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10

Let $(X_i, Y_i)$ for $i=1,2$ be i.i.d. and has uniform distribution on $[0, 1]^2$. Then $|X_1 - X_2|$ has PDF $$ f(x) = \begin{cases} 2(1-x), & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}. $$ This shows that the average distance is \begin{align*} \int_{0}^{1}\int_{0}^{1} 4(1-x)(1-y)(x^2 + y^2)^{1/2} \, dxdy &= ...


7

When you say "without calculus", then presumably you are talking about just using something like basic algebra or Euclidean geometry. However, as you have seen from the up-voted posted answers and comments, the exact answer involves natural logarithm. It is basically impossible to define natural logarithm without some sort of calculus and/or real analysis. ...


3

I have come across this notation in a different context (page 80, eq. (4.40)), where it was defined as $$ x^+ \equiv \max(x,0) $$ And correspondingly $$ x^- \equiv \min(x,0) $$ That is, the $+$ or $-$ indicates that the value is clamped to non-negative or non-positive numbers. Next to the other term in that equation, it seems likely that the same meaning ...


2

There are a number of 'measures of diversity' (see Wikipedia). Perhaps the simplest is called the Simpson Index (often denoted $\lambda$). It is the sum of squares of relative frequencies for the categories. In your case: $$\lambda = \sum_{i=1}^3 r_i^2 = .3^2 + .5^2 + .2^2 = .38.$$ Very roughly speaking, this is the probability that two people chosen at ...


2

This is a very broad area. Statistical tests can only rule out strings on the basis that they are "statistically unlikely to have been generated by a true random generator". Some keywords are NIST Tests, DieHard tests. Look at the Crypto and/or Theoretical Computer Science stackexchange sites and there will be posts tagged randomness or randomness testing. ...


2

Suppose we are tossing repeatedly a coin that has probability $p$ of landing heads. Then the probability of a tail is $1-p$, which we call $q$. Let $X$ be the number of tails before the first head. We want to find $E(X)$. Let $c=E(X)$. We condition on the result of the first toss. If we get a head on that toss, then $X=0$, and therefore $E(X)=0$. More ...


2

I would ask that "someone" for more details first why this should be true. Forget about the logarithm; if its arguments are well-defined (positive), the inequality holds (assuming it should hold for any $h$) if and only if $\Sigma\preceq I$. But take, e.g., for $\rho_1=\rho_2=\rho_3=1/4$ and $h=[1,1]^T$, $$ h^*\Sigma ...


2

The probability that two independent continuous random variables are exactly equal is 0. And, of course, normal random variables are continuous. Continuous random variables put 0 probability on any one point. A more interesting problem might be the probability that the two variables are within one unit of each other. For discrete random variables the ...


2

$$Pr[X \text{ occurs at least once in sixty years }]$$ $$ = 1 - Pr[X \text{ occurs never in sixty years}] = 1 - Pr[X \text{ does not occur in a given year}]^{60}$$ $$ = 1-(1-1/5 000 000)^{60}$$


2

$X$ is the count of Bernoulli trials before the first success. $A_1$ and $A_2$ are the events of a success and failure (respectively) on the first trial.   These events are mutually exclusive and exhaustive; they partition the outcome space. $\begin{align} \mathsf E(X) & = \sum_{x=0}^\infty x\mathsf P(X=x) & \text{by definition of expectation} ...


1

You can approximate the calculation of Walter by the expression: $$1-(1-1/5 000 000)^{60}\approx 1-e^{-1.2\cdot 10^{-5}}$$ The approximation is sufficient, because 60 is large enough. In both cases it is $1-0.99998800007$ The approximation is done by applying the relation: $$\lim_{n \to \infty } \left( 1+\frac{x}{n} \right)^n=e^x$$


1

One of the other answers gives an exact solution, but if your number of chances for the event to occur are somewhat large and the probability of the event occurring in one trial is very small, then on a computer you may run into underflow errors and/or inaccuracies with floating point arithmetic. A very good approximation, provided that the probability of ...


1

The issue that you rised with this question is in the area of robust statistics. In the case of estimating a parameter, it is called robust parameter estimation. There is a good book by Huber. I think this one will help alot. The idea is as follows. When you are estimating a parameter, the regular process first finds the log likelihood ratio of the density ...


1

Each time you simulate data from fitted parameters you find another estimate of the parameter. If the original 95% CI is valid, about 95% of these new estimates ought to lie in the original CI. You must be studying, or about to study, parametric bootstrapping. There are so many different formulations of this idea that I hesitate to get into a theoretical ...


1

S1=10*10=100 S2=75 p=75/100=3/4


1

Consider a point chosen randomly in $[0,10]^2$. We want to find the area of the points satisfying $|x-y| \le 5$ for $(x,y) \in [0,10]^2$. If you'll excuse the rude drawing, you can see that the set of satisfying points takes ${3\over 4}$ of the area. The answer to your other question is similar, but doesn't have the convenient fact that $5 = {1 \over 2} ...


1

A "quick and easy" route: Think of it as placing one restaurant first and then seeing where you can place the next. The two extremes are when you place the first restaurant in the middle (where the probability of a randomly selected position being within $5$ km is $1$) and then the first restaurant is at the edge (where the probability is $\frac{1}{2}$). ...


1

I think we had some interaction on Stack Overflow a few days ago regarding F-stats. The concept of the F-test is fairly widespread in statistics, so it is not an innovation by Michael Newville (lmfit developer). Basically there are two types of F-tests to perform. Both of which deal with comparing two models, because an F is simply a ratio of chi-squares ...


1

Since so many of your values are $(0,0)$, there probably isn't much harm (and maybe some benefit, since some people repeat terms more than others) in just asking for each tweet, "Did they mention NFL?" and "Did they mention NBA?". There are 4 possible combinations of yes/no for these questions. You want to know if the answers to the questions are independent ...


1

$\newcommand{\E}{\operatorname{E}}$I've always found this particular problem mildly icky in the way in which it mixes the discrete and the continuous and apparently has to be done piecewise. You need \begin{align} & \E(X_1\mid X_{(n)}) \\[10pt] = {} & \E(\E(X_1 \mid X_1=X_{(n)})\mid X_{(n)})\Pr(X_1=X_{(n)}\mid X_{(n)}) \\[2pt] & {} + \E(\E(X_1 ...


1

Just differentiate $P(T\leq t)$, it'll give you the density.


1

If you want to test whether there is a difference in population means, assuming these data come from two (nearly) normal populations, then you want a two-sample t test (do not assume variances equal). I notice a negative number -68.0 in your first sample. Is that a typo, or what you really intended? If it's real, then some would say you should use a ...


1

What you have here is a Markov chain with a given transition matrix. This matrix is the transpose of your table. The task is to find an eigenvector of the matrix for the eigenvalue $1$ and normalize it w.r.t. the $\|\cdot\|_1$-Norm (i.e. make sure all components sum to $1$; they will all be nonnegative) $$M = ...


1

Since $\Pr(Y<\theta)=1$, one must have $\operatorname{E}(Y)<\theta$, so $Y$ is not an unbiased estimator of $\theta$.


1

You can calculate the cumulants using the following recursion formula : http://upload.wikimedia.org/math/9/b/2/9b2b9e42c18cd141a6236fa04b6715a5.png where the $\kappa$ are the cumulants and the $m$ are the moments. The $p$-th order moment is given by $$m_p=\frac{1}{n}\sum_{k=1}^nx_k^p$$


1

You can obtain the square root of a matrix M using the Cholesky Decomposition, M = LL'. Then compute the inverse of L.


1

a) is correct. b) $$1-\sum_{n=1}^4P(\text{off after the }n\text{-th song})=1-\sum_{n=1}^4 \frac1{10}\left(\frac{9}{10}\right)^{n-1}$$ ci) There will be turned off after the $6$-th song if this song is from B and exactly one of the preceding $5$ songs is from B. These events have probabilities: $\frac{1}{10}$ and ...


1

Since one of the parts has standard deviation $0$ the return is not much complicated. We have an expected return of $\mu = 0.08w+0.02(1-w)=0.02+0.06w$ And the standard deviation will simply be $\sigma = 0.25w$. That's because $\sigma_{XY}=\sqrt{\sigma_X^2+\sigma_Y^2}$


1

An alternative method is to use a Bayesian approach (in which case, the interval estimate calculated is not a "confidence interval" but a "credible interval"). The idea is to treat the rate parameter $\lambda$ as a random variable. For observations $$X_i \sim \operatorname{Exponential}(\lambda)$$ the conjugate prior is Gamma distributed; i.e. the choice ...


1

A t-interval would be a very approximate procedure here. It is intended for use when the data are at least roughly normal, and the exponential distribution is very far from normal. (Such a procedure might be OK for really large samples.) CI based on gamma distribution. Here is a better way: If $X_1, X_2, \dots, X_n$ are a random sample from ...



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