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27

I think this is a slightly more intuitive way of looking at the question: Suppose $f (x)=k$. Once we have chosen $k$, there are $200$ possible values for $g (x)$, one of which is $k$, hence we get an answer of $\frac{1}{200}$.


13

is it possible that the event has in fact zero probability? Yes. For a situation where this always happens, assume that one observes a random number $x$ drawn from the uniform distribution on $(0,1)$. Then the probability to observe $x$ is zero. (Proof: For every interval $I\subseteq(0,1)$ the probability to observe a number in $I$ is the length of ...


12

Let me get you started: if $(X,Y)$ is i.i.d., $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$ hence $E(u(Y-X))=E(u(X-Y))$ for every $u$ such that the expectations exist. For $u(t)=t^3$ (or any odd function), this yields $E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3)$ hence... Thus, the answer is "Yes, provided $E(|X|^3)$ is finite" (otherwise ...


12

Assuming the selections are independent and uniform, there are $100 \times 200$ possible equiprobable pairs of which $100$ can be a match, so the probability is ${100 \over 200 \cdot 100 }$.


12

Start with $x_1=0$ and then just let $x_{n+1}=1$ whenever $f(n)/n\leq p$ and $x_{n+1}=0$ whenever $f(n)/n> p$. It's easy to prove by induction that we will always have $\left|\frac{f(n)}n-p\right|\leq\frac 1n$ and so $f(n)/n\rightarrow p$.


11

A direct proof: For any measurable set $A$, applying independence of the events $\{X\in A\}$ and $\{X\in A\}$ gives $$ \mathbb P(X\in A)=\mathbb P(X\in A)\mathbb P(X\in A)\implies \mathbb P(x\in A)\in \{0,1\}. $$ Hence $\mathbb P$ is a dirac measure. Therefore $X$ is a.s. constant.


11

The proof of this result has nothing to do with probability theory in general or with Borel-Cantelli lemmas in particular, it is only plain-old-deterministic real analysis: Let $(x_n)$ denote a real valued sequence such that $0\leqslant x_n\lt1$ for every $n$, then $\prod\limits_n(1-x_n)=0$ if and only if $\sum\limits_nx_n$ diverges. Proof: If ...


10

It can be shown that nonnegative random variables $X$ and $Y$ have the same distribution so long as $\mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha]$ is finite for all $\alpha\in(a,b]$, any $0\le a <b$. Setting $U=\log X$ and $V=\log Y$, define the functions $$ f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ...


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 ...


10

It turns out the numerical findings about $\mathbb{E}[N_3]$ by David E is not a coincidence, it is exact! $$\mathbb{E}[N_3] = \frac{1}{1-\sin(1)}$$ Let $X_1, X_2, \ldots $ be a sequence of random variables. For each $n$, we will assume $X_n$ take value from the set $\langle n \rangle \stackrel{def}{=}\{ 1, 2, \ldots, n \}$ with uniform probability. After ...


10

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


9

Assuming that by "number" you mean "integer," and by "random" you mean "uniformly at random," AND ASSUMING THAT THE CHOICES ARE INDEPENDENT, then the desired probability is $$\sum_{k=1}^{200} \Pr[X = k]\Pr[Y = k] = \sum_{k=1}^{100} \frac{1}{100} \cdot \frac{1}{200} + \sum_{k=101}^{200} 0 \cdot \frac{1}{200} = \frac{1}{200}.$$


9

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


9

This is false, when $\lim_{n \to \infty} p_n$ doesn't exist. For instance, consider $p_{2n-1} = \dfrac1{2^n}$ and $p_{2n} = 1-\dfrac1{2^n}$. This gives us $\sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N$, whereas $$\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$ However, the claim is true when $\lim_{n \to \infty} p_n$ exists. ...


8

Since $X_n$ and $X$ are both measurable (they are random variables after all) and the function $f(x,y) = \lvert x-y \rvert$ preserves measurability (due to its continuity), $\lvert X_n - X \rvert$ is measurable as well. Maybe it is better for your understanding if I argue in the following way. If $X_n$ and $X$ are measurable, then so is $X_n - X$. This is a ...


8

Hint: It is $E(X(X-1))=E(X^2)-E(X)$. And $Var(X)=E(X^2)- [E(X)]^2 \Rightarrow E(X^2)=Var(X)+[E(X)]^2$


8

I'll try to give you an example that shows why it is very useful. Suppose that the random variable $X$ follows the binomial distribution with parameters $n$ and $p$. Then the expected value of $X$ is given by $$ \operatorname EX=\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}. $$ But we know that $X=\sum_{j=1}^nY_j$, where $Y_1,\ldots,Y_n$ are independent and ...


8

Convergence in probability does not imply convergence almost surely: Consider the sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ (endowed with Lebesgue measure $\lambda$) defined by $$\begin{align*} X_1(\omega) &:= 1_{\big(\frac{1}{2},1 \big]}(\omega) \\ X_2(\omega) &:= 1_{\big(0, ...


8

Use the strong law of large numbers. Choose $B\in M$ so that $\lambda(B)\neq \mu(B)$, and consider the disjoint sets $$\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \lambda(B)\right\}\mbox{ and }\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \mu(B)\right\}.$$


8

You have some event, which you typically don't know when occurs, but that can/will occur some time in the future. The time that this event occurs is random, and it is a stopping time if, at any point in time, you know whether the event has occurred or not. A few quick examples. 1) Your own (a stopping time): Let $\tau$ denote the time that I'm ruined (i.e. ...


8

Since $\mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2], $ observe that $$ \mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0, $$ which implies that $X$ and $Y$ differ on a set of measure zero. For the weaker condition where $X$ and $Y$ are ...


8

Consider two normal distributions with the same variance and different means.


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

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} = ...


7

$\{X_k\}$ cannot be a nonnegative MTG with $X_0=0$. So, I assume it is just a MTG. Using the original Doob's weak inequality, for some $z >-r$ (and noticing that $\mathbb{E}X_n=0$): $$P\{\max_{1\le k\le n}{X_k}\ge r\}\le P\{\max_{1\le k\le n}{(X_k+z)^2}\ge (r+z)^2\}\le\frac{1}{(r+z)^2}\mathbb{E}(X_n+z)^2=\frac{1}{(r+z)^2}(\mathbb{E}X_n^2+z^2)$$ ...


7

For concreteness, assume that we have a biased coin that has unknown probability of landing "heads" and we want to give a $95\%$ confidence interval for $p$. Rrepeat the experiment of tossing the coin $n$ times. Let $\bar{X}$ be the sample proportion of heads. We will use $\bar{X}$ to estimate $p$. Note that $\bar{X}$ has mean $p$ and variance ...


7

Hint: Let $b\in\mathbb{N}$, $b\ge2$, $Y=\{0,1,...,b-1\}$ with fair probabilities assigned, and let $F:Y^\infty\to[0,1]$ be the map defined by $$F(x_1,x_2,x_3,...)=\frac{x_1}b+\frac{x_2}{b^2}+\frac{x_3}{b^3}+..., \text{for all}~(x_1,x_2,x_3,...)\in Y^\infty;$$ that is, $F=\sum_{n=1}^\infty\frac1{b^n}f_n$ where $f_n=\sum_{k=1}^{b-1}k\chi_{A_{nk}}$ and ...


7

It can be proved by induction that $x_n = \frac{1}{(n-1)!}$ $x_1$ and $x_2$ satisfy this, and by doing induction, $$x_{m+1} = x_{m}-\frac{1}{m}x_{m-1} = \frac{1}{(m-1)!} - \frac{1}{m}\frac{1}{(m-2)!} = \frac{1}{m!} $$ Thus you series reduces to the ordinary series $\sum_{n=0}^{\infty}\frac{1}{n!}$ for $e$


7

Note that $$\bigcup_{n=1}^\infty A_n \supset \bigcup_{n=1}^\infty A_n \cap \bigcup_{n=2}^\infty A_n \cap \ldots \cap \bigcup_{n=k}^\infty A_n \cap \ldots = \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n = \limsup_{n\to\infty} A_n$$ And use $A\subset B \Rightarrow P(A) \le P(B)$


7

Let $S_{100} = \sum_{k=1}^{100}X_k$ denote the winnings after 100 bets. Then $$E(S_{100})= 100E(X)= 25$$ and $$var(S_{100})= \sum_{k=1}^{100}var(X)= 100(27/16)\approx 168.75.$$ As a sum of $100$ independent random variables, $S_{100}$ is approximately normally distributed with mean $25$ and standard deviation $\sqrt{168.75} \approx 12.99$. The ...



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