}^a F(x)~f(x)~dx \rightarrow \int_{0}^{F(a)} t~dt= \frac{(F(a))^2}{2}$$


3

Take an element $i$. Let $p_i$ denote the probability that $i$ is not in the intersection of your two randomly chosen sets. It is in exactly $2^{n-1}$ subsets, hence the probability that it is in a randomly chosen subset is $\frac 12$. It follows that the probability that it is in both of two randomly chosen subsets is $\frac 14$. Therefore $p_i =\frac ...


3

The condition ensures that the integral $$\int_{\mathbb{R} \backslash \{0\}} (e^{iux}-1-iux 1_{\{|x|<1\}}) \, \nu(dx)$$ is well-defined. Since $|e^{iux}| \leq 1$, we have $$|e^{iux}-1| \leq 2$$ for all $x \in \mathbb{R}$. On the other hand, applying Taylor's formula $$f(x) = f(0)+f'(0) x + \frac{f''(\xi)}{2} x^2$$ (here $\xi \in (0,x)$ is some ...


3

The integral is defined as the supremum of functions of bounded functions with finite support. I would work directly from this definition to show the result. If your function had been bounded and of finite support, then your above analysis is almost complete. Each $f_n$ would be composed of a finite set of functions that as $n$ goes to infinity, ...


3

Given $\varepsilon > 0$, since $X_n \to X$ in distribution and $P(|X| < \infty) = 1$, it can be shown that $\{X_n\}$ is bounded in probability, that is, there exists $M > 0$ such that $$P(|X_n| > M) < \frac{\varepsilon}{2}$$ for all $n \geq 1$. Thus for every $c > 0$, by condition, there exists $N \in \mathbb{N}$ such that for all $n > ...


3

Hint: use the Borel-Cantelli lemma to show that $$P(X_n \ne 2 \text{ i.o.}) = 0.$$ (Independence is not needed,)


3

This is to establish the following for Borel set $B\subseteq [0,1]$, $$\begin{align} \mathsf P(Y\in B) & = \iint_{Y^{-1}(B)} f_{X,Y}(x,y) \mathrm dx\mathrm dy \\ & = \iint_{[0;1]\times B} f_{X,Y}(x,y) \mathrm dx\mathrm dy \\ & = \int_B \int_0^1 (x+y) \mathrm dx\mathrm dy \\ & =\int_B (y + \frac 1 2 ) \mathrm dy. \end{align}$$ Thus, the ...



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