Sheridan Grant
• Member for 5 years, 11 months
• Last seen more than a month ago

The third line is wrong, because you don't have $\text{E}[X|Y]$ in the second line. For example, if $Y$ is Bernoulli(1/2) and $X$ is 1 if $Y$ is 1 and -1 if $Y$ is 0, then $\text{E}[(X-\text{E}[X|Y])^... View answer 6 votes If the only values the data take on are$\{0,25,50,75,100\}$then the Beta isn't an appropriate model since it's continuous and the data are highly discrete. If this is the case, and you don't want to ... View answer Accepted answer 4 votes If you are interested in interpreting coefficients and significance/p-values, don't use stepwise regression. See this post. In fact, stepwise regression is basically always a bad idea in this day and ... View answer Accepted answer 3 votes Your basic idea is right; to formalize it, I would say you should just "de-trend" the data and then apply anomaly/changepoint detection. So, fit a time series regression model that accounts for the ... View answer Accepted answer 3 votes I presume you mean to ask which direction should you attempt in order to get to 2 in the smallest expected number of moves--you'll reach the finish with probability 1 eventually. You have to get to ... View answer 2 votes This sounds like a bad idea, though I can't exactly tell what the criterion for adding a predictor to the model was. How can a "predictor's odds ratio" be "changed" by 5% if it was ... View answer 2 votes "First-order condition" in optimization just means taking the first derivative and setting it to zero. The second derivative here is clearly negative, so under mild regularity conditions the first-... View answer 2 votes The use of hierarchical random effects here is warranted, I think, and it sounds like you agree. What we "recommend" as far as the fixed, effects, however, depends on the scientific question. If you ... View answer Accepted answer 2 votes Create a binary dummy variable that is 0 before and 1 after. Run the regression with the dummy variable and an interaction between x and the dummy variable. The interaction coefficient will tell you ... View answer 2 votes First, I'd change the axes to say "Months on Medication" and "Number of Patients." I'd report the number of zeros, and be able to verbally address the issue of zero-inflation. Definitely report the ... View answer 2 votes I'm not sure if there's a better term, but what you are doing is known as "achieving balance" between the two groups on the covariate var. It is closely related to the technique of "matching," which ... View answer 1 votes There are a few problems here some of which are just terminology. You don't have a distribution, you have data. You want to estimate a distribution from your data. An MLE isn't a distribution, it's a ... View answer 1 votes There seems to be a hangup in the comments over "outlier removal," which is often done without thinking and can be a bad idea. You're just looking for error detection--removing errors, I agree, is ... View answer Accepted answer 1 votes Here are a couple options: Do a grid search to find the point that discriminates best on your data: just test the accuracy on a large list of points between the means or medians of your data. Sounds ... View answer 1 votes Because$X$is discrete (it takes on discrete names as values), it doesn't have a density, but rather a mass function. The most straightforward way of estimating this mass function, or pmf, is to ... View answer 1 votes If the non-negative integers have true integer "meaning" (i.e. 3 represents a quantity three times what 1 represents), and particularly if they represent arrivals in a process, then Poisson regression ... View answer 1 votes With 48,000 observations, even if the data is pretty highly skewed the t test might do okay. As nba2020 suggested, you could bootstrap your sample and observe the distribution of means, though this ... View answer 1 votes Note that for each customer, the only information you have is the total number of visits ("has_visited" and "repeat_customer" are both functions of the total number of visits). Since you don't have ... View answer 1 votes Take the Bayesian perspective, supposing that$E[m(y)\mid\theta] = \theta$(posterior mean is unbiased). Then$$E[\theta m(y)] = E[E[\theta m(y) \mid \theta]] = E[\theta E[m(y)\mid\theta]] = E[\theta^... View answer 1 votes If there is just one category of interest, e.g. lemons, there's no problem with "binarizing" it and extracting a CI for the population proportion who like lemons. However, if you do this for each ... View answer Accepted answer 1 votes I'm assuming you've also observed some covariates for each customer beyond whether or not they canceled. Your suggested approach, then, would simply give you 4 groups of customers who canceled that ... View answer 1 votes Related answer here from just yesterday. If the distribution you observe is the distribution you wish to test, then binning forfeits information and will thus on average reduce your ability to (in ... View answer 1 votes I'm assuming you want a result of the form$P(d_{TV}(p,q) = \frac{1}{2} \sum_k |p(k) - q(k)| > \epsilon) < f(n,\epsilon)$. If your estimator is the empirical distribution, then there are general ... View answer 1 votes Mann-Whitney only tests for a difference in medians when you assume that the only difference in the distributions of the two samples is the location, and not the scale or shape, of the distribution, ... View answer 1 votes First, let me point out that Mann-Whitney does not test stochastic dominance. It tests the null hypothesis$P(X > Y) = P(X < Y)$. If$X$is uniform on$[0,1]$and$Y$is 1 with probability 0.9 ... View answer 0 votes The premise in the title of the question isn't correct. The estimated$b_0$and$b_1$are by definition functions of the observed data, which are functions of the$\epsilon_i$. Unless the question is ... View answer 0 votes You're right that this is a very difficult problem--it's hard to test independence in general without a lot of data. There are loads of test statistics you can construct and test to try to reject the ... View answer Accepted answer 0 votes Good question--models exist for exactly this situation, which is called a "top-$t$" ranking (we only observe the top$t\$ items of the potentially infinite ranking). Plackett-Luce tends to work well in ...