New answers tagged

-2

Please look at this work 1 which has investigated the interference from neighbor cells, and hence, investigated the distribution of distances between users. Hope it works for you.


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We can use Chebyshev's inequality on the random variable $\frac{1}{n}\sum_{i=1}^nX_i$, which states the following: $$\mathbb{P}(|Y-\mathbb{E}(Y)|\geq a)\leq\frac{1}{a^2} \text{Var}(Y)$$ Thus $$\mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2})\leq4 \text{Var}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{4}{n^2}n\cdot\frac{1}{12}=\frac{1}{3n}...


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Let's pick an example: I want to generate $3.1$ grams of ozone per hour. This is $0.31$ of the full-time output. What we'll do is, for each second, accumulate $0.31$ into an 'accumulator' and then when that value reaches $1$, we'll emit for a second, subtract $1$, and continue. So: At the start of second $0$, it's at $0$, we add $0.31$, getting $0.31$ ...


1

You can prove that the answer is $1/2$ by induction. Denote with $E_n$ the event that after $n$ draws the sum is even. Base case: $n=1$. You draw $1$ number from $\{1,2\}$. Obviously, the probability that the sum is even, is equal to the probability that you draw $2$, is equal to $1/2$, i.e. $$P(E_1)=\frac12$$ Induction step. Assume this holds for $n$, i....


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If a number can appear more than once, then this is easily seen to be $\frac12$. You can see this if you stop to catch your breath after you've drawn the first $n-1$ numbers. Then you realize that whether the total sum will become even or odd depends only on whether the last number (that you haven't drawn yet) is even or odd (the sum might end up having the ...


2

It seems that $\Phi$ is meant to be the cumulative distribution function of the standard normal distribution. Since you conclude from $\Phi$ being positive that $\Phi^{-1}$ is positive, I get the impression that you're mistaking $\Phi^{-1}$ to refer to the reciprocal of $\Phi$. Here $\Phi^{-1}$ denotes the inverse of $\Phi$. Since $\Phi(0)=\frac12$, $\Phi^{-...


1

Recall that $$\mathbb{P}[U \le u] = \max\{\min\{u,1\},0\})$$ Observe that $$\mathbb{P}[\max\{U_3, U_5\} \le u] = (\max\{\min\{u,1\},0\})^2$$ $$u = \max\{\min\{u,1\},0\}, u \in [0,1]$$ $$\mathbb{P}[\min\{U_1, U_2, U_4, U_6, U_7\} \le u] = 1 - (1 - \mathbb{P}[U_1 \le u])^5$$ $$E = \{\max\{U_3, U_5\} < U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$ $$ = \{\...


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Several ideas: (more "graphical") calculate the values of these curves on a grid, eg. let $x$ be from $-k$ to $k$ with some step $h=\frac{k}{n}$. Then calculate $y$ values for this set of $x$ coordinates and check in which interval it belongs. To compare two shapes: count matching "pixels". You can use some measure of difference (metric) between functions (...


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The ages of $A,B,$ and $C$ may be independent†, but the events of pairwise orders are not.   If you are told that $A$ is one of the two oldest children (because $A$ is older than $C$) it should raise your anticipation that $A$ is also older than $B$. († though, actually, they are not independent if they have the same mother; but that is ...


1

Let the children $A, B, C$ have ages $X, Y, Z$ respectively. For each child pair, $\{A,B\}, \{A,C\}, and \{B,C\}$ there are three age possibilities. For the pair $\{A,B\}$, the possibilities are as follows:$$1\ \ \ \ \ X \gt Y$$ $$2\ \ \ \ \ X = Y$$ $$3\ \ \ \ \ X<Y$$ Note that the distribution holds for the other children pairs. Now to answer the ...


1

The problem is that $P((X\gt Y)\cap(X\gt Z))=P(X\gt Y)P(X\gt Z)=\frac14$ isn't right. The probability that $X$ is greater than both $Y$ and $Z$ is just $\frac13$, the probability of one of three equivalent children to be the eldest. Thus the events aren't independent.


3

For fixed $M$, for large $N$ most of the $X_{(i)}$ have an almost certain value; only a fraction of order $N^{\frac12}$ near the values $i=k\frac NM$ with integer $k$ have an appreciable probability to take one of two different values. These only yield a contribution of order $N^{-\frac12}$ to the ratio, so we can disregard them in the limit $N\to\infty$ and ...


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Define events A and B s.t. $$Y = 1_{(X < b)} := 1_{B}$$ $$Z = 1_{(a < X)} := 1_{A}$$ Observe that Y and Z are independent iff A and B are independent. $$P(A \cap B) = \int_a^b 1 dx = b-a$$ $$P(B) = \int_0^b 1 dx = b$$ $$P(A) = \int_a^1 1 dx = 1-a$$ So for $x \in [0,1]$, A and B are independent iff $$b-a=b(1-a) \iff b-a=b-ba \iff -a=-ba \iff a =...


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Your problem can be interpreted as a special case of Bertrand's ballot theorem, which answers the question: In an election where candidate 1 receives the same number of votes as candidate 2, what is the chance that candidate 1 is never behind throughout the vote count? The proof creates a 1:1 correspondence between your merged set of $2n$ items and paths in ...


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It's obvious that $Y$ and $Z$ cannot be independent. If you observe $Y = 1$, then you know $X \in (0,b)$, and intuitively, this information affects the probability of observing $Z = 1$. Formally, $$\Pr[Z = 1 \mid Y = 1] = \frac{\Pr[(Z = 1) \cap (Y = 1)]}{\Pr[Y = 1]} = \frac{b-a}{b},$$ but $$\Pr[Z = 1] = 1-a \ne 1 - \frac{a}{b}$$ if $b \ne 1$.


3

No, $\overline{X_n}$ isn't uniformly distributed; but it's distributed symmetrically about $\frac12$, so you can nevertheless conclude that the probability is $\frac12$ even without the limit.


3

The maximum is less than $x$ iff each variable is less than $x$. For $0\le x< \theta$ $$P[X_{(n)}\le x]=P[X_1\le x,X_2\le x,...,X_n\le x]=\prod_{j=1}^nP[X_j\le x]\\=\prod_{j=1}^n\int_0^xf_{X_j}(t)dt=\prod_{j=1}^n\int_0^x{1\over\theta}dt=\prod_{j=1}^n\left({x\over\theta}\right)=\left({x\over\theta}\right)^n$$


1

The marginal distribution is indeed given by $$ f_Y(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\;dx $$ and since $f_{X,Y}(x,y)=\frac{1}{1-x}$ if $0<x<y<1$ and $f_{X,Y}(x,y)=0$ otherwise, it follows that if $0<y<1$ then $$ \int_{-\infty}^{\infty}f_{X,Y}(x,y)\;dx=\int_0^y\frac{1}{1-x}\;dx $$ since we must have $0<x<y$.


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The marginal distribution of a subset of a collection of random variables is the probability distribution of the variables contained in the subset. It gives the probabilities of various values of the variables in the subset without reference to the values of the other variables. - Wikipedia Thus, we want to integrate from $0$ to $y$ because it will give us ...


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That is entirely subjective and depends on how much you think each member of each category should get. For example, if you feel each regular player should get three times as much funding as someone who rarely comes, and each "average" player should get twice as much as someone who rarely comes we can come up with the following: Let $y_1$ be the number of ...


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Yes, you might want to check your local library for the solution to this task. My local library is THE place to be and I really think it could help you right now to check your local library for the solution to this task. Have a nice day


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In a uniform distribution, the probability that a points lands on a interval depends only on the length of the interval. Consider the intervals $A=[0,h]$ and $B=[\frac{\pi}{2}-h,\frac{\pi}{2}]$, which have the same length $h>0$. The probability that $p$ lands on $A$ is higher than the probability that $p$ lands on $B$, because $\sin(A)$ has length $\sin(...


1

I'll solve this in more generality, since questions like this keep coming up and we need a standard answer to mark them as a duplicate of. So I'll calculate the probabilities for exactly $k$ bins to be empty and for at least $k$ bins to be empty when $m$ balls are distributed uniformly randomly over $n$ bins. Your case is $n=m=N$, with various $k$ given in ...


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It is because $0\leq y\leq x$ is the support if $Y$. When you factor in the support of $X$, the support of the joint distribution is , $0\leq y, \max(1,y)\leq x$ $$\begin{align}f_{Y}(y) =&~ \int_{y}^\infty f_{X,Y}(x,y)\operatorname d x~\big[0\leq y\big]\\[2ex] =&~ \int_{\max(1,y)}^\infty \tfrac 1{x^3}\operatorname d x~\big[0\leq y\big]\\[2ex]=&~...


1

In the general case, neither of these is right. You have two inequalities for $x$ in $f(x,y)$, namely $x\ge1$ and $x\ge y$, so the lower limit of the integral is $\max(1,y)$. You appear to be implying that the lower limit $y$ was specified in some book or lecture or the like. If so, I suspect that this was done because you need the value for $Y=\frac32$, and ...


0

Since the distribution of $X$ is symmetric about $0$, we have $\mathbb E[X^n]=0$ for all odd $n$. For even $n$, we have $$\mathbb E[X^n] = \int_{-1}^1 \frac12 x^n\ \mathsf dx = \int_0^1 x^n\ \mathsf dx = \frac1{n+1}. $$ It follows that $$\mathbb E[XY]=\mathbb E[X^5]=0$$ and further $$\mathbb E[X]\mathbb E[Y] = \mathbb E[X]\mathbb E[X^4] = 0\cdot\frac15=0.$$ ...


3

Hint: We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required. You can do ...


2

Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$



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