# Tag Info

## New answers tagged statistics

0

An easy to read article with data: http://www.jmir.org/2015/6/e160/

1

The distribution function $F$ is the cumulative distribution function, not the probability mass function or the probability density function. It is defined by $F(x) = \mathbb{P}(X \le x)$. I am assuming $F(x-)$ denotes $\lim_{t \nearrow x} F(t)$, where the limit is for $t$ approaching $x$ from below. You may want to convince yourself that $F(x-) = ... 0 Yes$F(x-)$does mean approaching a limit$x$from below, hence the minus sign. Now for a piecewise function$f(x)$;$P(a \lt X \le b)$is defined to be$F(b)-F(a)$where$F(x) = P(X \le x)$In your question$F(1)=P(X \le 1)$not$P(X = 1)$, this is because$F(x)$is the cumulative distribution.$X$is the (discrete) random variable and$x$is the value ... 0 "Is this your homework, Larry?" I. Look at the scatter plot And find a relation such that$y_i\approx f(x_i), i=1,...,n$II. "You might want to watch out that front window Larry". Set a family of functions$F$(ex: linear fonctions) and a cost function$L$such that:$\sum_{i=1}^{n}L(y-f(x))$is a minimum for a given function$f\in F$., wher n is the ... 0 Your calculation is right, the$Y$as defined in the post does not have exponential distribution. The definition should have been $$Y=-\frac{1}{\theta}\log\left(\frac{X}{\theta}\right).$$ Then a calculation much like yours gives that$Y$has cdf$1-e^{-\theta y}$on the interval$(0,\infty)$. 0 The correct answer is equivalent to not having all ten not O-negative, i.e.$1-(1-0.07)^{10} \approx 0.516$. Any of the following Excel formulae will give this: =1-(1-0.07)^10 =1-BINOM.DIST(0,10,0.07,TRUE) =1-BINOM.DIST(0,10,0.07,FALSE) =1-BINOM.DIST(10,10,0.93,FALSE) =BINOM.DIST(9,10,0.93,TRUE) 0 a) Let$X$be the discrete random variable that expresses the number of the non-defective items in our sample. Assuming that the probability of getting a non-defective item remains the same, the probability of success is$p=1-0.1 =0.9$Now, considering the formula of the binomial distribution (with$p=0.9$and$n=10), we have: $$\Pr(X \ge ... 0 If a horse has 4-1 against odds of winning, then his probability of losing is 4/5 or 80 percent, from the formula you gave. Thus his probability of winning is 20 percent. 0 Forest + Coal = 21,929 + 124,595 = 146,524 Total = 335,770 as stated 146,534 of 335,770 car loads are either Coal or Forest 1*(146,534/335,770) = 0.4364 0.4364 = 43.64% Do the same thing for #9 except 1*(coal/(coal + forest)) 0 The set of all events here is X and hence P(X) = 1. We are given that P(A) = 0.2 and P(B) = 0.5, as well as P(A\cap B) = 0.12. Also, P(A) + P(A^c) = 1 and P(B) + P(B^c) = 1, as in fact for any set of events Y \subset X, we have that the total probability$$P(Y) + P(Y^c) = P(Y \cup Y^c) = P(X) = 1.$$Now, what is P((A\cap B)\cup A^c) in ... 0 Hint: Assuming that the sample size of 6 is small compared to the population of male students at the university, the probability distribution is approximately binomial. Also, you can use the fact that P(E)=1-P(E^c). 0 Hint: use the fact that E[E(Y|X)] = E[Y]. 0 Never mind, I understood how to resolve it. Find the cdf P(a+(b-a)Y\leq w) and then it just simply translates to P(Y\leq w-a/b-a, which tells us that the cdf G(W) = w-a/b-a. 0 The simplest way to do (a) is to find the probability of answering 0 or only 1 correctly, then subtracting that from 1. With four possible answer to each problem, the probability of answering one correctly by choosing at random is 1/4 and the probability of answering incorrectly is 3/4. So the probability of answering all 8 questions incorrectly (0 ... 2 If the true model is that you have homogeneity across the units, i.e. in the models$$y_{it}=X_{it}\beta_i+\alpha_i+u_{it}$$we have$$\beta_i=\betathen pool mean group estimator i.e. pooling everything and estimate \beta is up to a statistically vanishing (with growing (N,T)) error the same as estimating each equation separately and the averaging ... 0 The joint distribution of inter-arrival times is uniform on \{x_1+\dots + x_n<t\} conditionally on the fact that N(t) = n. Unconditionally, they are independent and have exponential distribution. There is no contradiction; you can try to verify this fact yourself, it's not that hard. Moreover, as @Did wrote, while the joint (conditional) distribution ... 0 Hint: N is distributed as P_1 + \dots + P_A, where P_k are iid \mathrm{Poisson}(\lambda). So by the law of large numbers, N/A \to \lambda, n\to\infty, in probability. 1 Not assume, but prove... By definition, \bar{A} and B are independent iff P(\bar{A}\cap B)=P(\bar{A})\cdot P(B). We can express B as B=(A\cap B)\cup(\bar{A}\cap B). (If this is not clear, look at a two-set Venn diagram.) So we can write P(B)=P(A\cap B)+P(\bar{A}\cap B). Then \begin{align*} P(A\cap B)&=P(B)-P(\bar{A}\cap B)\\ ... 0 Comments: (1) I suppose you may already know that this is closely related to the famous birthday problem. (2) If n is moderately large compared with very large k, then Y = n - X (the number of redundancies) may be approximately Poisson. (3) If k = 20 and n = 10, then your formula has E(X) = 8.0253. A simulation of a million such experiments ... 1 You have not given anywhere near enough information for me to give you a responsible answer. So I will make some very speculative assumptions, and give you an answer based on those assumptions. If you are not satisfied with the answer, maybe that will prompt you to provide more insight into what your data are like and what you really want to do. I assume ... 1 In my previous Answer I state that CIs for \sigma^2 based on S^2 when \mu is unknown are, on average, longer than CIs for \sigma^2 based on V when \mu is known. In a Comment you raise the interesting question how often and in what circumstances the former type of CI actually turns out to be shorter than the latter. This is an intricate ... 0Y_i=B_0+B_1X_i+\epsilon_i\hat{Y_i}=\hat{B_0}+\hat{B_1}X_i$$a)$$E[MSE]=E[\frac{\sum(Y_i-\hat{Y_i})^2}{n-2}]=\sigma^2=0.6^2E[MSR]=E[\sum(\hat{Y_i}-\overline{Y})^2]=\sigma^2+B_1\sum(X_i-\overline{X})^2=1026.36$$b)$$\sigma(\hat{B_1})=\sqrt{\frac{\sigma^2}{\sum(X_i-\overline{X})^2}}=\frac{0.6}{\sqrt{\sum(X_i-\overline{X})^2}}$$for the case ... 2 Consider a Poisson process of rate \lambda. Independently, each occurrence of the Poisson process is "special" with probability q = 1/(p+1). Your T is the waiting time until the first special occurrence. You can also consider this from a different point of view: the special and the non-special occurrences form independent Poisson processes with ... 0 You can use MGFs to obtain the distribution of the conditional sum X \mid N and show that this is gamma distributed with rate \lambda and shape N. Then we would compute the unconditional/marginal distribution X by summing over all N = 0, 1, 2, \ldots, weighted by the probability \Pr[N = n]. 1 The probability of rolling a 2 or 6 on a fair die is 1/3 and the probability of rolling an odd number (so 1, 3 or 5) is 1/2. Since these two events are independent, P(A) = 1/2 \cdot 1/3 = 1/6. In general with problems like these, you can sometimes split up the event A into more manageable parts (divide and conquer) that are independent and then ... 1 Use the quadratic formula or complete the square to solve the corresponding equality:$$0 = \tau^2 + 2\sigma\tau\rho - 3\sigma^2$$implies$$0 = (\tau/\sigma)^2 + 2(\tau/\sigma)\rho - 3,$$thus defining \kappa = \tau/\sigma, we have$$\kappa = -\rho + \sqrt{\rho^2 + 3},$$where the positive root is taken to ensure the ratio is positive. It follows that ... 1 The answer to your question depends on the distribution of the population. Suppose the population is normal with population mean \mu and population variance \sigma^2. Then what you suspect is true. In the usual case, both \mu and \sigma^2 are unknown. You find \bar X to estimate the population mean and then find S^2 = \frac{\sum(X_i - \bar ... 0 Notice that your initial region \Omega is the box (x,y) \in [0,1]^2. Your event Y/X \leq w introduces the constraint Y = wX (for each fixed w), which is a line passing through the origin. Define a region A which is the part of \Omega below the line Y = wX, and find the area of the region, e.g. by double integral over it (it should depend on ... 0 I'm assuming you are modelling the statistics as vectors because you expect some of the 2011 stats to be correlated to each other? If not, you could simplify you life and model each statistic separately, getting some sort of multivariate regression for each. One problem with a straight prediction vector is that each statistic has a different range, so if ... 0 As you probably know, for a random sample \vec{X}=(X_{1}, X_{2}, \ldots, X_{n}) with joint pdf f(\vec{x}; \theta), a statistic T = T(\vec{X}) is sufficient if the conditional pdf for \vec{X} given T is \theta-free". This is the definition of sufficiency. I think it is easier to (a) believe or show that the Fisher-Neyman Theorem completely ... 2 The first equality that you don't understand is indeed a MacLaurin series expansion of the characteristic function, using the link between the moments of a random variable and the derivatives of the characteristic function. See https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Moments . So Shalop is right, the \sigma is a ... 2 Your intuition is correct. That does not sound, and it isn't, right. Let x be the conditional probability that a hair is left at the scene given that somebody else did it. By the law of total probability, the probability that a hair is left at the scene is$$ 0.99 = 0.99 \times 0.1 + 0.01 \times x $$So x=89.1>1, which cannot be a probability. The ... 0 If there are 24 turns: P(Any player Levels up) = P(A Player Levels Up on turn 1) + P(A player levels up on turn 2) + ... P(A Player Levels Up on turn 24) What is the probability that a player levels up on his or her turn? Assuming they only level up with a blue level up card, then they have to land on a blue square AND draw a level up card. P(A player ... 0 I meant to ask what is the probability that Jeanie remembers at least one of the three errands? This is the complement of Jeanie forgetting all three errands, so the answer is: 1 - 0.1 * 0.1 * 0.1 = 0.999 This is the easy way of solving it. You could also add up the probabilities that Jeanie forgets one, two or all three errands, and you should get ... 0 Well, in the first place, the speed limit is not quite 1 standard deviation above the mean. It's 4.4 mph above the mean, which works out to 4.4/4.78 \doteq 0.92 standard deviations. So, use your z-table, your symbolic calculator, the web—whatever—to find the probability p that a normally distributed random variable is "to the left" of 0.92 ... 1 You have used indicators to do:$$\begin{align} \mathsf E[X] & = \sum_{j=1}^k \mathsf P[X_j{=}1] \\ & = \sum_{j=1}^k (1-\mathsf P[X_j{=}0]) \\ & = k (1-(1-\frac{1}{k})^n)\end{align}$$So continue:$$\begin{align}\mathsf E[X^2] & = \sum_{j=1}^k \mathsf P[X_j{=}1] + \mathop{\sum\sum}_{\substack{j\in\{1;k\}\\i\in\{1;n\}\setminus\{j\}}}\mathsf ... 1 Fori=1,\ldots,n,\;$let$R_i(n)$be the number of passengers on the bus when it leaves stop$n$that got on at stop$i$. So$R(n) = \sum\limits_{i=1}^n R_i(n).$Any passenger getting on a stop$i$must "survive"$n-i$stops to still be on the bus when leaving stop$n$. This has probability$(1-p)^{n-i}$. Thus,$R_i(n)$has a Poisson distribution with ... 1 It seems to me that you are in danger to misuse Cluster Analysis, which is a technique to discover structure in data and not to describe a structure that one just assumes to exist. To see what I mean, ask yourself the following questions for example: Why are you clustering into 3 groups? Not 4 or 9? When plotting the data (luckily there are 3 attributes ... 0 If$\inf_{x\in\mathcal X} f(x) \geq C > 0$is true, then $$\sup_{x\in\mathcal X} \left\vert \frac{1}{\hat{f}_n(x)} - \frac{1}{f(x)} \right\vert = O_P(a_n).$$ Proof: By assumptions, for each$\varepsilon>0$there exists$M>0$such that for all$nlarge enough P\left( \sup_{x\in\mathcal X} \left\vert f(x) - \hat{f}_n(x) \right\vert > ... 0 Basically, we have to use the fact that the total probability add up to 1 for a variable. So, from B to A the probability is 1-0.6=0.4 Probability of staying at A is 1-0.3 =0.7 1 Hint: The given information in the problem is filled out in the table below. Try filling in the rest of the table. \begin{array}{c|c|c||c}&\text{did bring backpack}&\text{did not bring backpack}&\text{total}\\ \hline \text{did bring hat}&\bullet&\bullet&50\\ \hline \text{did not bring hat}&\bullet&40&\bullet\\ \hline ... 1 If X_1,X_2\sim\operatorname{Exp}(\lambda) are independent with densities f and g then the density of X_1+X_2 can computed as \begin{align} h_2(t) &= (f\star g)(t)\\ &= \int_0^t f(s)g(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda^2 e^{-\lambda t} \int_0^t \ \mathsf dt\\ &= ... 0 If X and Y are independent random variables thenf_{X,Y}(x,y) = f_X(x)f_{Y}(y)$$It is not possible to write the given function f_{X,Y}(x,y) = \begin{cases} 2, \ ~~\text{if} \ 0<x<y,~~~ 0<y<1 \\ 0, \ ~~\text{elsewhere}\end{cases} on this form. If x<z<y<1 then we would need$$f_{X}(x)f_Y(z) = 2~~~\text{and}~~~f_{X}(x)f_Y(y) = ... 1 Hint: It is $$E[(2X+1)^2]=E[4X^2+4X+1]=4\cdot E[X^2]+4\cdot E[X]+1$$ And $$E[X]=\sum_{x \in A} x\cdot P(X=x)$$ $$E[X^2]=\sum_{x \in A} x^2\cdot P(X=x)$$A=\{-3,6,9\}$0 For a discrete distribution,$f(x_i)$, like in your example, you can use$E[g(X)] = \sum_i f(x_i) g(x_i)$. 1 Because the individual variables are not identically distributed, the calculation of the order statistic is not the same as if they were identically distributed. Define$W = \max(X,Y,Z)$. Then we want$\operatorname{E}[W]$and$\operatorname{Var}[W]$. To this end, we should try to calculate the CDF $$F_W(w) = \Pr[W \le w] = \Pr[\max(X,Y,Z) \le w].$$ But ... 1 I know that$f_W (x)=n\,f_X (x)(F_X (x))^{n−1}$That's only the case for identical distributions. For what you have use: $$f_W(w) = f_X(w) F_Y(w)F_Z(w) + F_X(w) f_Y(w)F_Z(w) + F_X(w) F_Y(w)f_Z(w)$$ This is the density where: one of the three equals$w$and the other two are less. 1 Let$\bar x = (x_1+\cdots+x_n)/nand recall from algebra that $$\sum_{i=1}^n (x_i-\theta)^2 = n(\bar x-\theta)^2 + \sum_{i=1}^n (x_i-\bar x)^2.$$ Then deal with the density: \begin{align} f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \propto {} & \prod_{i=1}^n\frac 1 \theta \exp\left( \frac{-1} 2 \left( \frac{x_i-\theta}{\theta} \right)^2 \right) = \frac 1 ... 4 I think I find this line of reasoning rather too speculative, I'm afraid. :-) First, you assert thatP(I) = 1/2$. I don't see any symmetry that justifies this kind of application of the principle of indifference; it's not as though the only difference between being selected for an interview and not being selected is the opportunity to go to the interview. ... 1$Continuous.\$ Assuming you are sampling from an continuous population with a density function of unknown shape, a relative frequency histogram (scale adjusted so the total area of histogram bars is unity) can give a reasonable idea of the shape of the density function, especially if you have a lot of data. Depending on the situation, you may need a few ...

Top 50 recent answers are included