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6

$\ln\left(\int_0^1 x\,dx\right)=\ln\left(\frac{1}{2}\right),$ but $\int_0^1 \ln x\,dx=-1$. So no, it's not that simple.


5

The intuition is that in an optimal strategy, the picker should be indifferent to what the guesser chooses. Suppose we just take $n=3$ for simplicity. Suppose the picker chooses $1$ with probability $p_1$, chooses $2$ with probability $p_2$, and $3$ with probability $p_3$. The selection of $p_1, p_2, p_3$ constitutes the picker's strategy. The ...


4

Some quick calculations, to showcase (presumably) what the OP has done. For the first game, the probability that the game ends on the $k$th turn is $\frac{(k-1)!}{(k+1)!}=\frac{1}{k(k+1)}$. Thus when $X$ is the number of turns, $$E(X)=\sum_{i=1}^\infty\frac{1}{i+1}=\infty$$ which is known to diverge. For game two, the probability that the game ends on the $...


4

If $E[\phi(X)]=E[\phi(Y)]$ for every $\phi$, the most you can conclude is that $X$ and $Y$ have the same distribution. This follows from taking $\phi(x) = I(x\in B)$ as $B$ ranges over an appropriately large collection of sets. To see why $X=Y$ need not be true, even if $X$ and $Y$ are defined on the same space, consider $X$ uniformly distributed on $[0,1]$ ...


3

For example, take $X_n$ s.t. $P(X_n = 2^n) = \frac{1}{2^n}$, $P(X_n = 0) = 1 - \frac{1}{2^n}$. Then a.s. all but finitely many $X_n$ are zeroes, so $\sum X_n$ converges a.s., but $\mathbb E (X_n) = 1$, so series of mean diverges.


3

If no such $\omega$ exists then $Y$ prescribed by $\omega\mapsto X(\omega)-\mathbb EX$ is a negative random variable. Then on base of $P(\bigcup_{n=1}^{\infty}\{Y<-\frac1n\})=P(\Omega)=1$ it can be shown that some $n\in\mathbb N$ exist such that $P(Y<-\frac1n)>0$ which leads in combination with the fact that $Y$ is negative to the conclusion that $...


3

$W(1)$ is a random variable and you cannot just suppose that it is equal to $a$. In fact, you know its distribution. If $W$ is a standard Brownian motion then $W(t) \sim \mathcal{N}(0,t)$ for every $t$. In particular, $W(1) \sim \mathcal{N}(0,1)$ which is why $\mathbb{E}[W(1)] = 0$.


2

Simple mistake you made: $Var(Z)=E[Z^2]-(E[Z])^2$ (and not with a $+$). So, $E[Z^2]=Var(Z)+(E[Z])^2$.


2

You can do this with any math software (such as Mathematica) or any reasonable language. I used Python. I attach a picture of the code, which you should try translating into your language / software of choice. There are three cells. The first loads the required required mathematical module ("numpy") and a plotting module. It also defines your transition ...


2

Since all the terms being summed are positive, Fubini's theorem allows you to shuffle them, so that $$ \sum_{y = - \infty}^{\infty} \sum_{x = -\infty}^{\infty} xP(X = x| Y = y)P(Y=y) = \sum_{x = - \infty}^{\infty} \sum_{y = -\infty}^{\infty} xP(X = x| Y = y)P(Y=y) $$ Then the innermost sum doesn't depend on $x$ and you can pull it out without worry. It's ...


2

Note that $$ X_{n-1} Y_{n-1} = (X_{n-1} - X_n) Y_{n-1} + X_n Y_{n-1} $$ and $$X_{n-1} Y_{n-1} = X_{n-1} (Y_{n-1} - Y_n) + X_{n-1} Y_n. $$ For the right-hand side of the first equation we can use \begin{align*} E[(X_{n-1} - X_n) Y_{n-1}] &= E[E[(X_{n-1} - X_n) Y_{n-1}| F_{n-1}]] \\ &= E[Y_{n-1}\ E[(X_{n-1} - X_n) | F_{n-1}]] \\ &= 0, \end{align*}...


2

Just as you did it: \begin{align} \sum^\infty_{n=1}(2n-1)P(X\geq n) &=\sum^\infty_{n=1}(2n-1)\Big(\sum^\infty_{j=n}P(X=j)\Big)\\ &=\sum^\infty_{j=1}\sum^j_{n=1}P(X=j)(2n-1)\\ &=\sum^\infty_{j=1}P(X=j)\Big(\sum^j_{n=1}(2n-1)\Big)\\ &=\sum^\infty_{j=1}P(X=j)j^2 \end{align} The last line follows from $\sum^j_{n=1}(2n-1)=2\frac{j(j+1)}{2}-j$


2

You are dealing with a probability space $(\Omega, \mathcal{F}, \mathbb{P})$; and your random variable $X$ is a measurable function $X\colon \Omega\to S$, from the sample space to the state space. Then, $\omega$ is just an element of $\Omega$. $X(\omega)$ is the realization of the random variable, $\mathbb{P}(\omega)$ is the probability of $\omega$.


2

$\omega\ $ stands for an outcome, i. e. for an element from the sample space $\Omega$.


2

If $(X,Y,Z)$ is uniformly distributed in $T=\{(x,y,z)\in[0,1]^3 : x+y+z=1\}$, then $(X,Y)$ is uniformly distributed in $T'=\{(x,y)\in[0,1]^2 : x+y\leqslant 1\}$, and $Z=1-X-Y$ a.s., thus $$\mathbb{E}f(X,Y,Z)=2\iint_{T'}f(x,y,1-x-y)\,dx\,dy.$$ A simplification by symmetry, $\mathbb{E}\max\{X,Y,Z\}=6\,\mathbb{E} Z[X\leqslant Y\leqslant Z]$, gives $$\mathbb{E}\...


2

Either you get a $1$ the first time with probability $\frac1{100}$, taking just one attempt, or you get a $0$ the first time with probability $1-\frac1{100}$ and then start over again, taking one attempt plus the number of attempts you would expect when restarting (this latter value being the same as the expected number when originally starting) So to find ...


2

First of all, note that Erfc does not denote the Error function but its "complement", i.e. $$\DeclareMathOperator{\erfc}{Erfc}\DeclareMathOperator{\erf}{Erf} \erfc(x) := \frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp (-y^2) \, dy = 1- \erf(x)$$ where $$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp (-y^2) \, dy$$ is the Error function. Since $$|2(-k)+1| = |...


2

Hint: When faced with a random variable $X$ and an event $A$, the following may help: $$E[X|A] = \frac{E[X1_{A}]}{P(A)}$$ where $1_{()}$ is the indicator function. In your case replace $X$ with $f(X,Y)$ and $A$ with $\{y \in \mathcal{Y}\}$. Let me know if you need more help.


2

No. As a counterexample, take $f(x)=e^x$.


1

Using a bit more machinery. First note that $$ X^2=\sum_{n=1}^\infty(2n-1)I(X\geq n)\tag{0} $$ with probability one where $I$ is the indicator function. To see that this identity is true note that when $X^2=k^2$ (so $X=k$) for an integer $k\geq 1$ the RHS is $$ \sum_{n=1}^k(2n-1)=\sum_{n=1}^k [n^2-(n-1)^2]=k^2 $$ as desired. Take expectations of $(0)$ (use ...


1

Let's try to put some order here: the general Expected Value of a continuous random variable $Z$, defined over a set $S_{Z}$ is given by: $$\mathbb{E}(Z) = \int_{S_{Z}}zp(z)dz$$ where $p(z)$ is $Z$'s density function. It is immediate to see that the best estimator for such Expected Value is the sample mean: $$\overline{Z} = \frac{1}{L} \sum_{l = 1}^{L}Z^{...


1

We first need to determine the range of possible passwords. The lowercase letters are in known positions and the w can be put in any one of those $5$ positions. There are then (since all characters are different) $25×24×23×22$ ways to assign the remaining lowercase letters. For the uppercase letters and digits there are $3$ positions where the uppercase ...


1

The antiderivative of $f(x)$ is $F(x)+c \;\; \forall c \in R\;$ and in this case $c=-1$ in order to get the proof. You have to integrate $xf(x)dx$ from $a$ to infinity so $x\geq a$, then $\int_a^{+\infty} xf(x)dx \geq \int_a^{+\infty}af(x)dx=a(1-F(a))$ $\int_0^{+\infty} xf(x)dx \geq \int_0^{a}af(x)dx+a(1-F(a))\;\;\; \forall a \geq 0$, so letting a tend to ...


1

Assuming you sample without replacement, and by "according to the weights" you mean that at the $j$-th draw, an item of weight $w$ is sampled with probability $\frac{w}{\sum_k n_k^{(j)} w_k}$, where $n_k^{(j)}$ is the number of items of weight $w_k$ at the moment the draw is made, then your variable follows Wallenius' noncentral hypergeometric ...


1

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1

The expected value is $\sum_{X=0}^{10}\binom{10}{X}Xp^{X}(1-p)^{10-X}=10p\sum_{Y=0}^9\binom{9}{Y}p^Y(1-p)^{9-Y}=10p$. by letting $Y=X-1$


1

The outlined solution is a standard technique in the theory of Markov chains, and to fully appreciate it with mathematical rigor, we need a deeper look. The main idea is that $X$ is completely determined by knowing the entire history of coin flips. More precisely, let $f$ be a function whose input is an entire history of coin tosses $y = (y_n)_{n\geq 1} \in ...


1

Let's rewrite the distribution of $N'=X_{(n)}$ a bit more precisely: $$P(N'=j)=\begin{cases}\dfrac{\binom{j-1}{n-1}}{\binom{N}{n}}&,\text{ if }j=n,n+1,\ldots,N \\ \\\quad 0&,\text{ otherwise }\end{cases}$$ Then we have \begin{align} E(N')&=\sum_{j=n}^N jP(N'=j) \\&=\frac{n}{\binom{N}{n}}\sum_{j=n}^N \frac{j}{n}\binom{j-1}{n-1} \\&=\...


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