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## Hot answers tagged variance

4

Your histogram looks roughly normal, which makes for an easy interpretation of standard deviation. In a normal distribution, 68% of observations are $\pm$ one standard deviation of the mean, 95% of observations are within $\pm$ two standard deviations of the mean, and 99.7% of observations are within $\pm$ three standard deviations of the mean. You can test ...

3

Because the tests look at different ways in which heteroskedasticity can manifest itself, and hence, a given data set may "look" heteroskedastic to one test and not so to another. A bit more specifically, the Breusch-Pagan test (BP) looks at whether squared residuals can be explained by observed regressors $z_i$, while the Goldfeld-Quandt test (GQ) relies ...

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If you also know the means of $X,Y$, you can use the definition of covariance: \begin{align}\operatorname{cov}(RX,RY)&=E[R^2XY]-E[RX]E[RY]\\&=E[R^2]E[XY]-E[R]^2E[X]E[Y]\\&=(\sigma_r^2+\mu_r^2)(c_{xy}+\mu_x\mu_y)-\mu_r^2\mu_x\mu_y\end{align} where $c_{xy}=\operatorname{cov}(X,Y)$.

2

Knowing tht the variance of the sum of independent random variables is the sum of their variances and that $Var(a+bX) = b^2Var(X)$, where $a$ and $b$ are constant: $$Var(y_t) = \sum_{i=0}^n0.5^{2i}Var(\epsilon_{t-2i})$$ You can solve the sum of the geometric series by using: $$\sum_{k=0}^{n-1}ar^k = a \frac{1-r^n}{1-r}$$

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I don't know what $t$ is doing in the OP's notation and so will ignore it here. The key is what is the meaning of "unrelated" in "random unrelated variables". If "unrelated" merely means different, then, as whuber points out in a comment on the OP's question, there is not enough information to answer the question. If "unrelated" means mutually independent ...

2

You're conflating two usages of bias. In the neural networks literature, "bias" is sometimes used to refer to adding a constant to a neuron. In the broader statistical literature, "bias" (of an estimator) is used to describe the difference between an estimator's expected value and its true value. https://en.wikipedia.org/wiki/Bias_of_an_estimator

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1) Exhaustive? Nope. 2) $\bar{x}$ is the MOM estimator and happens to be unbiased. Being unbiased and MOM do not go together, however. 3) The $n-1$ adjustment comes from taking the expected value of the MLE and finding that there’s an easy correction to get an unbiased estimator. When you take the square root to get what you might think is an unbiased ...

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I’m not sure what you are actually asking, but by the plug-in principle, the sample mean and sample variance are nonparametric estimators of the population mean and population variance for any distribution (provided they exist), and you can obtain their sampling distributions via bootstrapping.

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I do not know if you are still interested in this issue. I think it will be useful for your problem to look at the limiting result of the estimator mean squared error (for a penalty parameter approaching infinity). We can indicate with $\hat{\beta}_{r} = (X^\top X + \lambda I )^{-1} X^\top y$ the ridge estimator and with $\hat{\beta} = (X^\top X)^{-1} X^\... 1$\newcommand{\one}{\mathbf 1}$If we have$X\sim\mathcal N(\mu,\Sigma)$then the sample variance of a single draw can be computed as $$\frac 1{m-1} X^TSX$$ where$S = I - \frac 1m \one\one^T$is the matrix that projects into the space orthogonal to$\one. This is a quadratic form so we can compute its mean as $$\text{E}(X^TSX) = \text{tr}(S\Sigma) + \mu^T\... 1 When doing prediction from a linear regression, these are the values of the regressors for the point at which you are predicting the outcome. In a bivariate regression, when predicting y_i (the outcome of the point i), c_0=1 and c_1=x_i. 1 It's going to be a subjective one, but I think since there is no mention of sample in the question, it's more likely for it to refer to the population statistics. So, you'll divide by n, instead of n-1 as done in sample standard deviation formula. 1 I think you may be dealing with the numerator of an autocorrelation of lag k. First, let's use exactly the same notation for the two sums T_1 and T_2, and reverse the order of the product in the second sum:$$T_1=\sum_{i=1}^{n-k} (Y_i-\bar Y)(Y_{i+k}-\bar Y),T_2=\sum_{i=k+1}^n (Y_{i-k}-\bar Y)(Y_i - \bar Y).Now, let's look at the specific ... 1 Here is the GJR-GARCH model by Glosten et al. (1993): \begin{aligned} x_t &= \mu_t+\varepsilon_t, \\ \varepsilon_t &= \sigma_t z_t, \\ \sigma^2_t &= \omega + (\alpha+\gamma \mathbb{I}_{t-1})\varepsilon^2_{t-1} + \beta\sigma^2_{t-1}, \\ z_t &\sim D(0,1), \end{aligned} where \mu_t is the conditional mean of x_t, D is some distribution ... 1 jld has already shown in his comment that Var(Y)=0 because Y=0 and a constant must have variance zero. It should also be noted that the question is only possible when n=1. From the question, generalized to n: Let X_i ~ Unif(0, 1) s.t. X_1 + ... + X_n = 1 That is, each X_i is an uniform variable but the sum of all X_i is 1. Since the ... 1 (a) In many ANOVA designs there will not be enough replications for each 'cell' (combination of levels of the two factors) to do a useful test of homogeneity of variances. Do the ANOVA, find the residuals, and plot the them against data values. You can see serious heteroscedastisity visually, to warn you that P-values may not be correct. (b) What to do in ... 1 If the variance of the model errors is proportional to the expected value the power transforms may be appropriate When (and why) should you take the log of a distribution (of numbers)? ... your data plot doesn't suggest that remedy . If the variance of the errors changes at fixed points in time then GLS following the suggestion by TSAY might be appropriate ... 1 The answer is:var(m) = var( 10\% \times m_1 + 40\% \times m_2 + 50\% \times m_3) \\ = 0.1^2 var(m_1) + 0.4^2 var(m_2) + 0.5^2 var(m_3)$With$var(X_1) = var(X_2) = var(X_3) = s^2_R$, the residual variance. We can deduce that$var(m_1) = s^2_R / n_1, var(m_2) = s^2_R / n_2, var(m_3) = s^2_R / n_3$. From the information given in the exercise$n_1=...

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