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You are confusing the estimator for covariance with covariance itself. Covariance is defined as $$\operatorname{cov}(X, Y) = \operatorname{E}{\big[(X - \operatorname{E}[X])(Y - \operatorname{E}[Y])\big]}$$ Where $X$ and $Y$ are random variables and by calculating expected values we do consider their distributions and relative probabilities when calculating ...

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I interpret this to mean that you are unhappy if you have a data set like this. $$X, Y\\ 1,1\\ 2,3\\ 1,1\\ 0, -1$$ In this case, the $(1,1)$ is repeated, so you want to weight it double. However, that is covered by the formula. $$cov(X, Y) = \frac{1}{4}\sum_{i = 1}^4 (X_i - \bar X)(Y_i-\bar Y)\\ =\dfrac{(1-1)(1-1) + (2 - 1)(3 - 1) + (1 - 1)(1 - 1) + (0 - ... 6 There are some misconceptions in your question that I need to clear up before getting to the answer. The null hypothesis H_0 in a statistical test is always the claim you want to argue against. The alternative hypothesis H_1 is the claim you hope to be true. The null and the alternative need to be mutually exclusive (no overlap) collectively exhaustive (... 4 the data may strongly support \mu > 0, but does not constitute sufficient evidence for \mu \neq 0 I don't know if this reasoning helps or if it is 100% correct... You accept to be wrong \alpha-percent of the times in hypothetical repeats of the test; for example, you accept to be wrong 5% of the times in rejecting the null. In a two-tailed test you ... 2 For brevity, I will refer to your components and system using only the subscripts for the component number (and I recommend you use the same practice). For this kind of problem you should try to solve as much as you can analytically before using numerical solutions to evaluate particular integrals. The first thing to do is to derive the survival functions ... 1 It may be worth trying, especially if the counts are rather large so that you get something close to continuous. But if both counts are rather low (possibly even including zeros) then a binomial model would be most natural. Finally if the main response is a rather low count and the denominator ("exposure") is rather large then a Poisson model with ... 1 There is no unique way to compute p-values and the p-values based on different methods can be different. The only thing that is required is that the sampling distribution of the p-value is a uniform distribution if the null hypothesis is true. So it may occur that you reject H_0: \mu \leq 0 with an alternative hypothesis H_a: \mu > 0, but you can't ... 1 Proving non-existence is always hard. But yes, I would assume there is no simple recipe to determine your sample size for as complicated a setup as you have. So you will indeed need to simulate. Any first of all, you will need to decide what parameter you are interested in. The mean? The amount of zero inflation? Something else? Essentially, you will need to ... 1 First, due to the conditions on k values, we know that there are nk_1 points at -1, nk_2 points at 0 and nk_3 points at 1. For each such point, there's a respective row in X: Either (1,-1,1), (1,0,0) or (1,1,1). We therefore get$$X^TX= \begin{pmatrix} n & n(k_3-k_1) & n(k_3+k_1)\\ n(k_3-k_1) & n(k_3+k_1) & n(k_3-k_1)\\ ...

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We have \begin{align*} \frac{d}{d\hat\alpha}\sum_{i=1}^n(y_i-\hat\alpha-\hat\beta x_i)^2 &=\sum_{i=1}^n\frac{d}{d\hat\alpha}(y_i-\hat\alpha-\hat\beta x_i)^2\\ &=\sum_{i=1}^n2(y_i-\hat\alpha-\hat\beta x_i)\cdot\frac{d}{d\hat\alpha}(y_i-\hat\alpha-\hat\beta x_i)\quad\text{(chain rule)}\\ &=\sum_{i=1}^n2(y_i-\hat\alpha-\hat\beta x_i)\cdot(-1). \end{...

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