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4

$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$


4

As mentioned by @Nate, the key notion here is called coupling. Simply put, this means that there exists two random objects $G_p$ and $G_q$ defined on a common probability space $(\Omega,\mathcal F,\Pr)$ such that: $G_p$ is distributed as $A_p$ $G_q$ is distributed as $A_q$ $G_p\subseteq G_q$ with full probability Assuming these random objects exist, one ...


4

If $\varphi(y)=E[X\mid Y=y\,]$ is the non-random function of $y$, then $E[X\mid Y\,]$ is defined to be the random variable $\varphi(Y)$.


3

Here is the step I cannot prove for you: The best way of maximising the probability of winning is always to bet the maximum amount of money which is smaller than your goal. The reason behind is that the odds is against you so you should try to win as quickly as possible by making fewest bets possible. The intuition is the more money you bet, the more the ...


3

An answer depends also on your ultimate objective as well as your bet sizing and they are dependent. The objective could be to play until you have doubled your money, play until your adversary is bankrupt, etc. For comparison, if you bet one $1$ unit each round you can use the Gambler's Ruin solution where $M_1+M_2$ is your target: $$P_{ruin} = ...


3

Yes, the statement is true if the random variables are non-negative. Hint: By the triangle inequality and the positivity of the random variables, we have $$(X_n+X)-|X_n-X| \geq 0.$$ Now write $$2 \int X \, d\mathbb{P} = \int \liminf_{n \to \infty} (X_n+X-|X_n-X|) \, d\mathbb{P}$$ and apply Fatous lemma.


3

Yes. You know that the set $B+c$ is the inverse image of the measurable set, $B$, under the continuous function: $f(x)=x-c$, and since continuous functions are measurable, and the inverse image of a measurable set is measurable, $B+c$ is also measurable. Note: there is nothing particularly special about Lebesgue measure here other than it being a Haar ...


2

You have a typo in your CDF: the function you stated is not nondecreasing for all $x \ge 0$. Most likely, you should have a $-0.2$ instead of $+0.2$. I believe there is some misunderstanding of what exactly the solution calculated. I am guessing that the solution wrote $${\rm E}[X \wedge L] = \int_{x=0}^L S_X(x) \, dx,$$ where $L = 1000$ and $S_X(x) = ...


2

If $Z=E[X\mid Y]$ a random variable then it means $F_{Z}(z)=P(Z\leq z)=P(Y\in\{y\in\mathbb{R}:E[X\mid Y=y]\leq z\})$. If you want to think intuitively, it is very simple: to choose a random value for $Z$, first choose $Y$ randomly according to the probability of $Y$, you obtain $Y=y$. Then the value $Z$ takes is $E[X\mid Y=y]$. In fact, this is just a ...


2

There are fundamental differences between $E[X\mid Y=y]$ and $E[X\mid Y]$. You've got the fist one down, in that we are conditioning on a specific event. In the latter case, you are merely told the likelihood of various events, not which ones actually happen. Calculating the distribution of $E[X\mid Y]$ is exactly the same as calculating the distribution of ...


2

The equation $$\lim_{n \to \infty} \frac{1}{\sigma_n \sqrt{2\pi}} \int e^{-x^2/2\sigma_n^2} g(x) \, dx = \frac{1}{\sigma \sqrt{2\pi}} \int e^{-x^2/2\sigma^2} g(x) \, dx$$ shows that $\mu(f_n^{-1}(\cdot))$ converges in distribution to a normal distribution with variance $\sigma^2$. On the other hand, by assumption, $f_n$ converges in $L^2$ to $f$ and this ...


2

(This is not a full answer.) Conjecture: The CDF $F_A$ is the CDF of the random variable $\frac12(X+Y)$, where the random variable $X$ has distribution $P$, the random variable $Y$ has distribution $Q$ and the random couple $(X,Y)$ is maximally coupled (for example, if $P$ and $Q$ are square integrable then $E(XY)$ is maximal). Easy fact supporting the ...


2

You calculated the probability that one page has $3$ errors. The trials are independent so you need to calculate the probability that one page has one error. Then cube it in order to account for three such pages. Finally, you need to take care of the other seven pages. Multiply by the seventh power of the probability of not having one error. That is ...


2

@Matthias M.: Think of the seats like this: _ _ _ _ _ , _ _ _ _ _ _ _ _ _ _ You need 2 in the first five seats $\binom{5}{2}=10$ You have one left for the remaining 10 seats $\binom{10}{1}=10$ $$ \frac{\binom{5}{2}\binom{10}{1}}{\binom{15}{3}}=\frac{20}{91}$$


2

Here is an alternative proof: By the Markov property, we have $$\mathbb{E}^{x}\big( \mathbb{E}^{X_t} g(X_s) \big) = \mathbb{E}^x g(X_{s+t})$$ for any $s,t \geq 0$. Applying Fubini's theorem therefore yields $$\mathbb{E}^{x}(R_{\alpha}g(X_t)) = \int_0^{\infty} e^{-\alpha s} \mathbb{E}^x g(X_{s+t}) \, ds.$$ Consequently, we find $$\begin{align*} ...


2

The point in doing so in general is, that $x^* = \left<\cdot, x\right>\colon X \to \mathbb R$ is a continuous linear functional. Let $\xi \colon \Omega \to X$ be a random variable, we will show that $\def\E{\mathbb E}\E\def\<#1>{\left<#1\right>}x^*(\xi) = x^*(\E \xi)$ holds if $\E\|\xi\|$ exists. We start with a characteristic $\xi = ...


1

The CLT is indeed applicable here as an approximation to the CDF of a sum of binomial random variables (in particular, the Lindeberg-Feller CLT) as long as we assume that the $\mathrm{Var}(X_i)=p_i(1-p_1)n_i$ are bounded from above or grow slow enough that individual variances become small compared to the variance of the entire sum (see Lindeberg Condition). ...


1

Let $p_{jk}=2k/j^2$ be the one step transition probability. Then from state $m+1$ we can only get to state $m$ if we jump there in the next step, so $W_{m+1,m}=p_{m+1,m}=2m/(m+1)^2.$ Note that this expression satisfies the general formula for $W$ so we can start the induction here and continue to $W_{m+2,m}, W_{m+3,m},...$ Starting from state $j$ we ...


1

Start with the formula $$E^x\left[R_\alpha g(X_t)\right]=\alpha\int^\infty_0e^{-\alpha s}\int_t^{t+s}E^x[g(X_v)]dv\,ds.$$ Plugging in $t=0$ we get $$R_\alpha g(x)=\alpha\int^\infty_0e^{-\alpha s}\int_0^{s}E^x[g(X_v)]dv\,ds.$$ Subtracting and dividing by $t$ gives, $${E^x\left[R_\alpha g(X_t)\right]-R_\alpha g(x)\over t} = \alpha\int^\infty_0e^{-\alpha ...


1

I guess you mean the cumulative distribution function $F(k;n,p)$ of the (discrete) binomial distribution with number of trials $n \ge 0$ and success probability $0 \le p \le 1.\;$ The probability mass functions (PMF) is $$f(k;n,p) = \binom{n}{k}p^k(1-p)^{n-k} $$ and the CDF can be expressed with normalized incomplete Beta function $$F(k;n,p) = ...


1

Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have: $F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$. What distribution has this CDF?


1

$$ Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\ $$ The last equality is from the definition of the quantile function.


1

Is that correct? No, first because there is no proof here, second because in some settings the boundary case is positive recurrent, respectively transient.


1

a) If $\rho<1$, then $X$ is positive recurrent. Let's assume I have proved a) Can I say the following: $X$ is positive recurrent if and only if $\rho<1$ No, you cannot - you have only proved that $\rho <1$ is a sufficient condition for the positive recurrence of $X$, but you have not established its necessity. So you can say, $X$ is ...


1

Let's call the persons 1,2,3 and assume they enter in that order. The case you want to look at are 1-2 or 1-3 or 2-3 in section A while 3 in B or C, 2 in B or C, 1 in B or C. It results that you don't even need to differentiate between B and C, so let's call it BC with 10 seats. Case 1-2 in A, 3 in BC: 5*4*10 (1 has 5 seats, B has 4 seats available, C 10 ...


1

No. $p(3)$ is the probability of one page containing $3$ errors, not the probability of three pages containing one error each.


1

From this previous question(pdf of the difference of two exponentially distributed random variables) they derive for $Y\sim Exponential(\lambda)$ and $X\sim Exponential(\mu)$ letting $Z=X-Y$ a CDF of $Z$ $$P(X<Y)=P(X-Y<0)=P(Z<0)=F_{Z}(0)=1-\frac{\lambda}{\lambda+\mu}e^{-\mu (0)}=1-\frac{\lambda}{\lambda+\mu}=\frac{\mu}{\lambda+\mu}$$ in your ...


1

Suppose $(E_i)_{i\geq 1}\subseteq L$ is a sequence of pairwise disjoint sets from $L$ and let $E=\bigcup\limits_{i\geq 1} E_i$. Then $E\in \mathcal{F}_1\otimes \mathcal{F}_2$ and $$ \{y: (x,y)\in E\}=\bigcup_{i\geq 1}\,\{y:(x,y)\in E_i\} $$ with the RHS being a pairwise disjoint union. Thus $$ \nu_E(x)=\sum_{i\geq 1}\nu_{E_i}(x) $$ which is measurable since ...


1

If you consider $$f(x,y)=\frac{2}{3}(x+2y){\bf 1}_{[0,1]\times[0,1]}(x,y)$$ then this will answer your question.


1

As pointed out by saz, it is true if the random variables are non-negative. In general, it is not true: take $X_n$ a random variable taking the value $0$ with probability $1-n^{-2}$, $n^3$ with probability $1/(2n^2)$ and $-n^3$ with probability $1/(2n^2)$. Then $X_n\to 0$ almost surely, $\mathbb E[X_n]=0$ but $\mathbb E|X_n|=n$.



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