New answers tagged

1

Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.


0

Use the inverse transform method: With $X \sim N(0,1)$ and CDF $F$. $1.$ Generate a random number $u$ from $U$ in the interval $[0,1]$ $2.$ Compute the value $x$ such that $F(x) = u$ $3.$ Take $x$ to be the random number drawn from the distribution described by $F$


0

I solved this trivially. We can extend both ranges by employing $y_i = \xi_i / x$ in case of the range from $[0,x]$ and $y_i = \frac{\xi_i - x}{1 - x}$.


1

If the students have mean $\mu$ and variance $\sigma^2$, then the distribution for the average of $n$ students is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$. In this case, you want to know the probability that the sample mean $\bar{X}$ is within $\pm r$ of the true mean $\mu$. So, $$P(\mu-r \leq \bar{X} \leq \mu-r) = P(-r \leq ...


1

$(X,Z)$ are independent if forany $f,g$ $$\mathbb{E}(f(X)g(Z))=\mathbb{E}(f(X))\mathbb{E}(g(Z))$$ In your case : $$\mathbb{E}(f(X)g(Z))=\mathbb{E}\left(f(X)g\left(\frac{Y-pX}{\sqrt{1-p^2}}\right)\right)=\int_{\mathbb{R}^2}f(x)g\left(\frac{y-px}{\sqrt{1-p^2}}\right)p(x,y)~dx~dy$$ so this is a change of variable problem, you set : ...


0

HINT Note that in the exponential we have $$ x^2-2pxy+y^2=(y-px)^2+(1-p^2)x^2 $$ so that $$ f(x,y)=\frac{1}{2\pi\sqrt{1-p^2}} \mathrm e^{-\frac{(y-px)^2}{2(1-p^2)}-\frac{x^2}{2}}=\frac{1}{\sqrt{2\pi}\sqrt{1-p^2}}\exp\left(-\frac{(y-px)^2}{2(1-p^2)}\right)\times \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right) $$


1

2 cans of tomatoes, 3 of peas and 1 of beans $Y$ is the number of cans opened until the 2 cans of tomatoes are open.   We measure the probability of selecting 1 from 2 tomato cans and $y-2$ from the 4 others in some arrangement out of the ways to select $y-1$ from $6$ cans, times the conditioned probability of selecting the 1 tomato can from the ...


1

Clearly, the stopping criterion is that the second can of tomatoes is opened. If there are three cans of peas, then it is entirely possible that all three cans could be opened before the second can of tomatoes is opened, since the final can that is selected could be the second tomato can. The most direct way to compute the joint distribution is to consider ...


1

1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours. I define the random variable $Y_i$ =number of clients in the i-th hour of the opening hours $Y_i$=number of clients in the i-th hour of the opening hours for $i=1,2,3,4$ , since ...


1

1) The conditional distribution of the $8$ clients along the interval $[0,4]$ is Uniform, so the conditional distribution of the number of clients in a sub interval $(2,4]$ is Binomial with $p=\frac{2}{4}$ and $n=8$. Denote it by $X$, the event of interest is $\{X=4\}$, so $P(X=4) = \binom{8}{4}\frac{1}{2^8}$. The answers to 2 and 3 look OK to me.


1

I often don't know what to do, besides using somehow Borel-Cantelli Well, more often than not, you don't need to use anything else:) As for the question: $$ \sum_{n=0}^{\infty}P(|X_n|^p\geq n)=\sum_{n=0}^{\infty}P(|X_1|^p\geq n)\geq \int_0^{\infty}P(|X_1|^p\geq t)dt=E|X_1|^p=\infty. $$


0

I'm giving you a quick answer because I'm lazy, but hope this helps you! Treat the $m \times n$ matrix as a big vector of dimensions $mn$. So, treat, $X$ as a random vector $X: \Omega \to \mathbb{R}^{mn}$. Now, look the this wiki page: https://en.wikipedia.org/wiki/Multivariate_random_variable In short, $E(X)$ will be a matrix of same dimensions so that ...


2

You can solve it by inspection.   Just fill in the table: $$\begin{array}{l:l}x & \dfrac{2^x~\mathsf e^{-2}}{x!} & \displaystyle\sum\limits_{k=1}^x\dfrac{2^k~\mathsf e^{-2}}{k!} \\ \hdashline 0 \\ 1 \\ 2 \\ 3 \\ 4\\ 5 \\ \hline 6 \\ \vdots \end{array}$$ Stop when the third column exceeds $0.99$.   Do you have access to a spreadsheet ...


1

I started writing this before peterwhy posted his answer, and since its approach is a little different I decided to post it anyway. We consider the problem on the unit disk $\mathbb{D}$ of determining $\mathbb{P}(\text{$d(M,N)$ $\leq 1$})$. To that end, we shall use the law of total probability. Rather than consider a single partition of our sample, we ...


1

Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is $$P(R\le r) = \begin{cases} \frac{\pi r^2}{\pi\cdot 1^2} = r^2, & 0\le r\le 1\\ 1, &r>1\\ 0, &\text{otherwise} \end{cases}$$ The probability density function of $R$ is $$f_R(r) = \frac{d}{dr}P(R\le r) = \begin{cases} 2r, ...


-3

Unless you are given a specific sample space, the answer is 1/2, as there are uncountably many pairs of points distance r apart and uncountably many not. Without a sample space, the addition rule for probabilities no longer applies, so you have a 1/2 probability of the distance being each of less than, equal to, and greater than r.


0

The probability distribution of $Y|X^n=x$ converges to the probability distribution $Y|X=x$ for increasing $n$. Nevertheless, and that is the problem in this case, entropy is not continuous in a converging sequence of probability measures (see, e.g., this paper).


0

The variance is defined in terms of the transpose, i.e. say $X$ is a real-valued random variable in matrix form then its variance is given by $$ \mathrm{Var} (X) = \mathbb{E} \left[ (X - \mathbb{E}[X])(X - \mathbb{E}[X])^\top\right].$$ In your case this would results in $$\mathrm{Var} (X) = \frac1n \sum_{k=1}^n \left(X_k- \mathbb{E}[X] \right)\left(X_k - ...


4

Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$ Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}. $$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$. So the answer is "no" to that main question. It ...


0

a) Using the multinomial coefficient, we have $$P(Y_1 = i, Y_2 = j, Y_3 = k, Y_4 = l) = \binom{35}{i,j,k,l}\left(\frac{10}{100}\right)^i\left(\frac{35}{100}\right)^j\left(\frac{3}{100}\right)^k\left(\frac{52}{100}\right)^l$$ b) Instead of considering four categories, we know consider three categories where categories two and three have been collapsed ...


0

Definition. Let $X \sim \chi^2_{k}$ and $Y \sim \chi^2_{m}$ be independent. Then $$F = \dfrac{X/k}{Y/m}$$ follows the $F$-distribution with $(k, m)$ degrees of freedom. It follows that $$\mathbb{E}[F] = \dfrac{m}{k}\mathbb{E}\left[\dfrac{X}{Y}\right] = ...


1

I did not read the answer you linked to, but here is what I think. Since the covariance is a bilinear form, under a change of basis $$ \begin{pmatrix} X' \\ Y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} $$the covariance matrix will transform as $$ \Sigma' = S \; \Sigma \; S^T $$ where $ S $ is ...


1

Your problem is that $\Sigma^{-1/2}$ is not $\begin{pmatrix}1 & \frac{1}{\sqrt{\rho}} \\ \frac{1}{\sqrt{\rho}} & 1 \end{pmatrix}$, it is a matrix $T$ such that $$T \cdot T \cdot \Sigma = \Sigma \cdot T \cdot T = \begin{pmatrix}1&0\\0&1\end{pmatrix},$$ where $\cdot$ denotes matrix multiplication. One possible such $T$ is (credit to Wolfram ...


1

For a start: $$\begin{bmatrix}1 & \rho\\\rho & 1\end{bmatrix}^{-1}~=~\dfrac1{1-\rho^2}\begin{bmatrix}-1 & \rho\\\rho & -1\end{bmatrix}$$ So $$\begin{bmatrix}1 & \rho\\\rho & 1\end{bmatrix}^{-1/2}~=~\dfrac1{2}\begin{bmatrix}\dfrac 1{\sqrt{1+\rho}}+\dfrac 1{\sqrt{1-\rho}} & \dfrac 1{\sqrt{1+\rho}}-\dfrac 1{\sqrt{1-\rho}}\\\dfrac ...


1

In the theory of probability as developed by Kolmogorov, random variables just are measurable functions (provided that the underlying space is a probability space, which is just a space with a normalized finite measure). That's actually the key idea to Kolmogorov's theory. It allows us to rigorously encode everything that one did with elementary ...


0

If you read further in the paper, $E(X,Y)$ denotes the "energy function" of the random variables $X$ and $Y$, not an expectation - hence it is a scalar valued function of two inputs, no inconsistency there. Also note that $Pr(y|x)$ is supposed to denote the density of the conditional distribution in this paper, not the conditional distribution itself. I.e. ...


-1

Formally, they are the same: a random variable is a particular type of measurable function. However, there are important differences which may be more philosophical than mathematical. There is a (somewhat subjective) difference in the underlying domains. A random variable operates on a set of outcomes of a random experiment or process. A measurable function ...


0

The transformation "almost" works if $X$ is continuous which implies that $F_X(x)$ has no jumps. Consider consecutive values $y'<y$ in support of $Y$. Extend $F^{-1}_Y(u)$ by letting it map the values from the interval $(P(Y\le y'),P(Y\le y)]=(F_Y(y'),F_Y(y)]$ to $y$. Note that the interval is half-open. If $y$ is the minimum value in the support of $Y$, ...


0

Randomized-select is a recursive algorithm that partitions its input array using a random pivot, and proceeds to work on the partition containing the $i$-th smallest element of the original array. When the array has $n$ elements, the random pivot partitions the array into smaller arrays of size $k-1$ and $n-k$ for some $k$ between $1$ and $n$. The sum of ...


0

Hint: This very closely related to the Cantor Intersection Theorem. Read the wikipedia page on that, and think about how to adopt it to a probabilistic result.


0

Hint. Integrating by parts with respect to $x$ gives $$ e^{-y}\int_{0}^{y}xe^{-x}dx=e^{-y}\left(1-e^{-y} (1+y)\right)=e^{-y}-(1+y)e^{-2y} $$ integrating by parts once more with respect to $y$ $$ \int_{0}^{\infty}e^{-y}dy-\int_{0}^{\infty}(1+y)e^{-2y}dy=1-\frac34=\frac14. $$


2

$W$, the conditional expectation of $Y$ given $X$ will be a piecewise function partitioned on $X$'s enumeration. $$\begin{align}W~=~& \mathsf E(Y\mid X) \\[2ex]~=~& \sum\limits_{\omega\in X^{-1}(X)} Y(\omega)\cdot\mathsf P^{Y\mid X}(\omega) & :~ \mathsf P^{Y\mid X}(\omega) \mathop{:=} \Pr(Y{=}Y(\omega)\mid X{=}X(\omega)) \\[2ex] =~& \tfrac ...


1

What follows is definitely not the simplest approach but it does showcase generating function methods. Let $z$ represent men and $w$ represent women and $v$ women who are not next to a man then we get the generating function (this is like flipping coins) $$G(z, w, v) = (1 + vw + v^2 w^2 + v^3 w^3 + \cdots) \\ + (1 + w + v w^2 + v^2w^3 + \cdots) ...


7

We do it another way, using indicator random variables. Let the elements of $A$ be $a_1$ to $a_m$. For $i=1$ to $m$, define random variable $Y_i$ by $Y_i=1$ if $a_i$ is in the range, and $Y_i=0$ otherwise. Then $X_m=\sum_1^m Y_i$, so by the linearity of expectation we have $E(X_m)=\sum_1^m E(Y_i)=mE(Y_1)$. Thus $\frac{X_m}{m}$ has expectation $E(Y_1)$. ...


5

You had a good idea, but it's not quite right. You're counting not only seats occupied by women not surrounded by women but also all seats occupied by men. (You also didn't execute the idea correctly – for corner seats you calculated a conditional probability, and for interior seats I'm not sure exactly what you calculated.) Here are three correct ...


1

Part $1$ is much simpler than what you have. Note that $$P(|X_n| > \epsilon) \leq P(|X_n| > 0) = P(X_n = n) = n^{-\alpha} \to 0.$$ Part $2$ is correct. For Part $3$, consider the following: $$\|X_n\|_{L^r}^r = n^r \cdot n^{-\alpha} = n^{r - \alpha}.$$ Under what conditions does this go to $0$?


0

First of all, note that you have to show $$\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) = \color{red}{\infty} \tag{1}$$ in order to apply the Borel-Cantelli-lemma. Suppose that $(1)$ does not hold true, i.e. $\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) < \infty$. Since the random variables are identically distributed, this implies $$\sum_{n ...


1

Note that $\exp(-\lambda)$ is constant and can be pulled to the front of the sum. The remaining inner sum has the value $\exp\left(\frac{r - 1}{r} \lambda\right)$ (remember the representation of $\exp$ as power series). The outer sum can easily be calculated when you see that the inner sum does not depend on the index of the outer sum. To answer your ...


3

For (1) observe that firstly that $\sum_{i=1}^rc=rc$ and secondly that $\sum_{n=0}^{\infty}\frac{a^n}{n!}=e^a$. Apply this on $a=\frac{r-1}{r}\lambda$. alternative route and making use of symmetry: $P\left\{ I_{1}=1\right\} =\sum_{n=0}^{\infty}P\left\{ I_{1}=1\mid N=n\right\} P\left\{ N=n\right\} ...


1

Take $Y_n = Y$ uniform on $[-1,1]$. Of course $Y_n \to Y$ in probability (or any other sense). Take $X_n = Y$ but $X = -Y$. Since $Y$ and $-Y$ have the same distribution, $X_n \to X$ in distribution. But $X_n + Y_n = 2 Y$ does not converge in distribution to $X + Y = 0$.


3

No, we don't generally have $\mathbb E(Z) = \frac{n}{2k+1}$. For instance, for $n\le k$ we have $Z=0$ with probability $1$. To find the expected median for large $n$ for $k=1$, with $3$ bins, we can expand the multinomial coefficients around $m=\frac n3$ using Stirling's approximation: \begin{align} &\binom n{m-x,m-y,m+x+y}\\ ...


0

I am assuming both $T$ and $S$ are independent of $X \sim exp(\lambda)$! If not, you must have some joint distribution. Rough sketch: Suppose $f_T$ is the pdf of $T$. Then, $\begin{eqnarray} P(X >T+s \mid X>T) &=& \frac{ P(X >T+s) }{P(X >T)} \\ &=& \frac{\int_o^\infty P(X > t+s \mid T=t ) f_T(t)\,dt}{\int_o^\infty P(X ...


2

Independence follows from the fact that for every fixed $k\in\mathbb N$:$$P(X_N\in A)=\sum_{n=0}^{\infty} P(X_N\in A\mid N=n)P(N=n)=$$$$\sum_{n=0}^{\infty} P(X_n\in A)P(N=n)=\sum_{n=0}^{\infty} P(X_k\in A)P(N=n)=$$$$P(X_k\in A)=P(X_N\in A\mid N=k)$$


-2

This is not different from rolling a green and a red die ans using a coin flip to decide which of the dice you use and which you ignore. You still have a coin flip and a die. All that matters is that the $X_n$ are identically distributed (they need not be normally distributed and not independent of one anaother) and that $N$ is independent of all ...


3

The process is not well-defined since you didn't specify a distribution. It stands to reason that a patient randomly choosing a pill out of a bottle is more likely to pick a large one than a small one. For this answer, I'll assume that you meant to imply that the pills are chosen independently and with uniform distribution. Let there be $k$ large and $n$ ...


0

I guess both sentences are true. Namely, I think both of random variable can takes infinite number of values because a discrete random variable can only take on a finite or countably infinite number of values. Yes, they can be one or the other; but which is which ? What is the support of a Poisson random variable, of mean $\lambda$ ?   Is this ...


1

$$\mathsf H(Y\mid X) = - \sum_x p_X(x) \sum_y p_{Y\mid X}(y\mid x)\log p_{Y\mid X}(y\mid x)$$ or $$\mathsf H(Y\mid X) ~=~ \sum_{x,y} p_{X,Y}(x,y)\log \frac{p_{X}(x)}{p_{X,Y}(x,y)}$$ Hint: (the latter will be a sum of six terms.)


2

I think you mean "absolutely continuous" so both have a density. In that case, you have $$P(X < Y) ~=~ \iint_{\{(s,t)\mid s < t\}} f_X(s)~f_Y(t)~\mathrm d\,s~\mathrm d\,t$$ Notice the integrand is symmetric in is arguments (since $f_X(u)=f_Y(u)$). This will do the job.


0

So when you are given a sample of response variables Y and some predictor variables $X_1, ... , X_p$, I think you are asked to find the regression between Y and $X_1, ... , X_p$. On the very basics, we use simple linear regression $Y = a_0 + a_1X_1 + ... + a_pX_p + \epsilon$ where $\epsilon$ is the noise(error) term. Q1. The domain of Y should just be the ...


1

The expectation of the cumulative distribution function is independent of the distribution. The cumulative distribution function is uniformly distributed over $[0,1]$, so its expectation is $\frac12$.



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