# Tag Info

10

Use your original data. Claims about losing power when groups have unequal sizes make more sense in the context of designing the experiment. If you have enough resources to make $100$ observations, the greatest power will be when you allocate $50$ to the control group and the other $50$ to the treatment group. However, $150$ in the control group with $50$ in ...

5

When you use the values of the normal density then you get automatically the first four conditions satisfied $$P(X = x\vert \mu,\sigma,a,b,c) = \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sum_{y \in \Omega} e^{-\frac{(y-\mu)^2}{2\sigma^2}}}$$ where $\Omega = \lbrace a,a+c,a+2c,\dots b-c,c \rbrace$ all the values in the support. Maximum entropy The fifth ...

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The other answer here uses the normal density values at the exact points. Another similar method would be to take the normal probabilities across intervals centred on those points. In the latter case, taking the support to be $X = 1,...,m$ you get: $$p_X(x) = \frac{\Phi \Big( \frac{x - \mu + 1/2}{\sigma} \Big) - \Phi \Big( \frac{x - \mu - 1/2}{\sigma} \Big)... 4 If the aim is not just forecasting but trying to get a good handle on which predictors are most important in some sense, in my view PCA is likely to be a distraction and its use to turn out to be a detour. You would be better off applying some flavour of regression directly. That is unlikely to be trivially easy and selection of predictors is a deep and ... 4 Intro Let us start with some notation: We have m simple hypotheses we test, with each null numbered H_{0,i}. The global null hypothesis can be written as an intersection of all the local nulls: H_0=\bigcap_{i=1}^{m}{H_{0,i}}. Next, we assume that each hypothesis H_{0,i} has a test statistic t_i for which we can compute the p-value p_i. More ... 4 Partial results Below is a trial by comparing the sum of Pareto variables (with \alpha = 0.5) with a Levy distribution. The shifting and scaling are done based on the median and interquartile range. The convergence is not very fast but it does seem to work. ### function to get a scaled sample mean require(actuar) sum_sample = function(n,s) { a_n = n^(-... 3 There's no error. Start with X^\top X = (X^\top X+\lambda I) - \lambda I; premultiply both sides by (X^\top X+\lambda I)^{-1}. Simplify. Postmultiply by \beta 2 Consider the model$$Y=X\beta+\varepsilon\,,$$where Y is an n\times 1 response vector, X is an n\times p matrix of covariates (fixed) with full column rank, \beta is a p\times 1 vector of parameters and \varepsilon is an n\times 1 error vector. Also assume n>p. The residual sum of squares is then$$\text{RSS}=(Y-X\hat\beta)^T(Y-X\hat\...

2

By the law of iterated expectation, we can write $$E[(y - E[y|x])(E[y|x] - E[y])] = E[E[(y - E[y|x])(E[y|x] - E[y]) | x]]$$ Thus, to show that the LHS is 0, it suffices to show that the conditional expectation on the RHS is 0 for each fixed value of $x$, because $E[0] = 0$. So we just need to show that $$E[(y - E[y|x])(E[y|x] - E[y]) | x]$$ What is special ...

2

That's a legitimate concern. But since $\hat\beta_\lambda$ is a linear combination of the response $y,$ the explanation ought to go back to $y,$ thus: $$\hat\beta_\lambda = (X^\prime X + \lambda)^{-1} X^\prime y.$$ Recall that (conditional on $X$) the components of $y$ are independent (and therefore uncorrelated) variables with common variance $\sigma^2.$ ...

2

Both are correct based on which layout notation (i.e. numerator or denominator) is being used. Unfortunately, this is not explicitly stated in most sources and you need to infer it.

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You were on the right track. But the likelihood is simpler than what it looks in your post. If $\boldsymbol X_i=(X_{ij})_{1\le j\le n}$ for every $i$, then $\boldsymbol X_i$'s are i.i.d $N_n(\mu\mathbf1_n,\Sigma)$ with $$\Sigma=\sigma^2 I_n + \tau^2\mathbf1_n\mathbf1_n^T$$ So joint pdf of the $X_{ij}$'s is joint pdf of the $\boldsymbol X_i$'s, given by \...

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Let's elucidate this issue using a Venn diagram for the three sets at issue. Let $A$ be the orange circle, let $B$ be the violet circle and let $C$ be the green circle, and denote the probabilities of the intersection-areas in the diagram by $P_1,...,P_7$ (ignoring the area outside the three sets). From this diagram we can see that: \begin{align} \mathbb{... 2 Let’s define \Sigma as the covariance matrix. You did \dfrac{\Sigma}{\sigma_X\sigma_Y}. In other words, divide each entry of the matrix by the standard deviations. This is not correct for the diagonal elements. The diagonal elements should be divided by the variance of the corresponding random variable, so \sigma_X\sigma_X for the top left and \... 2 By "the skewness" I guess you mean the moment-based skewness calculated from third and second powers of deviations from the mean. But there are other ways of thinking about skewness that perhaps come closer to what you seek. You can measure skewness by a measure using mean, median and standard deviation (SD): (mean - median) / SD Like any other ... 1 I'll give my two cents, though it's far from a "perfect" solution (if one exists). First, as for the issue of varying dataset sizes, you can normalize the token frequencies by the document count, and choose a threshold over the normalized frequency rather than the absolute one. Second, as you noticed, using a threshold in such cases is problematic, ... 1 If you keep the parameters n and p of the distribution fixed at n=14 and p=\frac 1 4 then of course the variance will also be constant. You have to vary on of the parameters. I assume that you want p to vary. Then you can plot this variance using the curve function in R. See also its help page (?curve) for more details. curve(14 * x * (1 - x), ... 1 The two statements :\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, 1)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2)\right) $$and$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2\sigma^2)\...

1

Some empirical assertions if there are $n$ terms and $Y_n = \frac{\sum_1^n{X_i}}{\sqrt{\sum_1^n{X_i^2}}}$ with the $X_i$ iid $N(0,1)$: $Y_n$ has expectation $0$ and variance $1$ $Y_n$ has support on $[-\sqrt{n},\sqrt{n}]$ (unless $n=1$, in which case the support is just $\{-1,1\}$) The density for $Y_n$ on its support is proportional to $(n-y^2)^{(n-3)/2}... 1 Hint: you can rewrite$\sum x_i^2 = \bar{x}^2 + \sum (x_i -\bar{x})^2$. Note that the terms$n\bar{x}^2$and$\sum (x_i -\bar{x})^2$are independent chi-squared distributed variables. Then you can rewrite$Y = \frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$as $$\begin{array}{} Y^2 &= &\frac{\left(\sum{x_i}\right)^2}{{\sum{x_i^2}}}\\ &=& \frac{n^2\bar{... 1 Here is a list of point forecast accuracy measures. The entire textbook is very much recommended. Also relevant: MAPE vs R-squared in regression models What are the shortcomings of the Mean Absolute Percentage Error (MAPE)? Mean absolute error OR root mean squared error? There is a number of normality tests, the most common is the Shapiro-Wilks test. In 15 ... 1 With ANOVA you compare the variance of the difference between the group means and the variance of the differences within the groups. For the comparison of variance it doesn't in principle matter what distribution you have. If the distributions are the same then the distribution of the means has the variance of the distribution of the individuals divided by ... 1$$\bigg(\frac{s}{t}\bigg)^k\bigg(1-\frac{s}{t}\bigg)^{n-k}=\bigg(\frac{1}{t}s\bigg)^k\bigg[\frac{1}{t}(t-s)\bigg]^{n-k}=\bigg(\frac{1}{t}\bigg)^ks^k\bigg[\frac{1}{t}\bigg]^{n-k}(t-s)^{n-k}=\bigg(\frac{1}{t}\bigg)^ns^k(t-s)^{n-k}=\bigg(\frac{1}{t^n}\bigg)s^k(t-s)^{n-k}=\frac{s^k(t-s)^{n-k}}{t^n}$$Let me know if I have made any ... 1 Why can't I estimate \beta_{0} and \beta_{1} by optimizing the sum of residuals such that the sum is equal to zero ? You can find some \beta_0 and \beta_1 such that \sum_{i=1}^n \epsilon_i =0 but the solution won't be unique and it is also not meaningfull. The image below shows an example with two points. By tilting the line you can increase ... 1 It's a fairly loose concept, it applies in some situations and not others, depending on the assumptions & underlying distribution. But wouldn't you expect any future point to likely "be close to the mean" regardless simply because that's the expectation value of the rv? Yes, if all RVs come from exactly the same distribution. What does this ... 1 Are the position indices (quartiles, median, mean) of distribution in classes always different from those of the unit distribution? If so, for what reason? In theory, since the class distribution is a summary, they should never be the same [...] We are talking about comparing sample estimates to estimates obtaining aggregated data from a histogram. As you ... 1 Think about adding the powers of matrix. For example the following identity holds: \mathbf{A}^{a}\mathbf{A}^{b} = \mathbf{A}^{a + b} Applied to your problem: \mathbf{A}^{0.5}\mathbf{A}^{0.5}\mathbf{A}^{-1} = \mathbf{A}^{0} = \mathbf{I} 1 You can run an ANOVA and then do a post-hoc comparison. With only two observations per group there needs to be decent size differences between the groups. You can also run Dunnet's Test to compare all three against the control. See code below. There is no minimum rule sample size such as 3. All depends on the effect size you're trying to detect. library(... 1 First, the sample variance of a sample of size 2 may not be the best estimate of the population variance, but it does exist, so your model with 2 replications per level of a one-factor experiment should work according to the usual formulas. in R: y = c(16, 18); var(y) [1] 2 I input your data as follows (please proofread): x = c(16,18, 20,17, 25,27, 23,21) ... 1 the solution of weighted least squares problem with diagonal weight matrix W_m and working responses X_i^T\beta_m+U_i(\eta_m)/w_{m,j}. ... Working responses what does that mean? The working response is the response Y_{m,i} in the weighted least squares problem during the current step$$\min_{\beta_{m+1}} W_{m,i} (Y_{m,i} - X_i^T\beta_{m+1})^2$\$ This ...

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