The proper terminology is Cumulative Distribution Function, (CDF). The CDF is defined as $$F_X(x) = \mathrm{P}\{X \leq x\}.$$ With this definition, the nature of the random variable $X$ is irrelevant:...

Probably the best-known example of a statistical method "developed" from an astronomy problem was Gauss' use of least squares to generate an orbit for Ceres on the basis of Piazzi's observations. ...

PMFs are associated with discrete random variables, PDFs with continuous random variables. For any type of random of random variable, the CDF always exists (and is unique), defined as $$F_X(x) = P\{X ... View answer 6 votes Experimental designs are a combination of three structures: The treatment structure: How are treatments formed from factors of interest? The design structure: How are experimental units grouped and ... View answer Accepted answer 5 votes Pearson's \chi^2 test is useful for a sample of n observations cross-classified by two variables, say A and B. These tests test the null hypothesis that A and B are independent variables. ... View answer Accepted answer 5 votes Let's start from theory, and worry about "How To" later. Let's suppose that U \sim \mathrm{Uniform}(0, 10). Now, fix an integer, say i. The probability that i-\epsilon/2 < U < i+\epsilon/... View answer Accepted answer 5 votes If those are the choices, then Complex Analysis is probably the way to go. PDEs are interesting and useful, but not too much in statistics. Complex Analysis, on the other hand does pop up from time ... View answer 4 votes The usual factor analysis model is$$\mathbf{Y} = \mathbf{\mu}+ \mathbf{\Phi}\mathbf{L} + \mathbf{\eta},$$where \mathbf{Y} represents a collection of n observations of k random variables; i.... View answer 4 votes I don't know that it's called anything in Mathematics. In Statistics, it's an example of a trimmed mean, and particularly a 14% trimmed mean. View answer Accepted answer 4 votes Levenberg-Marquardt is a general (nonlinear) optimization technique. It is not specific to LS, although that is probably its widest use. Looking at your referenced paper, they are (mostly) fitting ... View answer Accepted answer 4 votes As I'm sure you are aware, ARMA is an acronym for AutoRegressive Moving Average (Stochastic Process). More fully, we use ARMA(p, q) where p is the order of the autoregressive portion and q the order ... View answer 3 votes To T or not to T -- is that the question? I would suggest backing off for a moment and asking yourself, "What IS the question?" Is the question, "Are the means of populations 1 and 2 the same?", ... View answer 3 votes Don't do hypothesis testing. It's only going to tell you what you already know: they are different. Quantify the differences in your response variable(s) across the environment. You can do it in a ... View answer 3 votes Friends don't let friends use Excel for serious statistics. Yes, that is a slam on the folks in Redmond. When they fix their pseudorandom number generator I'll take another look at Excel. Now that ... View answer 2 votes Starting with the case of two coins, we know that$$E\{T_k \}=m_k\theta_A+n_k\theta_B.$$If k=2, we can solve$$t_1 = m_1\hat{\theta}_A +n_1\hat{\theta}_Bt_2 = m_2\hat{\theta}_A +n_2\hat{\theta}...

The fact is that Alecos' answer is the easiest way to handle the problem, but the problem can be solved via Rao-Blackwell as well. Start with the joint density $$f(x_1,..., x_n | \theta) = e^{-\sum ... View answer 2 votes Under the stated conditions (\mu_X = \mu_Y = 0 and \sigma_X = \sigma_Y = 1), the likelihood function for a random sample of size n is$$L(\rho\; |\; X, Y) = \frac{1}{(2\pi[1-\rho^2])^{n/2}}\exp \...

You know that $S^2_e = \frac{1}{n-k}\mathbf{e}'\mathbf{e}$ is an unbiased estimator of $\sigma^2$. So, if you show that $Var(S^2_e) \to 0$ as $n \to \infty$ you've shown that $S^2_e$ is consistent (i....

I'm going to assume this is self-study, and make some suggestions rather than giving an explicit answer. The proof is in two steps: Find the joint distribution function of $U_{(1)}, U_{(n)}$; Show ...

The alternative hypothesis is $H_1: \mu \neq \,\$95,000.$The significance level is the same as the error rate of the confidence interval,$\alpha = 0.01$. You cannot recover an observed ... View answer 2 votes First off, why are you assuming equal variances in the two groups? Please don't say, "Because it's convenient." I seriously doubt that the group variances are equal, although in the case of equal ... View answer 2 votes It depends on what you are trying to accomplish. What distribution do you want the imputation to reflect? $$\mathrm{N}(10, 25)$$ or $$\mathrm{N}(10,\frac{25}{\sqrt{1000}})?$$ Your second task could ... View answer Accepted answer 2 votes It is pretty well known that $$\mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \mathrm{Cov}(X,Y)$$ or in terms of$\rho$,$\sigma^2_X$and$\sigma^2_Y$\sigma^2_{X+Y} = \sigma^2_X + \... View answer Accepted answer 2 votes Nope. What happened to your product term,$\sum_{i,j}x^{(i)}_j\cdot y_i$? You'll find this clearer if you write it in matrix notation rather than scalar notation. Your joint density is then$f(\...

You don't need anything but base R. Use the scan( ) function and pipe from LabVIEW to stdio.

It is a repeated-measures factorial design, albeit of a very simple sort. You have one "within-subject" factor (pre-test and post-test (after treatment) and two "between-subjects" factors (A and B). ...

The mean of the hypergeometric distribution can be interpreted as the finite sampling equivalent of $\mu = np$ from the binomial, taking $p=\frac{K}{N}$. The variance can be expressed as $\sigma^2 = \... View answer Accepted answer 2 votes Use Craig's Theorem. Consider the quadratic form on b. If two random variables are independent, then any univariate functions of those random variables are likewise independent. The quadratic forms ... View answer Accepted answer 2 votes Oy veh iz mir... The source is wrong. In fact, the source acknowledges that it is wrong immediately below the claim when it states that under$H_o:\rho=0$and when$(X, Y)$are jointly Normal,$T = ...