191

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 ...


126

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. ...


29

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 ...


29

The standard Cauchy distribution is derived from ratio of two independent normally distributed random variables. 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 ...


26

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 ...


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,...


23

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 ...


22

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 ...


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

Let $X$ be a standard normal random variable, and let $Y = -X$ (pointwise). Then both are a.c., but $X+Y$ is $0$ everywhere.


18

A random variable $R$ is said to be continuous if for every real number $t,$ the probability that $R$ equals $t$ is zero $P(R = t) = 0.$ A random variable $R$ is said to be discrete if there exists a countable set of values $t_1, \ldots, t_n, \ldots$ such that $P(R = t_i) > 0$ for all $i$ and $\sum\limits_i P(R = t_i) = 1.$ The Radon-Nikodym and Lebesgue ...


17

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 <- ...


16

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" ...


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

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 ...


15

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 ...


14

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 ...


12

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 ...


12

It's really not a statistics question. It's a real analysis question. For instance, it's almost the same as asking "what's the width of a point in line?" (the answer is zero, by the way) This is an interesting situation though. In mathematics the line is defined as a set of points. There are certain geometric constraints on the points, so that they form a ...


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

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. ...


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 ...


11

Wikipedia has a list of distributions supported on an interval Leaving aside mixtures and 0-inflated and 0-1 inflated cases (though you should definitely be aware of all of those if you model data on the unit interval), which ones are common would be hard to establish (it will vary across application areas for example), but the beta family, and the ...


11

It's really that bad, as detailed in many of Patrick and my writings. Think about it these ways, for starters: If you categorize a marker the loss of information is so great that you will need to go an collect more markers to make up for the loss. Why not instead get the most information out of a single marker? Then there is the issue that it is easy to ...


10

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 ...


10

"The concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible." A. Kolmogorov For continuous random variables, $X$ and $Y$ say, conditional distributions are defined by the property that they recover the original probability measure, that is, for all measurable sets $A\in\mathcal{B}(\...


10

You seem to want to use a fractional logit, i.e. a quasi-likelihood model for a proportion. The key here is that it is a quasi-likelihood model, so the family refers to the variance function and nothing else. In quasi-likelihood that variance is a nuisance parameter, which does not have to be correctly specified in your model if your dataset is large enough. ...


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 ...


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