4

Update/clarification. (with thanks to Nathan's pointing this out!) My original answer below was not a proof that $\|p^*\|_\infty/\|p^*\|_2^2\le2\sqrt n/5$. This is because the minimum I obtained was the minimum for the function $\sqrt n\|p\|_2^2-\|p\|_\infty$, so a direct substitution would lead to a misleading result of the desired ratio, as we have $\|p\|...


3

Since expectation (as an operator) is linear, the following statement is correct regardless of the independence between $X$ and $Y$: $$E(Z) = E(X − Y) = E(X) − E(Y) = 0.5 − 0.5 = 0.$$ However, if we do want to find the probability (density) function (pdf) of $Z = X - Y$, we can use the convolution formula to obtain this. Generally, the density of $Z = X - Y$ ...


3

Here is a counterexample: assume each $Y_i$ has distribution $\mathcal N(0,1)$, so that $\frac 1{\sqrt n}\sum_{i=1}^nY_i \sim \mathcal N(0,1)$. Suppose for the sake of contradiction that there is some $X\sim \mathcal N(0,1)$ independent of $(Y_1,\ldots)$ with $\frac 1{\sqrt n}\sum_{i=1}^nY_i \xrightarrow{L^p}X$. Since $X$ is independent of $(Y_1,Y_2,\ldots)$ ...


3

Solution 1. The range of $Y=\sin X$ with $X\in[0,2\pi]$ is $[-1,1]$ not $[0,0]$ because $\sin$ is not monotone. You can draw the graph of $\sin$ with $X\in[0,2\pi]$ to see the range. For $y\in[0,1]$, $$Y=\sin X\leq y\iff 0\leq X\leq\arcsin y\quad\text{or}\quad\pi-\arcsin y\leq X\leq 2\pi.$$ Thus, \begin{align*} F_Y(y)&=P(Y\leq y)=P(0\leq X\leq\arcsin y)+...


2

A summary of the brief discussion with Gabriel Romon: Given any independent, identically distributed random variables $(Y_k)_{k\in\mathbb N}$ with $\mathsf EY_k=0, \mathsf VY_k=1$, if $X$ is independent of $(Y_k)_{k\in\mathbb N}$, then according to the Central Limit Theorem,* $-X+\frac1{\sqrt n}\sum_{k=1}^\infty Y_k$ converges in distribution to (slight ...


2

The answer will depend on your parameters and your choice of noise. I'll give partial answer, using the following reference: Peigné, Woess, Stochastic dyamical systems with weak contractivity properties. I. Strong and local contractivity, which contains many helpful other references (Goldie...). First, I will exclude the value $0$ of the domain, so that $(...


2

In general no: take $N$ standard normal and $\varepsilon$ a random variable independent of $N$ taking the values $1$ and $-1$ with probability $1/2$. Let $X_1=N$ and $X_2=N\varepsilon$. Then $X_1$ and $X_2$ have a standard normal distribution and are uncorrelated. But $X_1+X_2=N(1+\varepsilon)$ is not Gaussian, as it takes the value $0$ with probability $1/2$...


2

As long as all expectations are defined (or always, if you allow infinite values). You have $|X+Y|\leq |X|+|Y|$ almost surely, and then you can take the expectation on both sides.


2

Are $\hat{X_1}$ and $\hat{X_2}$ independent? If so then Let $U=$ sum and $f_U(u)=\int\limits_0^1 I_{(0,1)}(x-u)dx$ So $f_U(u)=u$ for $0\le u \le 1$ and $f_U(u)=1-u$ for $1\le u \le 2$. $\hat{Y}=\frac{U}{2}$ with $f_{\hat{Y}}=4y$ for $0\le y\le 0.5$ and $=4(1-y)$ for $0.5\le y\le 1$..


2

Suppose that $X,Y$ are random variables and consider $Z = XY$ \begin{align} F_Z(z) & = P(Z \le z) \\ & = P(XY \le z) \\ & = P(X \le z/Y | Y > 0)P(Y>0) + P(X \ge z/Y | Y < 0) P(Y<0) \\ & = \int_0^\infty \int_{-\infty}^{z/y} f_{X,Y}(x,y) \ dx dy + \int_{-\infty}^0 \int_{z/y}^\infty f_{X,Y}(x,y) \ dx dy \\ f_Z(z) & = \int_0^\...


2

If $X$ and $Y$ are allowed to be discrete random variables rather than only continuous random variables such as normal random variables, then when $X$ and $Y$ are independent Bernoulli random variables with parameters $p$ and $q$ respectively, $XY$ is also a Bernoulli random variable with albeit with a different parameter $pq$. If you want symmetric ...


1

This is an old question, but a good one, and one that has some interesting connections across topics that might seem unrelated- the uniform distribution on the n-sphere has a relationship derived in information geometry to the Jeffreys prior for multinomial probabilities- taking the square root of probabilities results in a diffeomorphism between the unit ...


1

You are missing several additional inequalities satsifesd by $m$ and $w$. The given inequalities for $x$ and $y$ are equiavlen to to the following: $0 \leq m \leq 1$, $0 \leq w \leq \frac 1 2$, $w \leq \frac m 2$, and $m \leq w+\frac 1 2$. The third inequality comes from $x \leq y$. The last one comes from $y \leq 1$. [Always make it a point to check if ...


1

The functional form of the transformed joint density is correct, but the support is incorrect. All you have to do to rectify this is to first note that the support for $(X,Y)$ is bounded by the lines $$X = 0, \quad X = Y, \quad Y = 1.$$ Because the transformation is linear and invertible, the resulting support in $(M, W)$ space is also bounded by three ...


1

The mean of uniform distribution on $\{1,2,...,n\}$ is $\sum\limits_{k=1}^{n}\frac k n=\frac {n(n+1)} {2n}=\frac {1+n} 2$. Hence, the mean of $Y$ given $X$ is $\frac {1+X} 2$. Now take the mean of this: $EY=E(\frac {1 +X} 2)$. Can you complete this?


1

Given that the continuous random variable $X$ has probability density function $$f(x) = ax - bx^2, \quad 0 \leq x \leq 2$$ and 0 elsewhere, we have $$1 = \int_{-\infty}^\infty f(x) dx = \int_0^2 f(x) dx = \int_0^2 (ax - bx^2) dx = \left[\frac{a}{2}x^2 - \frac{b}{3}x^3\right]\bigg|_0^2 = 2a - \frac{8}{3}b.$$ Given also that $E[X] = 1$, we have $$1 = E[X] = \...


1

Let $p_n$ be the probability that A wins, given that his current bankroll is $n$. We are asked for $p_3$. We know that $p_0=0$ and that $p_8=1$. Since the probability that A wins any game is $\frac23$, we know that $$p_n=\frac23p_{n+1}+\frac13p_{n-1},\ 0<n<8$$ If you substitute $n=1$, you'll be able to express $p_2$ in terms of $p_1$. Then ...


1

$(1 - (1-0.3)\cdot (1-0.4\cdot 0.4)\cdot (1-0.2))\cdot 0.1 = (1-0.7\cdot 0.84\cdot 0.8)\cdot 0.1$ The point being, you need at least one of the parallel paths to work which is the opposite of all of the parallel paths not working.


1

Oh, it converges; the integrand diverges at $0$, but in a harmless way, like $\int_0^\epsilon x^{-1/2}dx=2\epsilon^{1/2}$ for $\epsilon\ge0$. The Gamma function $\Gamma(s):=\int_0^\infty x^{s-1}e^{-x}dx$ satisfies $\Gamma(1/2)=\sqrt{\pi}$. You can easily show $\Bbb Ee^{tX}$ is $(1-2t)^{-1/2}$ for $t<\tfrac12$, otherwise it's infinite. Needless to say, ...


1

Recall that exponential distributions are memory-less. Therefore, at the time when the first of Jake and Naomi get done with their service, Paul would go to the server that becomes empty. Say he goes to server 1 at time $t=s$. Then, due to memoryless property, $\mathbb{P}[X_2>t+s|X_2>s] = \mathbb{P}[X_2>t]$. Thus, we are comparing the distributions ...


1

Welcome to MSE! The idea is that they have a function, $\Phi$, which computes the probability of a standard normal variable falling below a stated value. So, the entire computation is transforming the random variable that you start with, which they assume to be Gaussian with a mean, $\mu$, and a standard deviation, $\sigma$, to a standard gaussian which has ...


1

I would use the probability density function, $f(x)$, to calculate the expectation using $$ E(x) = \int x f(x) dx $$ We can get $f(x)$ from the cumulative density function by differentiating in each region: $$ f(x) = \left\{ \begin{array}{ll} 0 & \quad x<-3 \\ \frac{1}{6} & \quad -3\le x<0 \\ 0 & ...


1

The mathematical proof shared by Jonathan Christensen in the answer below is great. Here is my intuitive interpretation: I was also deeply confused when every "simple" explanation out there references the Poisson distribution which is intuitively not right because the underlying process should be a Binomial. I too initially thought that the chi-...


1

The easiest way to solve the problem is to do a drawing of the transformation function and realize that $$\mathbb{P}[Y>y]=F_X\left(\frac{1}{y}\right)-F_X(y)=e^{-y}-e^{-1/y}$$ thus $$F_Y(y)=1+e^{-1/y}-e^{-y}$$ derivating you get your density...and observe that $y \in(0;1]$ $$f_Y(y)=\left[\frac{e^{-1/y}}{y^2}+e^{-y}\right]\cdot\mathbb{1}_{(0;1]}(y)$$ this ...


1

What's the probability that the first and second error both occur on page #89? Assuming no correlation, it's $(0.01)^2 = 0.0001$. What's the probability that any two specific errors occurs on page #89? It's $0.0001$. Assuming that exactly two of the ten errors occurs on page #89, how many different ways could this occur? That's ${}_{10}C_{2} = 45$. So, the ...


1

Turns out I found the answer. For anyone interested in the future, there is this wonderful paper by Hoeffding himself. The proof of the fact I wanted can be found in Theorem 4.


1

It is known that $$ \Gamma (x) \ge \sqrt {2\pi } x^{x - 1/2} e^{ - x} $$ for all $x>0$ (cf. http://dlmf.nist.gov/5.6.E1). Because of Stirling's formula, this is a rather sharp lower bound. This gives the simple bound $$ \frac{{2^{2^{ - m} (2^m - 2^n )(m - n)} e^{ - 2^{n - m} } }}{{\Gamma (2^{n - m} )}} \le \frac{1}{{\sqrt {2\pi } 2^{(n - m)/2} }}. $$ ...


1

While you can use a hypergeometric distribution, it is not necessary in this case because there is only one token you are interested in. This simplifies the probability model as follows. Consider a bowl of $n$ tokens, one of which is winning. We first compute the probability that you will win on exactly the $x^{\rm th}$ draw. To do this, imagine that ...


1

About the statement $$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+...+(-1)^k\frac{1*3*7*...*(2k-1)}{x^{(2k+1)}})}\xrightarrow{x \to +\infty} 1$$ In fact, it holds true, even for the general form like this one $$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k})}\xrightarrow{...


1

I think the answer by TheSimpliFire was only almost correct but not quite because it identified the wrong $p$ on the $(\|p\|_\infty,-\|p\|_2^2)$ Pareto frontier. Here's my attempt, which shows that $\lim_{n\to\infty}\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\sqrt{n}\|p\|_2^2}=\frac12$, approaching it from above. Let $\Delta^n$ denote the $(n-1)$-dimensional ...


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