180

The title of this question suggests a fundamental misunderstanding. The most basic idea of correlation is "as one variable increases, does the other variable increase (positive correlation), decrease (negative correlation), or stay the same (no correlation)" with a scale such that perfect positive correlation is +1, no correlation is 0, and perfect negative ...


122

It depends on what sense of a correlation you want. When you run the prototypical Pearson's product moment correlation, you get a measure of the strength of association and you get a test of the significance of that association. More typically however, the significance test and the measure of effect size differ. Significance tests: Continuous vs. ...


73

You're right on both counts. See Frank Harrell's page here for a long list of problems with binning continuous variables. If you use a few bins you throw away a lot of information in the predictors; if you use many you tend to fit wiggles in what should be a smooth, if not linear, relationship, & use up a lot of degrees of freedom. Generally better to ...


26

The standard Cauchy distribution is derived from the ratio of two independent Normal Distributions. If $X \sim N(0,1)$, and $Y \sim N(0,1)$, then $\tfrac{X}{Y} \sim \operatorname{Cauchy}(0,1)$. The Cauchy distribution is important in physics (where it’s known as the Lorentz distribution) because it’s the solution to the differential equation describing ...


24

That's an interesting question. My research group has been using the distribution you refer to for some years in our publicly available bioinformatics software. As far as I know, the distribution does not have a name and there is no literature on it. While the paper by Chandra et al (2012) cited by Aksakal is closely related, the distribution they consider ...


23

Is there a sharp discontinuity at your thresholds? For instance, suppose you have two patients A and B with values 3.9 and 4.1, and another two patients C and D with values 6.7 and 6.9. Is the difference in the likelihood for cancer between A and B much larger than the corresponding difference between C and D? If yes, then discretizing makes sense. If not,...


21

Aggregation is substantively meaningful (whether or not the researcher is aware of that). One should bin data, including independent variables, based on the data itself when one wants: To hemorrhage statistical power. To bias measures of association. A literature starting, I believe, with Ghelke and Biehl (1934—definitely worth a read, and suggestive of ...


21

In addition to its usefulness in physics, the Cauchy distribution is commonly used in models in finance to represent deviations in returns from the predictive model. The reason for this is that practitioners in finance are wary of using models that have light-tailed distributions (e.g., the normal distribution) on their returns, and they generally prefer to ...


20

Nominal vs Interval The most classic "correlation" measure between a nominal and an interval ("numeric") variable is Eta, also called correlation ratio, and equal to the root R-square of the one-way ANOVA (with p-value = that of the ANOVA). Eta can be seen as a symmetric association measure, like correlation, because Eta of ANOVA (with the nominal as ...


20

You arguably don't need measure theory to understand continuous random variables at all; those are just the random variables which are absolutely continuous with respect to Lebesgue measure. For most intents and purposes, the Riemann integral is sufficient in that case. After all, most commonly used probability densities have very nice regularity properties. ...


19

Look at this paper: Chandra, Nimai Kumar, and Dilip Roy. A continuous version of the negative binomial distribution. Statistica 72, no. 1 (2012): 81. It's defined in the paper as the survival function, which is a natural approach since neg binomial was introduced in reliability analysis: $$S_r(x)=\begin{cases}q^x & \text{for}\ r=1 \\ \sum_{k=0}^{r-1}...


18

It is a slight exaggeration to say that binning should be avoided at all costs, but it is certainly the case that binning introduces bin choices that introduce some arbitrariness to the analysis. With modern statistical methods it is generally not necessary to engage in binning, since anything that can be done on discretized "binned" data can ...


16

Looks like you're also looking for an answer from a predictive standpoint, so I put together a short demonstration of two approaches in R Binning a variable into equal sized factors. Natural cubic splines. Below, I've given the code for a function that will compare the two methods automatically for any given true signal function test_cuts_vs_splines <- ...


15

I've seen the following cheatsheet linked before: https://stats.idre.ucla.edu/other/mult-pkg/whatstat/ It may be useful to you. It even has links to specific R libraries.


15

Probabilities are models for the relative frequencies of observations. If an event $A$ is observed to have occurred $N_A$ times on $N$ trials, then its relative frequency is $$\text{relative frequency of }(A) = \frac{N_A}{N}$$ and it is generally believed that the numerical value of the above ratio is a close approximation to $P(A)$ when $N$ is "large" ...


14

Assuming proportional hazards (as in a Cox model) and the hazard ratio for a 1 mg increase in nicotine smoked a day is 1.02, then this tells you that persons smoking 11 mgs were 1.02 as likely to die in the monitored time period than persons smoking 10 mgs. The same applies to 12 vs 11 mgs etc. If the units of your continuous covariable are too small for ...


14

The result can be proven with a picture: the visible gray areas show that a uniform distribution cannot be decomposed as a sum of two independent identically distributed variables. Notation Let $X$ and $Y$ be iid such that $X+Y$ has a uniform distribution on $[0,1]$. This means that for all $0\le a \le b \le 1$, $$\Pr(a < X+Y \le b) = b-a.$$ The ...


14

Let $(\Omega,\mathscr{F},P)$ be the underlying probability space. We say that a measurable function $X:\Omega\to\mathbb{R}$ is an absolutely continuous random variable if the probability measure $\mu_X$ over $(\mathbb{R},\mathscr{B})$ defined by $\mu_X(B)=P\{X\in B\}$, known as the distribution of $X$, is dominated by Lebesgue measure $\lambda$, in the sense ...


11

There is, as far as I know, no taxonomy of variables that captures all the contrasts that might be important for some theoretical or practical purpose, even for statistics alone. If such a taxonomy existed, it would probably be too complicated to be widely acceptable. It is best to focus on examples rather than give numerous definitions. Number of days is ...


11

A part of this answer that I've learned since asking is that not binning and binning seeks to answer two slightly different questions - What is the incremental change in the data? and What is the difference between the lowest and the highest?. Not binning says "this is a quantification of the trend seen in the data" and binning says "I don't have enough ...


11

That is not a necessary result, but it is certainly plausible. If you turn a quantitive predictor into a single categorical predictor you lose a lot of information; with the categorical predictor you only know whether an observation is below or above a certain threshold (e.g. the mean or median), while with a quantitative predictor you also know how much ...


11

If you're talking about an interaction in a general linear model (e.g., ANCOVA), and if your categorical moderator has a reasonably small number of levels, you can plot separate regression lines for each level of the moderator. If you want these on the same plot, superimpose them, code by color or line type, and provide a legend. One of your plot's axes will ...


11

This seems like a job for survival analysis, like Cox proportional hazards analysis or possibly some parametric survival model. Think about this problem in reverse from the way you're explaining it: what are the predictor variables associated with earlier distances to quitting? Quitting is the event. The distance covered might be considered equivalent to ...


11

Actually nobody says that such event is impossible. Probability equal to zero is not the same as impossibility (check here, here and here). Probability that $X=x$ is equal to zero for continuous variables because chance of such event happening is infinitely small since there is an infinite number of real numbers. Also from purely mathematical point of view ...


10

I tried finding a proof without considering characteristic functions. Excess kurtosis does the trick. Here's the two-line answer: $\text{Kurt}(U) = \text{Kurt}(X + Y) = \text{Kurt}(X) / 2$ since $X$ and $Y$ are iid. Then $\text{Kurt}(U) = -1.2$ implies $\text{Kurt}(X) = -2.4$ which is a contradiction as $\text{Kurt}(X) \geq -2$ for any random variable. ...


10

Instead of throwing away data by categorizing, you could consider fitting your continuous predictor as a spline function with a specified number of knots or with the number of knots chosen by cross-validation. That will use up no more degrees of freedom than categorization. If you are willing to envision up to 8 categories, it's not clear that categorization ...


10

A guess at your point of confusion: Zero probability does not mean an event cannot occur! It means the probability measure gives the event (a set of outcomes) a measure zero. As @Aksakai's answer points out, the union of an infinite number of zero width points can form a positive width line segment and similarly, the union of an infinite number of zero ...


9

When nnd is 0 a unit change in gonad is associated with a $(\exp(-1.5718 ) - 1)*100\% \approx -79 \% $ decrease in the odds of fullyspawned. For every unit increase in nnd this effect of gonad increases by $(\exp( 0.6407 ) - 1)*100\% \approx 90 \%$. So, when nnd is 1 the odds ratio for gonad is $1.9^1 \times .21 \approx .4$, that is, a unit change in ...


9

If you want a correlation matrix of categorical variables, you can use the following wrapper function (requiring the 'vcd' package): catcorrm <- function(vars, dat) sapply(vars, function(y) sapply(vars, function(x) assocstats(table(dat[,x], dat[,y]))$cramer)) Where: vars is a string vector of categorical variables you want to correlate dat is a data....


9

It's wrong because - as the answer explained - there are discrete atoms at 0 and 2. By that cdf, you can wait exactly 0 time with positive probability (similarly with 2). Because of that, the waiting time is mixed, not continuous. Presumably you've been given definitions of all three. How are continuous r.v.s defined? If it's not immediately clear from ...


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