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0

In a deck of cards, for the aces to be equidistribute they need to be located in each of the modular positions. That is to say we need an ace at each of $0\mod4, 1\mod4, 2\mod4$ and $3\mod4$. If we consider a pack of cards with only 2 types of cards - 4 of type A and 48 of type , we can see that there are $\binom{52}{4}$ ways to deal the cards, and we need ...


0

"In the end, only the expectation of Y is needed, not its distribution." This a typical famous university exam question. As Did already mentioned $E(X^n)=0$ for n being odd. Now for $n$ even ($2m$ say and WLOG assume $\sigma^2=1$), $$E(X^{2m})=2\int_0^{\infty} \frac{x^{2m}\exp(-x^{2}/2)}{\sqrt{2\pi}}dx$$ Substitute, $y=x^2/2$ to make the above as a $Gamma$ ...


1

Six ways to get $1,3,6$: $$ \begin{array}{c|c|c|c|c|c} \overbrace{ \begin{array}{rc} \text{first die:} & 1 \\ \text{second die:} & 3 \\ \text{third die:} & 6 \end{array}} & \overbrace{ \begin{array}{rc} \text{first die:} & 1 \\ \text{second die:} & 6 \\ \text{third die:} & 3 \end{array}} & \overbrace{ \begin{array}{rc} ...


0

Let $0$ represent one orientation and $1$ represent the other orientation. Each arrangement of the $n$ magnets then corresponds to an $n$-bit binary string. Two adjacent magnets repel if they are in opposite orientations and attract if they are in the same orientation. Thus, a block of magnets corresponds exactly to a block of identical bits. For $n=8$, for ...


0

The similar and more simple situation is to throw two dice. We have 36 possible outcomes: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 Here for example there are two ...


0

Yes, you can colour the dice, but that actually makes it more, not less clear that $6+3+1$ counts six times and $3+3+3$ counts once: Now you have the six partitions $6R+3Y+1B$, $3R+6Y+1B$, etc., whereas you still have only the one partition $3R+3Y+3B$. There's nothing to rearrange; since the numbers are all $3$, nothing changes if you swap the colour labels ...


2

A closed form expression is provided in the following paper. SV Amari, RB Misra, Closed-form expressions for distribution of sum of exponential random variables, IEEE Transactions on Reliability, 46 (4), 519-522.


3

We have to consider three cases separately: $S_{k-1}(\omega)<0$: Then $S_k(\omega) \leq 0$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$ $S_{k-1}(\omega)=0$: Then $|S_k(\omega)|=1$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$ $S_{k-1}(\omega)>0$: Then $S_k(\omega) \geq 0$ and therefore ...


0

The two random variables are indipendent. Intuitively if x doesn't influence the value of y, so the image of x under f doesn't influence the other variable. If the two variables are independent then E[g(X)*f(Y)] = E[g(X)]*E[f(X)] which is essentially the fubini theorem (remember the the definition of expectation).


1

The result is true, and has nothing to do with topology or countability. Let ${\cal C}$ be the collection of subsets of $E^I$ of the form $\pi_i^{-1}(E)$ for some $i\in I$ and $E\in{\cal E}$. Then use the fact that $X^{-1}(\sigma({\cal C}))=\sigma(X^{-1}({\cal C}))$.


0

To solve question 2 you don't need Markov chains, only first step analysis. In the first step there are 3 possibilities: player $A$ wins, the game starts over, the deck goes to player $B$. Letting $p_A$ be the probability that player $A$ is the eventual winner, and taking these 3 cases into account gives equation (1) below. With $p={10\choose ...


0

The number of blocks is one plus the number of adjacent ends with equal polarities. Start reading the magnets from the left. What is the probability that the $k+1$th magnet is oriented so that the $k$th and the $k+1$th magnet repel each other, given the orientations of the first $k$ magnets?


-1

1 and 4 are right. 2 and 3 are incorrect.


0

The most basic difference between probability mass func and probability density function is that pmf concentrates on a certain point for eg if we have to find a probability of getting a number 2. Then our whole concentration is on 2. Hence we use pmf however in pdf our concentration our on the interval it is lying. For eg negative infinity <= X <= ...


0

By Strong Law of Large Numbers, without any assumptions on the variance of $\epsilon_i$, $$ \frac{1}{n}\sum^n_{i=1}\epsilon_i \rightarrow \mathbb E[\epsilon_i]=0 \quad \text{almost surely} $$ So since $\lim_{n \rightarrow \infty} n\beta^n = 0$ and $n^{-1}b_n \rightarrow 0$ almost surely, we have that $$ \beta^n b_n\rightarrow 0 \quad \text{almost surely} $$ ...


1

The answer is $1/2^m$ or $0$, depending on whether $y'$ is in the range of $A$ or not. Try the case $m=1$ to get a feel for this.


1

You didn't specify a distribution; I'll assume uniform distributions throughout. $y'$ may or may not be in the image of $A$. If it is, then since $A$ has full column rank, there's exactly one $y\in\mathbb F_2^m$ such that $Ay=y'$, so the probability is $2^{-m}$. If it isn't, then the probability is $0$. If instead $y'$ is picked randomly, the probability ...


1

Question 1: Your understanding of $A\cap B$ and your calculation of $P(A\cap B)$ are correct. You basically did use $P(A\mid B)$, intuitively. To expand your calculation, since $A$ and $B$ are clearly independent events by the nature of coin tossing, $P(A\mid B)=P(A)$. Therefore $$P(A\cap B)=P(A\mid B)\cdot P(B)=P(A)\cdot P(B)=\frac 12\cdot \frac 12= ...


4

An alternative intuition is to see it as 4 groups of 13 with 4 friends all in different groups. The first can be in any group, the 2nd has 39 permissible slots out of 51, and so on. Thus Pr = $\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$


3

The solution is arguably simpler if you distinguish between permutations within the hands. There are $4!$ ways of distributing the aces over the $4$ players, and $13^4$ possible positions for the aces in the hands. That leaves $48!$ possibilities for filling the remaining $48$ slots. The size of the sample space in this perspective is just $52!$, so the ...


2

Rather than trying to rephrase, I think a great choice is the explanation by F. Scholz here:


1

I always thought it was easier to think in terms of the multinomial coefficient : you define it for $k_1,\cdots,k_d \ge 0, k_1 + \cdots + k_d = n$ and then you let $$ \binom n{k_1,\cdots,k_d} \overset{def}= \frac{n!}{k_1! \cdots k_d!}. $$ It counts the number of ways to have $n$ distinguishable balls and put $k_1$ of them in the first box, $k_2$ in the ...


0

This is not true $X \sim \text{Poi}(\mu)$ $Y \sim \text{Poi}(\lambda)$ $$P(X = Y - k) = \sum_{j = k}^\infty \frac{e^{-\lambda} \lambda^j}{j!} \frac{e^{-\mu} \mu^{j-k}}{(j-k)!} = f(\mu)$$ $$ f'(\mu)= \frac{e^{-\lambda} \lambda^k}{k!} (-e^{-\mu}) + \sum_{j = k+1}^\infty \frac{e^{-\lambda} \lambda^j}{j!}\bigg( \frac{-e^{-\mu} \mu^{j-k}}{(j-k)!}+ ...


1

Proof: Consider $X$ and let $A \in \mathcal{B}$ where $\mathcal{B}$ is the Borel sigma algebra on $\mathbb{R}$. Then \begin{align} P(X \in A) & = P(X \in A, Y \in \mathbb{R}) &\qquad (1)\\ & = P(\mathbf{X} \in A \times \mathbb{R}) &\qquad (2) \\ & = \int_{A \times \mathbb{R}} f(x,y) d\lambda^2 &\qquad (3) \\ & = \int_A ...


0

ok thanks a lot for that corindo, do you believe we can do a similar transformation for the three-dimensional case: $$dX_t^1=(X_t^2-2X_t^1)\,dt+\sigma_1 \,dW_t^1,\quad X_0^1=x_1$$ $$dX_t^2=(X_t^1+X_t^3-2X_t^2)\,dt+\sigma_2 \,dW_t^2,\quad X_0^2=x_2$$ $$dX_t^3=(X_t^2-2X_t^3)\,dt+\sigma_3 \,dW_t^3,\quad X_0^3=x_3$$ My final aim is to simulate the ...


0

First decide whether you mean the line segments only, or the whole lines through the two points. Then the following method should lead you to the answer: Generate four points at random. There are three ways to pair those points together, and the probability that the lines/line segments intersect will be $0$ or $\frac13$ or $\frac23$ or $1$, depending on ...


0

For any sequence of sets $(A_n)$ the $\limsup$ is defined by $$\limsup A_n =\bigcap_{n=1}^\infty\left(\bigcup_{j=n}^\infty A_j\right)=\bigcap_{n=1}^\infty B_n.$$ Since the $B_n$ are decreasing, if we assume the measure is finite we have $$\mu(\limsup A_n)=\lim_{n\to\infty}\mu(B_n) =\limsup_{n\to\infty}\mu(B_n).$$But $A_n\subset B_n$, so ...


1

First, I wouldn't modelise the markov chain like that , I would consider four states 1 = "player A play", 2 = "player B play", 3 = "player A had won" and 4 = "player B had won" The transition matrix would be $$M = \begin{pmatrix} 9/52 & 10/13 & 3/52 & 0 \\ 10/13 & 9/52 & 0 & 3/52 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 ...


0

I think I've figured out a proof. Suppose $\lim\sup_n \mu( A_n) > \mu (\lim\sup_n A_n)$. Then $\exists$ some rational $r$ with $\lim\sup_n \mu( A_n) > r > \mu (\lim\sup_n A_n)$(two real numbers $a, b$ satisfy $a > b$ if and only if there is some rational r with $a > r > b$). So $\exists N ∈ \mathbb N^+$ such that $sup_n \mu (A_n)$ = ...


1

Assuming that $Var(\epsilon_1)=\sigma_{\epsilon}^2$ we have $$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$ Hence, $\beta^tb_t$ is $L^2$-bounded and uniformly integrable so that you can pass the limit inside the expectation... Edit: In the first step, you need to show that $\beta^tb_t$ converges a.s. (to $0$). This ...


2

For (b), we want $E(X^2-4XY+4Y^2)$, which is $E(X^2)-4E(XY)+4E(Y^2)$. However, we cannot say in general that $E(XY)=E(X)E(Y)$. To calculate $E(XY)$, recall perhaps that $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)$. We know the covariance, and $E(X)$ and $E(Y)$, so we know $E(XY)$. For (c), as you know the $\pi$ makes no difference. We use the formula ...


0

I'm not sure I understand completely your question but did you notice that for any event $A$, $$ \mathbb{P}(A) = \mathbb{E}(\mathbb{1}\{A\}), $$ where $1$ is the indicator function ?


0

Find the worst case distribution, and everything else must have a lesser probability. Since all that matters is whether a point is inside or outside the ball of radius $k \sigma$ centered on $\mu$, we should make sure all of the probability mass inside the ball is at $\mu$, and everything outside the ball is exactly on the boundary $|x - \mu| = k \sigma$; ...


2

The point mass at $0$ contributes an impulse of area p to the pdf $f_X(x)$. $$f_X(x) = p\delta(x) + f(x)$$ where f(x) is the part of $f_X(x)$ on (0,a]. When you convolve g with that impulse, you get a copy of g back multiplied by p. That adds on to the convolution of the rest of f with g since convolution is linear. $$h(z) = \int_0^{\infty}[p\delta(x) + ...


0

Let $\mathcal{C} = \{ B \in \text{Borel of } \Bbb{R}^2 | \{ \omega \in \Omega | (X,Y)(\omega) \in B\} \in \mathcal{F} \}\}$. Note that $\mathcal{C}$ is a $\sigma$-algebra. 1)$\mathbb{R}^2 \in \mathcal{C}$ 2) $B \in \mathcal{C} \Rightarrow B^c \in \mathcal{C}$ 3) $B_i \in \mathcal{C} \Rightarrow \cup_{i=1}^\infty B_i \in \mathcal{C}$ Now since the sets ...


0

It's useful to view Chebyshev's inequality as more of an application of Markov's inequality which for a nonnegative random variable $X$ and $\alpha > 0$ is given by, $$ \begin{align} P(X \geq \alpha) \leq \frac{\text{E}(X)}{\alpha} . \end{align} $$ (Notice how we arrive at Chebyshev's inequality by applying Markov's inequality to the event $\{(X - ...


0

The transformation you suggested gives: $dA_t=\frac{1}{\sqrt{2}}(\sigma_1 dW_t^1+\sigma_2 dW_t^2)$ $dB_t=\frac{1}{\sqrt{2}}(2X_t^2-2X_t^1+\sigma_1 dW_t^1-\sigma_2 dW_t^2)$ How is this supposed to help me?


6

The intuition is that if $g(x) \geq h(x) ~\forall x \in \mathbb R$, then $E[g(X)] \geq E[h(X)]$ for any random variable $X$ (for which these expectations exist). This is what one would intuitively expect: since $g(X)$ is always at least as large as $h(X)$, the average value of $g(X)$ must be at least as large as the average value of $h(X)$. Now apply ...


1

Hint: 0000000 0000001 0000010 0000100 0000101 0001000 0001001 0001010 0010000 0010001 0010010 0010100 0010101 0100000 0100001 0100010 0100100 0100101 0101000 0101001 0101010 1000000 1000001 1000010 1000100 1000101 1001000 1001001 1001010 1010000 1010001 1010010 1010100 1010101 0000011 0000110 0001011 0001100 0001101 0010011 0010110 0011000 0011001 ...


1

To me it means: The further away the random variable from the mean is, the more seldom it becomes. k gives you the number of standard deviations (if taken as a natural number) and the probabilty will automatically be limited by $1/k^2$. My intuition why this is a meaningful statement for all random variables is the following: The measure of the whole room ...


2

Square integrable variables are not any random variables. They are in fact pretty regular ! Once you know that your variable has a variance, it's natural that the distance to the mean of your variable can be controlled in probability by this variance. Chebyshev's inequality is probably the simplest way to achieve that.


1

Let's define $\varphi(s) \equiv \mathbb{E}(e^{-s X_1})$ and $\zeta(s) \equiv \mathbb{E}(s^N)$. The Laplace transform of $Y$ is then, $$ \begin{align} \mathbb{E}(e^{-s Y}) &= \mathbb{E}(e^{- s \sum_{i=1}^{N} X_i}) \\ &= \mathbb{E}[\mathbb{E}( e^{- s \sum_{i=1}^{N} X_i} \mid N)] \\ &= \mathbb{E} \left [ \prod_{i=1}^{N} \mathbb{E}(e^{- s X_i}) ...


0

I'm very interested in this distribution as well, unfortunately I believe TenaliRaman's answer is only an approximation. Perhaps it only holds when the distribution is sufficiently concentrated to be approximately Gaussian..? I tried the following in R: require(gtools) library(MASS) n <- 7000 C <- c(0,1,0.2,0.5) alpha <- rep(0.2,length(C)) x ...


0

You can simply try the transformation $$A_t = \frac{X_t^1 + X_t^2}{\sqrt{2}}, \quad B_t = \frac{X_t^1 - X_t^2}{\sqrt{2}}$$ The equations will get much simpler.


0

There is a trick for this one, which alleviates you of having to use the joint distribution of $B_t$ and $\sup B_t$: Start with using Girsanov to construct a change of measure $$ \frac{d\mathbb Q}{d \mathbb P}\bigg|_{\mathcal F_t} =\exp(c B_t -\frac{1}{2}c^2t) $$ so that $$ \hat B_t = B_t -c t \qquad (=X_t) $$ is a $\mathbb Q$-BM. Denote by $\mathbb ...


1

A measure on a product space need not be a product measure. Suppose $$ X = \left. \begin{cases} 0 & \text{with probability }1/4, \\ 1 & \text{with probability }3/4, \end{cases} \right\}\text{ and }Y = \left. \begin{cases} 0 & \text{with probability }1/3, \\ 1 & \text{with probability }2/3. \end{cases} \right\} \tag 1 $$ The following two ...


3

That "Wikipedia" definition is for the special case where the covariance matrix is $\sigma^2 I$. In that case they are equivalent. A measure $\mu$ on $\mathbb R^n$ has density $\rho$ with respect to Lebesgue measure $\lambda^n$ iff $\mu(A) = \int_A \rho(x) \; d\lambda^n(x)$ for all Lebesgue measurable sets $A$.


0

Let $X$ be the number of failures before the first success. Then $X$ takes on only non-negative integer values. For such a random variable $X$, we have $$E[X]=\sum_{i=1}^\infty \Pr(X\ge i).$$ For a proof, please see Wikipedia. If the probability of failure is $q$, then $\Pr(X\ge i)=q^i$. In our case $q\le 1/2$, so $$E[X]=\frac{q}{1-q}\le 1.$$ One can also ...


3

Using the martingale property, $$ \begin{align} \text{E}(W_n) &= \text{E}[\text{E}(W_n | W_{n - 1})] \\ &= \text{E}(W_{n - 1}) \\ &= ... \\ &= \text{E}(W_1) , \end{align} $$ and so you have a uniform bound on both $\text{Var}(W_n)$ and $\text{E}(W_n)^2$.


0

In (A), assuming you're talking about that normal variable: Saying $\mu(B)=P(X^{-1}(B))$ is the definition of "$\mu$ is the distribution of $X$". And now the definition of "$X$ is normal with mean $0$ and variance $1$" is "if we let $\mu$ be the distribution of $X$, as defined above, then $\mu(B)=\int_B f\,d\lambda$, where $f$ is as above". Regarding the ...



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