New answers tagged

1

Here is a trick to map an instance of your problem to the framework of the paper "Distributed Stochastic Optimization via Correlated Scheduling" (IEEE Trans Netw. April 2016): http://ee.usc.edu/stochastic-nets/docs/distributed-optimization-ton.pdf Let's perform a notation shift: \begin{align} (Y_1,Y_2) &\leftrightarrow (\omega_1, \omega_2) \quad \mbox{[...


0

Let $a, c$ be the points in $A$ and $C$ with the greatest distance (i.e. $d(a,b) = d(A,C)$). Let $b$ be any point in $B$. Then $d(a,c) \leq d(a,b) + d(b,c)$ because $d$ is metric. But we know that $d(a,b) \leq d(A,B)$ because the latter is the maximum of all such distances. The same applies to $d(b,c)$. Thus, $d(A,C) = d(a,c) \leq d(a,b) + d(b,c) \leq d(A,...


0

The idea of linear regression is to infer from dataset the dependence of some random variable in other variable, what the term regression mean in this context is that from the dataset we regress to some kind of dependency that brought us to that data.


3

It's not true that $Y$ depends on $X$ and not vice versa. If $Y=X^2$, then $X=\sqrt Y$. Edit in response to the comments: Two random variables are either dependent or independent; there is no such thing as one variable being dependent on another. You may be confusing this with the concept of one variable being a function of another. Indeed, in $Y=X^2$ you'...


0

Generally finding UMVUEs can be really tedious. Just read a little bit about the concept of 'Ancillary statistics' and 'Basu's Theorem' which greatly simplifies the math. You will comfortably handle the problem. To give you an idea If T and V are the unbiased estimator and complete sufficient statistic of a parameter b respectively, then S=T/V is referred ...


0

I want to check if a coin is fair(lands 50% of the times on each side. I assume that delta of 0.1% between them is fair). How many flips do i need in order to be 99% confident that the coin is fair? For large n you can apply moivre laplace. So you approximate the binomial distribution by the normal distribution. Let $p=\frac{X}{n}$, where $X\sim Bin(...


0

Great question! So let's look at the binomial distribution. We can represent the standard deviation with the equation $\mu_x = \sqrt{npq}$ and $p$ in our case is the probability of heads. This is $.5$. Same with tails. $n$ is the number of tosses. So $\mu_x = \sqrt{.5 * .5 * n} = .5*\sqrt{n}$ Our mean here is zero and our standard deviation is $.5\sqrt{n}$. ...


0

There is a bus that departs from a bus stop every quarter hour from 6 am until midnight. You arrive at the bus stop between 7.10 am and 7.20 am. Note if you arrive before $7{:}15$ you only have to wait until then, but if you arrive in the later part of the interval you have to wait until $7{:}30$ (at least ten more minutes). Let $T$ be the time in minutes,...


0

First problem: Let $Y$ be the sum. Then $E(Y)=(16)(-47.1)=-753.6$ and $Y$ has standard deviation $\sigma=(4)(3.2)=12.8$. Note that $90.88=(7.1)\sigma$. The Chebyshev Inequality says $$\Pr(|Y-E(Y)|\ge k\sigma)\le \frac{1}{k^2}.$$ So the probability the absolute value of the difference is $\ge 90.88$ is $\frac{1}{(7.1)^2}$, and therefore our required ...


0

Yes.   But rather that thinking of it as "counting arrangements where green marbles are in row", think of it as "ways to split a row of green marbles into groups"; then find places for those groups. There are $\tbinom{20}{20}$ ways to group twenty items into one group.   Though I'd rather count that as $\binom{19}{0}$ ways to put no split among ...


0

I will assume $Z$ is a column vector. Let $P_{\perp} = I - X(X^TX)^{-1}X^T$. This is the orthogonal projector to the orthogonal complement of column space of $X$ and is idempotent and symmetric. Let $\hat{\beta} = (X^TX)^{-1}X^TZ$. This is the vector of coefficients obtained from regressing $Z$ on $X$. I will assume $Z$ does not lie completely in column ...


0

Comments: This is a very perceptive question. And the answer of @GrahamKemp illustrates an important convolution formula that works quite generally for sums of independent random variables. For example, suppose you put a lead weight into each die just beneath the corner where faces 1, 2, and 3 meet in order to bias the die in favor of 4, 5, and 6. Perhaps ...


0

You have the conjecture that the expected count of tosses until the third consecutive head is: $$\small\rm \mathsf E({HHH}) = 2 + \mathsf P({HH})\mathsf E({HHH}\mid { HH})+\mathsf P({HT})\mathsf E({HHH}\mid {HT})+\mathsf P({TH})\mathsf E({HHH}\mid {TH})+\mathsf P({TT})\mathsf E({HHH}\mid {TT})$$ Your logic flops on the miss-identification of the ...


1

Sure you may do that. Call the result of the first die $X$ and the second die $Y$ (presuming you can identify the die). $$\begin{align}\mathsf P(X+Y\geq 10) ~=&~ \sum_{x=1}^6\mathsf P(X=x)\mathsf P(Y\geq 10-x) \\[1ex]=&~\sum_{x=4}^6\;\sum_{y=10-x}^6 \frac 1{36} \\[1ex]=&~\sum_{k=1}^3\;\sum_{h=1}^{h}\frac 1 {36} \\[1ex]=&~\frac{1+2+3}{36}\\[...


1

Hint: Consider $(0,1)$ endowed with Lebesgue measure $\lambda$ (restricted to $(0,1)$) and $$X_n(\omega) := n 1_{(0,n^{-1})}(\omega).$$


0

One of your errors is when you say $E(HHH|T)=1+E(HHH)$, as you should have $E(HHH|T)=E(HHH)$, and I also doubt your simulation of $E(HHH|HH)\approx 5$ I would have thought you should have: $E(H)=E(H|T)=2$ $E(HH)=E(HH|T)=6$ $E(HHH)=E(HHH|T)=14$ suggesting $E(HH|H)=1+\frac12\times 0+ \frac12 \times E(HH|T) = 4$ $E(HHH|HH)=1+\frac12\times 0+\frac12 \...


0

Here's my take on answering the question using Did's response above. The nice thing about his proof is that it only relies on equality of the CDFs at the integers and equality of the first moments. Let $X$ be Poisson with CDF $F_X$ and $Y$ be any random variable with CDF $F_Y$ such that $F_Y(n)=F_X(n)$ for every nonnegative integer n. Then, $F_X⩾F_Y$ ...


3

The probability of a tied result on a fair die: $\mathsf P(A=B) ~=~ \tfrac 1 6$ By symmetry: $\mathsf P(A>B)~=~\mathsf P(B>A)$. By total probability: $~\color{blue}{\mathsf P(B>A)}~$ $\color{blue}{=~ \tfrac 12(1-\mathsf P(A=B)) ~\\=~ \dfrac 5{12}}$ Note: if you wanted a tie or greater: $~\mathsf P(B\geq A) ~$$=~ \mathsf P(A=B)+\mathsf P(B>A)...


1

For better understanding I would like the to rephrase the question. If two random variables have normal marginal densities are they jointly normally distributed? The answer is NO. This enlightening counterexample appears in the famous book "Counterexamples in Probability and Statistics" by Joseph Romano and Andrew Spiegel. Let $$g(x,y)=(1/2\pi)\exp\...


1

It is certainly not true. Suppose $X\sim N(0,\sigma^2)$ and $$ Y = \begin{cases} \phantom{-}X & \text{if } -c<X<c, \\ -X & \text{otherwise.} \end{cases} $$ Then $Y\sim N(0,\sigma^2)$. The correlation between $X$ and $Y$ depends on $c$, and for one special value of $c$ it is $0$, and for all others it is not. (That it is not normally ...


1

No, for an extreme example, take $X \sim N(0, 1)$, and consider the random vector $(X, -X)$, clearly, $X$ and $-X$ are (perfectly) correlated while they are not jointly normally distributed. A simple way to verify this is by noting $X + (-X) \equiv 0$ is not normally distributed. On the other hand, given $(X, Y)$ are jointly normal, then $X + Y$ must also ...


1

Yes, you are right. What you are describing is what referred to in machine literature as non-robustness of LSE to outliers. The reason for this name is that, if you have a noisy point in your data, far away from the rest, the squared nature of the penalization tends to move the whole model towards that single noisy outlier. A more robust alternative to mean ...


2

This is one of those things that is hard to give a single answer on. At the end of the day there are modeling drawbacks indeed, but there are so many theoretical advantages that it is rather hard to resist. I think the best point I can make is the one I will make first: absolute differences (probably the most obvious idea to consider) have a serious problem....


2

Find the sample mean $\bar X$ and the sample SD $S_X$ of your observations in $(0,1).$ Then make the transformation $Y_i = (X_i - \bar X)/S_X.$ The result is that $\bar Y = 0$ and $S_Y = 1.$ Keep track of the original values $\bar X$ and $S_X,$ and you can reclaim the original $X_i$s from the $Y_i$s. This transformation is somewhat similar to the '...


0

The win outcome is defined when there is a majority. So at least $N=3$ trials would be needed. In some situations there is no majority. The possible win situations can be seen in the following table \begin{array}{ll} \text{Number of trials ($N$)} & \text{Number of required outcomes to win} \\ 3 & \text{3 or 2} ...


0

Since you want multifarious methods, out of $36$ possible outcomes, $A=B$ in $6$. By symmetry, $15$ of the remaining $30$ must have $B>A$, thus $P(\;set\;score) = \frac{15}{36}$ and P(not set score) $= 1 - \frac{15}{36}$


1

Originally you have a set of quiz grades: {$q_1, q_2, q_3, ..., q_n$}, then 10 is added to each quiz grade to get a new set: {$q_1+10, q_2+10, q_3+10, ..., q_n+10$} If the mean was originally 40, the new mean will be 50. This can be seen since: $$\frac{\sum_{i=0}^n q_n+10}{n} =\frac{\sum_{i=0}^n q_n}{n} + \frac{\sum_{i=0}^n 10}{n} = Original\ Mean\ + \frac{...


4

You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with $\color{red}x$ are the outcomes where the second dice has a greater outcome than the first dice. $ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{...


3

The straight forward, but somewhat tedious solution is to simply list the outcomes: First throw 1 then throw 1, no score First throw 1 then throw 2, score First throw 1 then throw 3, score etc. Then it's just a matter of counting scores vs no scores (every outcome is equally probable). I get $15$ outcomes that scores and $21$ that doesn't (there's $36$ ...


0

This seems close: $$ I(A(a)) = \frac{n^2}{||A(a)||^2_F}$$, where $||A(a)||_F$ is the matrix entrywise 2-norm. For the example in the question: $ I(A(0)) = 3$ and $ I(A(1)) = 9/5 = 1.8$


0

Roll the die $n$ times and for $i\in\{1,\dots,6\}$ let $E_i$ denote the event that number $i$ does not show up. Then you are looking for: $$1-\Pr(E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6)$$ Now attack this using inclusion/exclusion and symmetry.


1

First, use inclusion/exclusion principle in order to count the number of combinations: Include the number of combinations with at most $\color\red{6}$ values showing: $\binom{6}{\color\red{6}}\cdot\color\red{6}^n$ Exclude the number of combinations with at most $\color\red{5}$ values showing: $\binom{6}{\color\red{5}}\cdot\color\red{5}^n$ Include the ...


0

If $h:\mathbb R\to\mathbb R$ is Borel-measurable, then $h(Y)$ is measurable with respect to the $\sigma$-algebra generated by $Y$: $$\sigma(Y) = \{ Y^{-1}(B):B\in\mathcal B(\mathbb R)\}. $$ It follows from the definition of conditional expectation that $$\mathbb E[X h(Y)\mid Y] = h(Y)\mathbb E[X\mid Y]$$ with probability one.


4

Somewhat more generally, let $X$ be any random variable whose moment generating function $M(z) = \mathbb E[e^{tX}]$ is analytic in a neighbourhood of $0$. This says that the series $\sum_{j=0}^\infty \mathbb E[X^j] t^j/j! $ has positive radius of convergence, i.e. $|\mathbb E[X^j]| \le C D^j j!$ for some constants $C, D$. Then we have uniqueness in the ...


6

The moment generating function of a Poisson random variable $X$ with parameter $\lambda$ is $$ \mathbb{E}\left[e^{tX}\right]=\sum_{n=0}^\infty \frac{e^{-\lambda}e^{nt}\lambda^n}{n!}=e^{\lambda(e^t-1)} $$ which converges for all $t$, and in fact is the restriction to $\mathbb{R}$ of an entire function. In particular, the power series for the moment generating ...


2

I found a polynomial that works, but I am not sure if it the "best" one. Let the $n$ points be $p_1,p_2\dots p_n$ with $p_j=(a_{j,1},a_{j,2}\dots a_{j,d})$. And suppose we want to find a polynomial $P$ with $P(p_j)=b_j$. Notice that the polynomial $P_j=(x_1-a_{j,1})^2+(x_2-a_{j,2})^2+\dots+(x_d-a_{j,d})^2$ is only zero at the point $p_j$ Therefore the ...


4

In most cases the data set will be a true set if you view it as a set of observations. In the example of temperatures, there are only a few different temperatures, but each one corresponds to a different day. Your data set consists of ordered pairs (day, temperature on that day) and no pair is repeated. The only way you get repetition is if you observe ...


9

Arthur is right; the term "data set" usually means multiset. For example, a bivariate data set just means a multiset of elements of $\mathbb{R}^2$. Intuitively, a multiset in $X$ is like a finite subset of $X$, except that repetitions are allowed. (Order still doesn't matter.) For example, the following are multisets in $\mathbb{N}$: $$\{1,2,2\} \quad \{2,1,...


5

A "data set" in statistics does indeed allow repetitions and in that sense is different from a "set" in set theory. It wouldn't make much sense otherwise: for instance, if you take the average daily temperature of each day for a year, there are only going to be a couple of dozen values (or a few dozen, in Fahrenheit), and the concept of average or mean, ...


1

We would like to compare $\mathbb{E}\{\frac{X}{Y}\}$ with $\frac{\mathbb{E}\{X\}}{\mathbb{E}\{Y\}}$ assuming positive RVs $X$ and $Y$. For simplicity, first assume $X=c$ is a constant. It is then easy to generalize. From the Jensen's inequality we know that for a convex function $f$ we always have $$f(\mathbb{E}\{Y\})\le\mathbb{E}\{f(Y)\}$$ Hence, knowing ...


0

Comment on related issues. In a balanced one-factor analysis of variance, one has the model $$Y_{ij} = \mu_i + e_{ij},$$ for groups (independent samples) $i = 1, \dots, g,$ replications $j = 1, \dots, n$ observation within each group, and $e_{ij} \stackrel{indep}{\sim} Norm(0, \sigma).$ Notice that all treatment groups are assumed to produce normal data, ...


0

just draw the line : $y = x-z$ . and and integrate the density function. you know $y = x -z$ intersect with line $y=0$ is in $x=z$ , and with line $x=1$ in $1-z$. so the solution is double the sum of the triangle: $2*\dfrac{(1-z)^2}{2}$. or the integral: $$ \int_{z}^{1} \int_{0}^{x-z}2dxdy = (1-z)^2 $$


0

The posterior density can be written as $$\frac{\exp(-\theta)\cdot 1(\theta>x)}{P(\theta>x)},\; x>0$$ The posterior mean is $$ \exp(x)\int^{\infty}_x \theta \exp(-\theta)d\theta$$ This integral with the proper normalising constant is $1$-CDF of a $\text{Gamma}(2,1)$ variable.


0

Short answer If you have a calculator, the invNorm function on the TI-84 or similar commands on other calculators will take a probability $p$ as input and output the upper bound of the interval (starting from the left tail) by the number of standard deviations away it is from the mean (a negative output is left of the mean and a positive output is right of ...


1

The question asks you to verify that $E[\bar{X}^2]\neq \mu^2$. The expansion $E[\bar{X}^2]$ gives: $$ E\Big[\bar{X}^2\Big]=\frac{1}{n^2}E\Big[\Big(\sum_iX_i\Big)^2\Big]=\frac{1}{n^2}E\Big[\sum_iX_i^2+2\sum_{i<j}X_iX_j\Big]. $$ Independence only gives $E[X_iX_j]=E(X_i)E(X_j)=\mu^2$ for $i<j$. Independence doesn't make expectations of the cross terms go ...


1

$$ E\bar{X}_n^2 = var(\bar{X}_n)+E^2\bar{X}_n = 1/n+\mu^2 >\mu^2. $$ So the simplest unbiased estimator would be $$ \bar{X}^2_n-1/n $$


2

The first question is: Under what circumstances is $$ \frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet $$ the same as $$ \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?} $$ The next question is: Why is $$ \frac \partial {\partial\theta} \...


3

Let us rewrite the product rule as follows: $$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$ Yours is just the generalization to $n$ factors, but is handled in the exact same way.


0

Here are three rather complementary recommendations: Introductory Mathematical Statistics: Principles and Methods by Erwin Kreyszig This classic is an elementary introduction which presents in the first part Descriptive statistics, followed by Probability Theory and the main part is Statistical Inference. E. Kreyszig has the talent to make ...


1

Let $f(X)=\log(|\det(I+XA)|)$; we calculate $Df_X$ in a point $X$ s.t. $I+XA$ is invertible, that is, $-1$ is not an eigenvalue of $XA$. $Df_X:H\in M_{p,k}\rightarrow tr(HA(I+XA)^{-1})=tr((I+XA)^{-T}A^TH^T)$ or $Df_X(H)=<(I+XA)^{-T}A^T,H>$ (the scalar product over the matrices). In other words, the gradient of $f$ is $\nabla(f)(X)=(I+XA)^{-T}A^T$, ...



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