# Tag Info

1

Pretty much nothing. The covariance is a very weak measure for the dependence of two random variables. To see an example: Let $X \sim \mathcal{N}(0, 1)$ and $Z$ be Rademacher distributed, so that $X$ and $Z$ are independent. Set $X_n = X$ and $Y_n = Y = ZX$. Then $X$ and $Y$ are uncorrelated standard Gaussians, but their sum isn't Gaussian. If $X$ and $Y$ ...

1

I think this is related to monotone likelihood ratios $-$ https://en.wikipedia.org/wiki/Monotone_likelihood_ratio $-$ which does imply FOSD.

1

In general, to sample from a random variable $X$ with CDF $F$, you introduce the "quantile function" $Q$. This is formally defined as $$Q(y)=\inf \{ x : F(x) \geq y \}.$$ When $F$ is invertible, this is its inverse. When $F$ is not invertible, it is the "right generalization", in the sense that it correctly accounts for regions where $F$ is horizontal and ...

0

We deal with your question about why we multiply by $2$. We are interested in finding the cumulative distribution function $F(y)$ of $Y$. The only interesting part is when $-1\lt y\lt 1$, because $F(y)=1$ if $y\ge 1$ and $F(y)=0$ when $y\le -1$. It is geometrically perhaps a little easier to find $G(y)=\Pr(Y\gt y)$. Then the probability that $Y\le y$ is ...

0

The question in its present form is unclearly stated and some assertions in the question are incorrect. Based on comments under the question, here is what I think is meant: Two Poisson processes have respective rates $\lambda_0$ and $\lambda_1$; They are independent. (This was in comments, not in the original question.); Each time an arrival occurs, the ...

0

Consider that integrating against $\delta(y-f(x))$ will pick out $P_X(x)$ for all $x$ such that $f(x)=y$ for the given (fixed) $y$. It follows that (in the countable case) $$\int P_X(x)\delta(y-f(x))\, dx=\sum_{f(x)=y} P_X(x)$$ i.e. $P_Y(y)$ is given by the sum of the contributions $P_X(x)$ for all the $x$ that map to $y$.

0

An explicit derivation of Robert Israel's claim is fairly straightforward by induction: \begin{align*} \operatorname{E}[X^{2m}] &= \int_{x=-\infty}^\infty x^{2m} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \end{align*} and with the choice $$u = x^{2m-1}, \quad du = (2m-1) x^{2(m-1)} \, dx, \quad dv = x e^{-x^2/2} \, dx, \quad v = -e^{-x^2/2},$$ we ...

0

Extended comment. Here is a simulation experiment (in R), that suggests $Var(X^3) = 15.$ x = rnorm(10^6); y = x^3 var(y); mean(abs(y) < 40) ## 15.02577; 0.999341 summary(y) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## -1.020e+02 -3.077e-01 1.118e-08 -2.277e-04 3.101e-01 1.409e+02 hist(y[abs(y)< 40], ...

1

HINT If $X = \max(X_1,\ldots,X_n)$, then $X < c$ iff $X_i < c$ for all $i$. Probabilities multiply over independent events.

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HINT: Write $\langle x_n:n\in\Bbb Z^+\rangle\sim\langle y_n:n\in\Bbb Z^+\rangle$ if and only if there is a finite permutation $\sigma$ such that $y_n=\sigma(x_n)$ for each $n\in\Bbb Z^+$. Check that $\sim$ is an equivalence relation on $S$. Show that an event is exchangeable if and only if it is a union of $\sim$-equivalence classes. Since the ...

2

Hint: Write $F_X(x) = E[I\{X \le x\}]$ and use Fubini's theorem.

0

To answer the first question, it seems this procedure is called the Serial Approximate Entropy Test, according to section 2.12 of the NIST Statistical Test Suite for the Validation of Random Number Generators and Pseudo Random Number Generators.

1

You have been given that $$f_{X,Y}(x,y) = \begin{cases}4xy & : 0 \leq x \leq 1,\; 0 \leq y \leq 1\\ 0 & : \text{ elsewhere}\end{cases}$$ You wish to know where $f_{X,Y}(x, z-x)$ is supported (ie: not zero) with respect to $x$, for values of $z$ where $0\leq z\leq 2$.   (Since $z=x+y$, then $0+0\leq z\leq 1+1$.) $$f_{X,Y}(x,z-x) = ... 2 You could require f to only be continuous, but the idea is to have as few test functions as possible to satisfy the definition. Another way to say this is, the above possible values for \{...\} are equivalent to the requirement that the equality holds for continuous functions. Here is a proof of that for the continuous bounded characterization. On one ... 1 While sample set could also be worthwhile terminology, it is worth recalling that any space in mathematics is always a set with some sort of operations defined. For example, in the theory of stochastic processes, we could let \Omega = \mathcal{D}(\mathbb{R}_+, \mathbb{R}), where \mathcal{D}(\mathbb{R}_+, \mathbb{R}) is the Skorokhod space of real-valued ... 2 By definition:$$\begin{align} F_Z(z) & = \mathsf P(\min \{X,Y\}\leq z) \\[1ex] &= 1-\mathsf P(\min\{X,Y\}>z) \\[1ex] &= 1-\mathsf P(X>z)\mathsf P(Y>z) \\[1ex] & = 1-(1-F(z))^2 \\[1ex] & = 1 - (z+1)^{-4} \end{align}$$You can take it from here. 2 I think this satisfies the requirements:$$X_n = \begin{cases} \dfrac{1}{n}\mu_0, & \text{with probability $\dfrac{n-1}{n}$} \\[2ex] \left(n-1+\dfrac{1}{n}\right)\mu_0, & \text{with probability $\dfrac{1}{n}$.} \end{cases}$$We have E(X_n)=\mu_0 and E(\sqrt{X_n})=\sqrt{\mu_0}\left(\dfrac{n-1}{n^{3/2}} +\dfrac{\sqrt{n-1+1/n}}{n} \right) ... 2 A random variable X on \Omega is no more and no less than a function X:\>\Omega\to{\mathbb R} satisfying the technical condition that it is measurable: For any x\in{\mathbb R} the set \{\omega\in\Omega\>|\>X(\omega)\leq x\} belongs to {\cal F}. This guarantees that for any two given values a, b the probability$$P[a\leq X(\omega)\leq ...

1

The actual definition is that a function $X:\Omega\to\mathbb R$ is called a random variable when $X$ is $(\mathcal F, \mathcal B(\mathbb R))$-measurable. In other words, $X^{-1}(B)\in\mathcal F$ for any $B\in\mathcal B(\mathbb R)$, where $\mathcal B(\mathbb R)$ is the Borel $\sigma$-algebra on $\mathbb R$. Now, it is sufficient that ...

2

The random variable is not limited to values less than $x$. For instance, I can show you that the function $$X(t) = \frac{1}{t}$$ is a measurable function on $(0, 1)$. Here's how. Let's look at $$\{ t \mid X(t) \le 11 \}$$ That's the set of all points in the domain for which $X(t) = 1/t$ is less than 11, which is exactly $$A = \{t \mid \frac{1}{11} ... 0 The format of this site is not optimally suited for multiple questions. Anyway, here are some answers: At least if N \ge 3 there will be no lower bound. Basically, you can start with any such random variable with N-1 points and split one of the masses (i.e., probabilities sitting on points) into two points arbitrarily close to the original point ... 0 As far as I know yes, the statistics of the disturbances are assumed to be known. And not just that, the standard approach falls into the so called Linear Quadratic Gaussian problem (LQG) where the system is assumed to be perfectly linear and both the disturbance w and the measurement noise v are Gaussian white noise with known covariance matrices ... 0 The error lies in the evaluation of var(bar) - when you scale the random variable by factor k (which is 1/n in your case), the variance is scaled by k^2 (i.e. 1/n^2)$$var(ybar) = \color{red}{var}(Σ (yi/n)) = (1/\color{red}{n^2})* Σ(var(yi))= (1/\color{red}{n^2})* (n(σ^2 + w^2)) = (σ^2 + w^2)/n$$0 You should be doing$$\bar{x}\mp t_{0.975}(n-1)\frac{\hat{\sigma}}{\sqrt{n}}$$In this case, \bar{x}=9.6, t_{0.975}(24)=2.064, and \frac{\hat{\sigma}}{\sqrt{n}}=\frac{\sqrt{22.4}}{\sqrt{25}}\times\frac{\sqrt{25}}{\sqrt{\color{red}{25-1}}} since we require an unbiased estimate of the population variance, based on the sample variance. This gives us the ... 0 Let x_1, x_2,..., x_{25} be a sample from the independent stochastic variables X_i, i=1,2,...,25. We assume that X_i is N(\mu,\sigma) for i=1,2,...,25 for some unknown \mu and \sigma. Given is the sample mean \overline{x}=9.6, and the sample variance s^2=22.4. For future notation, we set n=25 as the number of observations. We then have ... 0 The idea behind is OK but the following expression is wrong:$$P(g | X=x) = \frac{P(X=x | g)P(g)}{P(X=x)}.$$Here is the right approach. Say, first, that the a black random point (denoted by X) lies in the interval [x,x+\Delta x]. Then we have$$P(g\mid x\le X \le x+\Delta x)=\frac{P(\{x\le X \le x+\Delta x\} \cap g)}{P(x\le X \le x+\Delta ...

0

You have the right idea but the wrong Bayes formula. You need to replace $P(X=x|g)$, which is indeed $0$, by $f_g(x)$ where $f_g$ is the pdf of $N(-1,1)$. Similarly, $P(X=x)=0$ but you replace it by $f_g(x)P(g)+f_b(x)P(b)$. This is called a gaussian mixture.

1

Whoever wrote this shouldn't have written $\Pr(X,Y)$. One can write $\Pr(X=x)$, and the capital $X$ is the random variable, and the whole expression is a function of $x$, which could be anything in the range. When writing like that, the expression $$\sum_{x=1}^3 \Pr(X=x)$$ means $$\Pr(X=1)+\Pr(X=2)+\Pr(X=3)$$ and you don't see lower-case $x$ in there ...

3

The notation is indeed lacking at some points. $X$ and $Y$ are $\mathbb{R}^p$ resp. $\mathbb{R}$-valued random variables. In this context they must be functions on a common set $\Omega$, so $X: \Omega \to \mathbb{R}^p$ and $Y:\Omega \to \mathbb{R}$ are measurable functions. $\Pr(X, Y)$ denotes the joint distribution of $X$ and $Y$. If $\mathbb{P}$ denotes ...

1

First the finite case: Let $V$ be a non-negative-definite symmetric real $n\times n$ matrix. The finite-dimensional version of the spectral theorem implies that $V$ has a non-negative-definite symmetric square root with real entries, which let us call $V^{1/2}$. Let $Z\in\mathbb R^{n\times 1}$ be a column vector whose entries are i.i.d. random variables the ...

0

You have the right idea. In case one, write $$\mathbb{P}\{ 0 \leq X \leq y + \mu \} = \int_{0}^{y + \mu} \frac{1}{\lambda} e^{-\lambda x} \, dx = \lambda^{-2}(e^{- \lambda(y + \mu)} - 1).$$ Thus, $$F_{Y}(y) = \mathbb{P}\{Y \leq y\} = \lambda^{-2}(e^{- \lambda(y + \mu)} - 1) \quad \quad (y \geq \mu).$$ I am assuming you know how to compute the PDF from ...

1

The theorem is called Wiener-Khinchin Theorem. This is quite surprising a result. Not to reinvent the wheel, here is the relationship you are looking for with explanation on Wolfram.

1

Your probabilities are $\pi_1, 1-\pi_1-\pi_3$ and $\pi_3$ since the sum of the probabilities is $1$ Then$E(X)=\pi_1+2-2\pi_1-2\pi_3+3\pi_3=2$ and the result follows immediately

2

If we ae told that $X$ and $Y$ are jointly Normal, then we know that $$E[X|Y] = E[X] + \rho \frac{\sigma_X}{\sigma_Y}( Y-E[Y])$$ in which in your case reduces to $$E[X|Y] = \rho Y$$ In general, if we only know that the variables are marginally Normal, then I don't think there's much to say. Calling $E[X|Y]=g(Y)$ we know that write $$E[XY] = E[E[XY|Y]] ... 1 The pair (X,Y) is uniformly distributed in a triangle. That does not imply that either X or Y is uniformly distributed. Draw the triangle, and you'll see why X is more likely to be in a short interval near 1 than in an interval of the same length near 0, so X will not be uniformly distributed. Similarly, Y is more likely to be in a short ... 1 This is nonsense as it stands. If X_i has mean 0 and Y_i is independent of X_i, then X_i Y_i has mean 0. If you want to get a random variable of mean \eta, just add the constant \eta. That is: Suppose X_1, \ldots, X_n are independent {\cal N}(0,1) random variables, and {\bf Y} = (Y_1, \ldots, Y_n) is a vector-valued random variable ... 0 I asked for clarification because the way I interpret the question neither of the possible answers is correct. First, you cannot find correlation if either X or Y has all equal elements. In that case you would be trying to divide by a zero standard deviation. Second, I'm not sure what it means to say 'all other values take a constant value'. It seems to ... 0 Suppose there exists A such that \sum_{n=1}^\infty \mathbb{P}(A\cap E_n)<\infty. By the Borel-Cantelli Lemma,$$\mathbb{P}\left(\limsup_n (A \cap E_n)\right)=0.$$Note that \limsup_n (A \cap E_n) = A \cap G. Thus, \mathbb{P}(A \cap G)=0, so \mathbb{P}(G)=\mathbb{P}(G \setminus A) \le 1-\mathbb{P}(A)<1. Conversely, if P(G)<1, then let ... 2 Q1: You can prove it directly, without the Vitali convergence theorem. A hint to get you started: Let's let M_n = \max_{1 \le k \le n} X_k. Now fix \epsilon > 0 and write$$E\left[\frac{M_n}{n}\right] \le E\left[\frac{M_n}{n}; M_n \le \epsilon n\right] + \sum_{k=1}^n E\left[\frac{M_n}{n}; M_n > \epsilon n, M_n = X_k\right].(The inequality is ... 1 Since \begin{align} & D_n^2 = \max_{n^2 \leq k < (n + 1)^2 } \left|S_k - S_{n^2}\right|^2 \\ \leq & \sum_{k = n^2}^{(n + 1)^2 - 1} \left|S_k - S_{n^2}\right|^2 \\ = & \sum_{k = n^2 + 1}^{(n + 1)^2 - 1} \left|S_k - S_{n^2}\right|^2 \end{align} It follows that \begin{align} & E\left(D_n^2\right) \leq \sum_{k = n^2 + 1}^{(n + 1)^2 - ... 4 Set\begin{align*} A &:= \left\{ \limsup_{n \to \infty} \frac{S_n}{n} = \infty \right\} \\ B &:= \left\{ \liminf_{n \to \infty} \frac{S_n}{n} = - \infty \right\}. \end{align*}$$Hints: By symmetry, we have \mathbb{P}(A) = \mathbb{P}(B). By Hewitt-Savage's 0-1-law, \mathbb{P}(A)=\mathbb{P}(B) \in \{0,1\}. Deduce from the first two steps and ... 0 I would say P(X_i>t)=e^{-\lambda t}, random variables  x_i  are independent, \Rightarrow P(X>t)=(e^{-\lambda t})^m=e^{-m\lambda t} \Rightarrow P(X\le t) = 1-e^{-m\lambda t} See For rate = 3 0 You are asking for the distribution of Z = \min(Y_i). The system "dies" when the first subsystem "dies." If the reliability function of Y_i is$$1 - F_i(t) = R_i(t) = P(X_i > t),$$then can you write the formula for the reliability function for t > 0? Next, do you see how to express the reliability function R_Z(t) of Z in terms of the ... 2 (\Leftarrow) Let S_n=\sum_{k\le n}X_k. Then n^{-1}S_n is a reversed MTG and using the Doob maximal inequality$$\mathbb{E}\sup_{n}\frac{|X_n|}{n}\le 2\mathbb{E}\sup_{n}\frac{|S_n|}{n}\le \frac{2e}{e-1}(1+\mathbb{E}|X_1|\ln^+|X_1|)\le \frac{2e}{e-1}(1+\mathbb{E}|X_1|\ln(1+|X_1|))<\infty$$where \ln^+(\cdot)=\ln(\cdot)\vee 0. (\Rightarrow) ... 2 Along the same lines as Batminovski's answer but with a different flavoring: X has a Rayleigh distribution (familiar to wireless communications engineers as the distribution of the amplitude of a sinusoidal signal that has undergone Rayleigh fading) while Y has the distribution of \cos(\Theta) where \Theta \sim U[0,2\pi) is a uniformly distributed ... 5 The output of a function is uniquely determined by its input. However, for a measurable function as a random variable, we don't know the input: only after a realization can we know x. 4 The philosophical setup is that \Omega is a sample space, \mathscr{F} a collection of events that could happen. A measurable X is called a random variable because we are pretending that we do not know which \omega will occur. I am going to flip a coin and let X be the indicator of heads. I model it as X(\omega_1) = 1 and X(\omega_2) = 0 (i.e. ... 2 Physics Argument: Consider a two-dimensional ideal gas with molecular mass m at temperature T. The velocity X- and Y-components of a gas molecule, called V_x and V_y, respectively, are independent with the same \mathcal{N}\left(0,\alpha^2\right)-distribution, where \alpha:=\sqrt{\frac{k_\text{B}T}{m}} and k_\text{B} is the Boltzmann ... 3 Take U=XY and V=X then Y=U/V. (Note: V^2 >U^2) The Jacobian will be  J= \left[ {\begin{array}{cc} 0 & 1/v \\ 1 & -u/v^2 \ \end{array} } \right]  i.e. the |Det(J)|=1/v by which you'll multiply the joint density function. Since X and Y are independent,$$f_{X,Y}(x,y)=\frac{1}{\alpha^2 \pi} ...

1

Sufficient statistics should be $M = \text{card}\{i: Y_i > 0\}$, $S = \sum_i Y_i$ and $T = \sum_i Y_i^2$, with likelihood function  \eqalign{L(Y) &= {n \choose M} \Phi\left(\dfrac{50-\mu}\sigma\right)^{n-M} \prod_{i: Y_i > 0} \dfrac{\exp(-(Y_i-\mu)/(2\sigma^2)}{\sqrt{2\pi \sigma^2}}\cr &= {n \choose M} ...

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