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How to Calculate $\frac{1}{\sqrt{n}}x^{\top}y$ given random vectors $x$ and $y$?

Assuming $X$ have finite second moment, $$Z_n:=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i Y_i$$ has zero mean and variance $\mathbf{Var}(Z_n)=\mathbf{E}[X^2]$. So, By the central limit theorem, $Z_n$ ...
Sangchul Lee's user avatar
1 vote

MIT Statistic For Applications course Question 1

It's Markov's inequality. thanks Matthew Towers.
tom tom's user avatar
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2 votes
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How to Determine Independence of Events Using Probability

Independent has a very specific meaning in probability, namely events $A$ and $B$ are independent if and only if $Pr(A|B)=Pr(A)$. A consequence of this, which comes from the rule $Pr(A|B)=\frac{Pr(A\...
Red Five's user avatar
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2 votes
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PDF of $Y=g(X)$ when $X\sim N(0,1)$. $g(X)$ is a piecewise function where each part is constant.

Your expression for the CDF of $Y$ is close, but should be $$F_{Y}\left(y\right)=\begin{cases} 0, & y<-1\\ 1/2, & -1\leq y<1\\ 1, & y\geq1 \end{cases}.$$ Since $Y$ is discrete, it ...
AOS's user avatar
  • 171
1 vote
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Expected value and average value problem

If $$1-\Pr\left\{ 0.64\leq X\leq0.66\right\}$$ $$=1-\Pr\left\{ \dfrac{0.64-\mu}{\sigma/\sqrt{n}}\leq\dfrac{X-\mu}{\sigma/\sqrt{n}}\leq\dfrac{0.70-\mu}{\sigma/\sqrt{n}}\right\},$$ then $$\Pr\left\{ 0....
AOS's user avatar
  • 171
0 votes

Suppose $Y\backsim N(0, 1)$ and $X\backsim N(0, Y^{-2})$. Show that $X$ has the standard Cauchy distribution.

$X\sim N(0,1/Y^2)$ implies $X\sim Z/|Y|$ where $Z\sim N(0,1)$ is independent of $Y$. Therefore using $$\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^{\infty}\frac{u^{a-1}}{(1+u)^{a+b}},\ \Gamma(p)\...
Letac Gérard's user avatar
1 vote

Suppose $Y\backsim N(0, 1)$ and $X\backsim N(0, Y^{-2})$. Show that $X$ has the standard Cauchy distribution.

If $$X \mid Y \sim \operatorname{Normal}(0, Y^{-2}),$$ then the variance of $X \mid Y$ is $1/Y^2 > 0$, and the conditional density of $X \mid Y$ is $$f_{X \mid Y}(s \mid t) = \frac{|t|}{\sqrt{2\pi}}...
heropup's user avatar
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2 votes

$\mathbb{E}[X^2]\leq k \mathbb{E}[X]^2$, upper bound second moment from first moment

Despite your "It's easy to show ...", Bernoulli random variables with $\mathbb P(X=1)=p$ are a counter-example: $$\dfrac{\mathbb{E}[X^2]}{ \mathbb{E}[X]^2} = \dfrac{p}{p^2} = \dfrac1p$$ and ...
Henry's user avatar
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5 votes
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Find the pdf of $Y=\frac{X}{X+1}$

The first step is to think about the support of $Y$. If $X \in [0,1]$, then what is the range of $Y = f(X) = X/(X+1)$? When $X = 0$, we have $Y = 0$. But when $X = 1$, then $Y = 1/2$. Intuition ...
heropup's user avatar
  • 137k
1 vote

What to call a sequence of Bernoulli trials with different probabilities?

It is Poisson Binomial Distribution : In Probability theory and Statistics, the Poisson Binomial Distribution is the Discrete Probability Distribution of a sum of independent Bernoulli trials that are ...
Prem's user avatar
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1 vote

What to call a sequence of Bernoulli trials with different probabilities?

If $p$ is fixed and all the experiments are Bernoulli trials then we can call it a Binomial Distribution. If $p$ is not fixed, then it is more useful to consider the Bernoulli trials as separate ...
Red Five's user avatar
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1 vote
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Uniform distribution problem solving

Consider one end (left end which I'll consider to be the back end) of the car to be at $x$ meters away from $0$. if say $3<x<5$, then the remaining space is $x$ meters behind the car(which is ...
Mr.Gandalf Sauron's user avatar
0 votes

1D Brownian motion: stopping time and stopping state not independent

Given $X_{T}=-3$, it means that state 1 has not been hit by time $T$. Hence, by the reflection principle, $$P(T\le t|X_{T}=-3, X_0=0)=1-P(hitting\; 1\; by\; time\; t)\\ =1-2\int_1^\infty \frac{1}{\...
toronto hrb's user avatar
3 votes
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Where am I wrong in this variance calculation?

$$E^2(X+Y) = (E(X+Y))^2 = (E(X)+E(Y))^2 = E^2(X) + 2E(XY) + E^2(Y)$$ The term with the 2 will cancel out and what remains is the variances you need.
Vincent Batens's user avatar
1 vote
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Variance of affine transformation of random variable

I think you mixed up the tranposes in your second line. Assuming that $X$ is a random column vector, where each element has finite variance. Denote $E[X] := \mu$, then the covariance matrix is defined ...
minginator's user avatar
1 vote
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Equivalent definitions for Support of random variables

Proof of $G_X \subseteq S_X$ under the assumption that $p_d$ is continuous: Suppose $x \in G_X$. If $x \notin S_X$ then $x \notin U_X$, so there exists $\epsilon >0$ such that $P_X((x-\epsilon, x+\...
geetha290krm's user avatar
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0 votes

Creating a martingale given

You are correct, the filtration by construction makes the process adapted. You also need to check the integrability condition of the definition. To check the martingale property, you need to use the ...
Oscar's user avatar
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2 votes

More HH than HT after tossing $n$ coins

In this reddit post, user bobjane asserts the following fact, which implies that player 2 wins more often than player 1. Claim: For all $n\ge 3$, let $S_1$ be the set of sequences in $\{\mathrm{H,T}\}...
Mike Earnest's user avatar
2 votes
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Probability that difference between any two numbers is at least $5$

Make $4$ blocks of chosen-unchosen with chosen shown as bullets, plus a short block of one chosen, viz four boxes like $\boxed{\bullet\circ\circ\circ\circ}$ and one $\boxed{\bullet}$ $150-21 = 129$ ...
true blue anil's user avatar
0 votes
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Sigma field generated by gaussian random variable

This answer is based on the first response on @user469053. Please tell me if it is correct. Let $A\sim \mathcal{N}(0,1)$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mathbb{P})$, with $A(\omega) = \omega$....
Bigggie's user avatar
  • 183
0 votes

Sigma field generated by gaussian random variable

Let's start with a general construction of how to construct independent copies of random variables. Let $A,B$ be random variables on (possibly different) probability spaces $(\Omega_0,\mathcal{A}_0,\...
user469053's user avatar
  • 2,057
1 vote

Sigma field generated by gaussian random variable

$\sigma(X)=\sigma(Y)$ iff $X$ is measurable function of $Y$ and conversely. So any independent gaussian variables $X$ and $Y$ are such that $\sigma(X)\neq\sigma(Y)$.
Will's user avatar
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0 votes

Expectation of a Standard Normal Random Variable

I see the question is already answered, so I'll present a trick that can help later. I'll break the problem into two sections, even and odd powers, this allows us to use symmetry arguments. Odd Power ...
amongus's user avatar
0 votes

"Random" generation of rotation matrices

If you have MATLAB, I suggest using the code from What Is a Random Orthogonal Matrix? by the late Professor Nicholas Higham (n in code represents the dimension). <...
Greenhand's user avatar
1 vote
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Concentration Inequality for the bounded random variables

Without loss of generality we take $b=1.$ Let $X_1,\ldots,X_n$ independent random variables valued in $[0,1]$ Consider $Z=X_1+\cdots+X_n$ , $m=E(Z)$ and $a>1.$ We show that $$\Pr(Z>am)\leq \...
Letac Gérard's user avatar
1 vote
Accepted

A random point $X$ from $(0,1)$ is selected. After $X$ is selected, $Y$ is selected from $(0,x^2)$. Find the marginal p.d.f. $f_2(y)$

There is no reason to believe that the joint density of $(X,Y)$ should be uniform on its support. We have a hierarchical model: $$X \sim \operatorname{Uniform}(0,1), \\ Y \mid X \sim \operatorname{...
heropup's user avatar
  • 137k
1 vote
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Find a continuous increasing function $T:(0,1)\to (0,\infty)$ s.t. $Y=T(X)$ has p.d.f $g(y)=\frac{2}{(y+1)^3}$

You were nearly done: $$g(y)=\frac{d}{dy}[T^{-1}(y)]$$ $$\implies \frac{-1}{(y+1)^2}-\frac{-1}{(0+1)^2}=T^{-1}(y)$$ $$\implies \frac{-1}{(T(x)+1)^2}+1=x$$ $$\implies T(x)=\frac1{\sqrt{1-x}}-1.$$
Anne Bauval's user avatar
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0 votes

Tightness of random variables and Borel-Cantelli Lemma

It's a definition of the term "tight". There is no "fact" in the definition, so there is nothing to "prove". A sequence $(X_n)_{n=1}^{\infty}$ is said to be tight if the ...
James Martin's user avatar
0 votes

Summing Laplace random variables

A standard Laplace variable has the form $X_1-Y_1$ where $X_1$ and $Y_1$ are iid with density $e^{-x}1_{(0,\infty)}(x)$. The sum of $n$ iid standard Laplace variables has the form $S_n=X_n-Y_n$ where ...
Letac Gérard's user avatar
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The distribution of a random variable under P

So, in probability theory, a (real-valued) random variable $T:\Omega\to\mathbb R$ is defined as a measurable function between a probability space $(\Omega,\mathcal F, \mu)$ and the space $(\mathbb R,\...
Stratos supports the strike's user avatar
1 vote

Summing Laplace random variables

Late to the party, but as I struggled over this myself over the weekend, here's my humble answer (assuming the Laplace rvs are independent): $$ \begin{array}{ccl} f_{S_{N}}\!\left(s\right) & = &...
RSMax's user avatar
  • 11
4 votes

Variance of the difference of two random variables

Note that $X+Y=10$. So $D=X-Y=X-(10-X)=2X-10$. This gives $var (D)=4 var (X)$. So $Var(D)=4\cdot np(1-p)=4\cdot 10\cdot \frac{1}{2}(1-\frac{1}{2})=10$.
geetha290krm's user avatar
  • 37.1k
3 votes
Accepted

Coupling of Random Variables Where Probability of Movement is Dependent on Position

In words, I'll define the following coupling: at each time step, we find which of the two particles is more likely to move right. Then we first decide if it moves right or left. If the more likely ...
stochasticboy321's user avatar
1 vote

The "turning-point fraction" of a random sample from a discrete distribution must have expectation less than 2/3?

Requested from comments and building on Misha Lavrov's answer: This uses linearity of expectation on each of the $n-2$ triplets $X_{i-1}, X_i, X_{i+1}$ with $1<i<n$. If $q_3= \sum\limits_x \Pr[...
Henry's user avatar
  • 157k
5 votes
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The "turning-point fraction" of a random sample from a discrete distribution must have expectation less than 2/3?

We can use the linearity of expectation argument in the discrete case as well, we just have to be a little bit more careful. For each $i$ with $1<i<n$, there are three cases: The values of $X_{...
Misha Lavrov's user avatar
1 vote

Convergence in distribution of empirical cdf almost surely.

While the existing answer provides a valid route to proving the desired result via the Glivenko-Cantelli theorem, I would like to offer a more simpler approach that avoids some of the technicalities ...
Anacardium's user avatar
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0 votes

Random vector $(X, Y)$ has a uniform distribution on the unit circle.

They are not independent: A mathematical reasoning can be: $f_{x,y} = \frac{1}{\pi}$ (given) From here we can find $f_{x} = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f_{x,y} \, dy = \frac{2\sqrt{1-x^2}}{\pi}...
HopelessBhai's user avatar
0 votes

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

Another way: \begin{gather*} f_{S_{N-1}|N}(s|n)=ns^{n-2}\int_1^{1+s}dx=ns^{n-1},\quad E[S_{N-1}|N]=\frac{n}{n+1}\\ f_{S_N|N}(s|n)=\int_{s-1}^1nx^{n-2}dx=\frac{n(1-(s-1)^{n-1})}{n-1},\quad E[S_N|N]=\...
Speltzu's user avatar
  • 543
0 votes

Random vector $(X, Y)$ has a uniform distribution on the unit circle.

Let $$ A=\left[-\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right]\times\left[-\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right] $$ be a square of area two that is inscribed in the circle with radius one. Simple ...
Kurt G.'s user avatar
  • 14.3k
1 vote

Random vector $(X, Y)$ has a uniform distribution on the unit circle.

Regardless of whether we're talking about just the boundary or including the interior of the circle, it is very easy to see that the components are not independent by considering a region between the ...
ConMan's user avatar
  • 24.3k
0 votes

Sum of Independent exponential random variables 1.0

If $\theta _2>\theta_1$ (say) then for $s$ small enough $$E(e^{s U})=\frac{\theta_1}{\theta_1-s}\frac{\theta_2}{\theta_2-s}=\frac{1}{\theta_2-\theta_1}\left(\theta_2\frac{\theta_1}{\theta_1-s}-\...
Letac Gérard's user avatar
0 votes

Local time of a random walk $=$ collision times of random walks

I'm just going to give the proof myself. Write $S^i_n = \sum_{j=1}^n X^i_j$ for $i = 1,2$. Actually, it's not hard to establish that the following holds: $$ Z_1 := (X^1_1 - X^2_1, X^1_2- X^2_2 ,\dots ,...
Nazono Sumiko's user avatar
2 votes

Change of Variables and Expectation of Random Variable

This equation is wrong: $$\mathfrak{f}_y(y)=\frac{d}{dy}(f_y(y))=\frac{d}{dy}\left(f_x\left(\frac{y-B}{A}\right)\right)=f_x(x)\cdot \frac{1}{A}$$ You accidentally turned $(y-B)/A$ into $x$ in the last ...
Noble Mushtak's user avatar
1 vote
Accepted

Expected number of passengers in a bus, if both bus and passengers arrival time have a Poisson distribution.

This solution is incorrect, because this statement is wrong: I've also worked out that the expected length of interarrival interval is 5 minutes. If you choose random interarrival intervals and take ...
Noble Mushtak's user avatar
1 vote
Accepted

Random variable $X$ it is distributed evenly over the segment $[-\pi, 0]$. Find the distribution

The CDF (cumulative distribution function) of $X$ is $$ \mathbb P(X<x)=\frac{x+\pi}{\pi}\,,\quad x\in[-\pi,0]\,. $$ The PDF (probability distribution function) is the derivative of this. To find ...
Kurt G.'s user avatar
  • 14.3k
0 votes

Determining whether a random variable is more likely to be positive or negative

This is not the case as written. Since the probabilities decrease away from the centre on both sides, it doesn’t matter that $|k_1|\lt|k_2|$; almost the entire mass on the positive side can be ...
joriki's user avatar
  • 238k
1 vote
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Derivative of expected value

I will assume $A$ is a constant. Let us write $$ D = D(z) = \frac{\mathrm{d}}{\mathrm{d}z} \mathbb{E}\left[ e^X (A - Y(X)) \mathbf{1}_{\{ Y(X) \leq z \}} \right]. $$ Since it is possible that $D$ does ...
Sangchul Lee's user avatar
3 votes

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

Your idea can work as follows. For $i\in \{2,3,\dots \}$, we can show that for $\color{blue}{t\in (0,1)}$: $$\color{blue}{F_{X_{N}|N=i}(t)=1-\left ( 1+ \frac{1}{i-1} \right ) (1-t) +\frac{1}{i-1} (1-t)...
Amir's user avatar
  • 4,764
2 votes

How is the mean of a high dimensional random variable defined?

If $X = (X_1, X_2, \ldots, X_n)^T$ then the mean of $X$ is defined to be $$ ( \mathbb{E}[X_1], \mathbb{E}[X_2], \ldots, \mathbb{E}[X_n])^T $$ where $$ \mathbb{E}[X_k] = \int_{\mathbb{R}} x_k \...
Thành Nguyễn's user avatar
3 votes
Accepted

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

This step is wrong: $$\int_0^1\mathbb P[X\leq t\cap X> 1-Y|Y<1, Y=y](i-1)y^{i-2}\ \text dy= \int_0^1(t-1+y)(i-1)y^{i-2}\ \text dy$$ You can't say $\mathbb P[X\leq t\cap X> 1-Y|Y<1, Y=y]=t-...
Noble Mushtak's user avatar

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