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I will here try to answer the "why": You have decided to fit a linear model to your observations using OLS method, and with only one dependent variable you now have the slope, $\hat{\beta}$, for that variable. Pay attention to the "hat", which means that $\hat{\beta}$ is just the sample estimate of the underlying (unknown) population ...


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Just to add to the discussion. Assumption 1. Linearity This assumption says that we believe the true population model is linear in parameters (not in explanatory variables). Thus: $$ y = a_0 + a_1 x + a_2 x^2 + a_3 ln(x) +e$$ Is a linear model. However, you can also interpret it in a slighly different way. the linearity assumption says that the conditional ...


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Condition 1 You seem to understand this one. This assumption can be written as $f(x_i) = \beta^\mathrm{T}x_i$ and in matrix form as $\mathbf{f} = \mathbf{X}\beta$ Condition 2 In short, you need that $\mathbb{P}(y\mid x) = \mathcal{N}(\mu(x), \sigma^2)$ where $\mu(x)$ is a function of $X$. This assumption is met when the residuals are normal since $y = f(x) + ...


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I'm going to answer to test my understanding since the literature on the topic is huge and lots of people on this forum can give better explanations. I assume we are talking about the ordinary linear regression, not the generalized linear models. b) Normality - The residual values must be normally distributed ( But, some sources say that the Response ...


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Gavin's useful answer can be generalized by the following in the situation where you may have more than one coefficient of interest. Let's say these coefficients are called c1, c2, c3, c4 for example. Replace i <- which(names(beta) == "Solar.R") with i <- which(names(beta) %in% c("c1", "c2", "c3", "c4"))...


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The 'linear' in 'linear regression' means linear in the parameters, which isn't necessarily what people normally mean by 'linear' outside of statistics. (To help clarify the issues, it may help you to read through this CV thread: How to tell the difference between linear and non-linear regression models?) The linearity at issue isn't really an assumption, ...


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I assume you are talking about OLS/linear regression. Using OLS implies already that one assumes that there is a linear relationship. Why? Because you explain the response variable by a linear combination of the regressors. Hence, using OLS if you don't think that there is a linear relationship between the explanatory variables and the response variable ...


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He basically just verified if the data is more or less linear Granted I know nothing about the course in question and its purpose, in real life assumptions are rarely if ever fully satisfied so realizing that the data is more or less linear may be enough to proceed provided you are aware of the limitations. I suppose many of the 6-months-DS-without-stat/...


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Keep in mind that the data you got in any context is just one possible realization of infinitely many. Hence, it is not really desirable to hit R^2=1 for a given dataset (keyword overfitting). Rather, you want to estimate parameters that provide for different samples on average reasonable results. From my point of view the instructor should have validated ...


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I have just solved the question so, here is the answer, I am changing the question as find the probability of W≤0.6 because I think that value serves better for purposes of teaching and will eliminate any possible confusions in the question. First of all what we should do first is to assign probabilities. If $ W = |X-Y|$ and $Pr(W <= 0.6)$ we are looking ...


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You can’t just add up the standard errors. You need to work with the variances and covariances of the coefficient estimates, then take the square root to get back to the standard error scale. The formula for the variance of a weighted sum of correlated variables is the key, as the estimates of the intercept and slope are correlated random variables, not ...


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Comment. Maybe some simulation results (from R) will help you get a mental picture of this problem. set.seed(213) X = runif(10^6); Y = runif(10^6, 0,2) W = abs(X-Y) mean(W <= .5) [1] 0.437011 # aprx 7/16 = 0.4375 hist(W, br=20, prob=T, col="skyblue2") abline(v=.5, col="orange", lwd=3, lty="dotted") Using fewer points ...


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Hi: I won't do all the work so this is not a complete answer but take the first case where W = (Y - X) > 0. Then, if W < 0.5, then that means that Y < 0.5 + X. So, the double integral goes from 0 to 1 on X and from 0 to 0.5 + X on Y. So, one gets $\int_{0}^{1} \int_{0}^{0.5 + x} \frac{1}{2} ~dy~dx $ The other case is similar except that (X - Y) &...


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Based on your comments, it sounds like your problem is how to determine the initial segment of your data that we can "reasonably" fit a straight line to. I would suggest a simple local search strategy. Start with a proportion $p_0=1$. For each iteration $i$, fit a straight line to the initial $p_i$ proportion of your data. Assess which proportion $...


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Your answer to question 2 is correct, provided here are actually individuals with $y \geq 85$. The easiest way to see this is that this truncation would at the very least decrease the mean value of $y$, so even if you cooked up the data in such a way that $\beta_1, \beta_2$ from such a regression stayed exactly the same, then $\beta_0$ would still have to ...


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Davidson and MacKinnon in ("Econometric Theory and Methods") address head on the flaw you point out in Wooldridge's presentation: "At this stage as long as we say nothing about the unobserved quantity $u_{t}$, equation (1.01) [$y_t=\beta_0+\beta_1x_t+u_t$] does not tell us anything. In fact, we can allow the parameters $\beta_0$ and $\beta_1$ ...


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You could create a dummy variable where people in group 1 get a zero and people in group 2 get a 1. You could do something more fine grained as well, say, maybe the minutes between tests if you think that is important. By observational study, do you mean that this is not an "experiment" (e.g., random assignment to treatment and random selection)? ...


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Yes there is. The group specific intercept for "Spice" is 0.5 (intercept of baseline category, i.e. "Nutmeg") - 0.1 (difference between baseline and category indicated by the dummy variable) = 0.4. If you would have swapped coefficients in the way you describe, this would be your baseline intercept. That is, your intercept would be 0.4 in ...


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I think that it's helpful to consider what you're telling your model is true. In the case of specifying Item as a random factor, you are essentially saying that the words used in the study are random selections from a larger body of possible words (which is probably true). You're also telling your model that, right now, each word has its own unique intercept ...


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When there is an intercept, in-sample $R^2 \ge 0$, so there is no risk of an imaginary root. Even $R^2=0$ is so unlikely in real (or even simulated) data that I would consider it to be practically impossible. You can use your equation without fear of an imaginary root or dividing by zero. If I had an awful fit, however, I would be skeptical of any inference. ...


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Since $k$ is fixed and $\dfrac{n}{n-k} \rightarrow 1$, it's not W.L.O.G to consider the estimator, $\hat{\sigma}^2 = \dfrac{1}{n}e^Te$ instead. After some matrix algebra, you can write, $$ \dfrac{1}{n}e^Te = \dfrac{1}{n}\epsilon^T\epsilon - \dfrac{1}{n}\epsilon^TH\epsilon$$ where $\epsilon$ is the vector containing the errors, $\epsilon_i$ and $H = X(X^TX)^{-...


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