51

If you "know" that the coin is fair then we still expect the long run proportion of heads to tend to $0.5$. This is not to say that we should expect more (than 50%) of the next flips to be tails, but rather that the initial $500$ flips become irrelevant as $n\rightarrow\infty$. A streak of $500$ heads may seem like a lot (and practically speaking it is), ...


43

A distribution that is normal is not highly skewed. That is a contradiction. Normally distributed variables have skew = 0.


41

I agree with X'ian that the problem is under-specified. However, there is an elegant, scalable, efficient, effective, and versatile solution worth considering. Because the product of the sample mean and sample size equals the sample sum, the problem concerns generating a random sample of $n$ values in the set $\{1,2,\ldots, k\}$ that sum to $s$ (assuming $...


36

No it is not uniform You can count the $36$ equally likely possibilities for the absolute differences second 1 2 3 4 5 6 first 1 0 1 2 3 4 5 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 5 4 3 2 1 0 1 6 5 ...


25

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

The citation is true. When you plug $x=0$ to the PDF function, you do NOT get the probability of taking this particular value. The resulting number is probability density which is not a probability. The probability of taking exactly $x=0$ is zero (consider the infinite number of similarly-likely values in the tiny interval $x\in[0,10^{-100}]$). To ...


21

Using only the most basic axioms about probabilities and real numbers, one can prove a much stronger statement: The difference of any two independent, identically distributed nonconstant random values $X-Y$ never has a discrete uniform distribution. (An analogous statement for continuous variables is proven at Uniform PDF of the difference of two r.v.) ...


19

If it has long right tail, then it's right skewed. It can't be a normal distribution since skew !=0, it's perhaps a unimodal skew normal distribution: https://en.wikipedia.org/wiki/Skew_normal_distribution


19

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 ...


17

The law of large numbers doesn't state that some force will bring the results back to the mean. It states that as the number of trials increases the fluctuations will become less and less significant. For example, if I toss the coin 10 times and get 7 heads, those two extra heads seem pretty significant. If I toss the coin 1,000,000 times and get 500,...


16

Since the probability element of $X$ is $f(x)\mathrm{d}x,$ the change of variable $y = x\sigma + \mu$ is equivalent to $x = (y-\mu)/\sigma,$ whence from $$f(x)\mathrm{d}x = f\left(\frac{y-\mu}{\sigma}\right)\mathrm{d}\left(\frac{y-\mu}{\sigma}\right) = \frac{1}{\sigma} f\left(\frac{y-\mu}{\sigma}\right) \mathrm{d}y$$ it follows that the density of $Y$ is $...


14

A fraction per day is certainly not negative. This rules out the normal distribution, which has probability mass over the entire real axis - in particular over the negative half. Power law distributions are often used to model things like income distributions, sizes of cities etc. They are nonnegative and typically highly skewed. These would be the first I ...


14

Since it looks like self-study question, I'll start with a hint: Think of $X_1-X_2$ with $X_1, X_2 \sim N(470, 70^2)$. What distribution does $X_1-X_2$ follow? How to interpret $X_1-X_2$?


14

If you had been looking for the distance to the next value above, and if you inserted an extra value at $1$ so this always had an answer, then using rotational symmetry the distribution of these distances $D$ would be the same as the distribution of the minimum of $n+1$ independent uniform random variables on $[0,1]$. That would have $P(D \le d) = 1-(1-d)...


14

The question is under-specified in that the constraints on the frequencies \begin{align}n_1+2n_2+3n_3+4n_4&=100M\\n_1+n_2+n_3+n_4&=100\end{align} do not determine a distribution: "random" is not associated with a particular distribution, unless the OP means "uniform". For instance, if there exists one solution $(n_1^0,n_2^0,n_3^0,n_4^0)$ to the above ...


13

It could be a log-normal distribution. As mentioned here: Users' dwell time on online articles (jokes, news etc.) follows a log-normal distribution. The reference given is: Yin, Peifeng; Luo, Ping; Lee, Wang-Chien; Wang, Min (2013). Silence is also evidence: interpreting dwell time for recommendation from psychological perspective. ACM International ...


12

There are a dozen of continues probability distributions There are an infinite number of continuous probability distributions. The ones that have been discussed enough to be named and included in the space of a couple of pages are nevertheless sufficient to fill numerous books (and indeed they do - see, for example, the many books by Johnson, Kotz and other ...


12

On an intuitive level, a random event can only be uniformly distributed if all of its outcomes are equally likely. Is that so for the random event in question -- absolute difference between two dice rolls? It suffices in this case to look at the extremes -- what are the biggest and smallest values this difference could take? Obviously 0 is the smallest (...


12

For simulation purposes, a Weibull distribution may work well. Allow me to explain why and to say something about the limitations. A plot of the original (unexponentiated) residuals immediately suggested a Weibull distribution to me. (One reason this family comes to mind is that it includes Rayleigh distributions, which are Weibull with shape parameter $...


12

You may be interested in the Earth mover's distance, also known as the Wasserstein metric. It is implemented in R (look at the emdist package) and in Python. We also have a number of threads on it. The EMD works for both continuous and discrete distributions. The emdist package for R works on discrete distributions. The advantage over something like a $\...


12

This can be done with a minimum of computation, relying only on (a) simple algebra and (b) basic knowledge of distributions associated with statistical tests. As such, the demonstration may have substantial pedagogical value--which is a fancy way of saying it's worth studying. Let $Z=X/(X+Y),$ so that $$Z - \frac{1}{2} = \frac{X}{X+Y} - \frac{X/2+Y/2}{X+Y}...


11

The general case of the cube of an normal random variable with any mean is quite complicated, but the case of a centered normal distribution (with zero mean) is quite simple. In this answer I will show the exact density for the simple case where the mean is zero, and I will show you how to obtain a simulated estimate of the density for the more general case....


11

A random variable which is not actually random, and doesn't change by chance, is by definition a constant. But, it is still a RV. Since the RV definition is a superset of constant RV definition, I believe there is no conceptual opposite.


10

Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $\times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed. The code ...


10

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $\mathbb E[X]=500$ is of the form $$...


10

Not unless you already know what the distribution is. And likely not even then, unless it's a normal or lognormal distribution (which can be completely described by those two values). You can calculation the mean and standard deviation from any set of numbers sampled from any distribution, so you cannot recreate a distribution based on them alone. They ...


10

Although this question relies heavily on Python, the answer does appear to benefit from some statistical reasoning. This function creates "training" and "test" datasets of points $(x_i,y_i)$ for a regression model $$y_i = w_0 x_i + w_1 x_i^2 + \varepsilon_i \sigma$$ where $\varepsilon_i$ are independent variables with standard Normal distributions. The ...


9

For purely positive quantities a log-transformation is indeed the standard first transformation to try and is very frequently used. It is also done if for regression you want a multiplicative interpretation of coefficients (e.g. doubling/ halving of blood cholesterol). Of course it will not always make a distribution more normal, e.g. take samples from a N(...


9

Correction: the Jacobian of the transform is $|V|$, not $V$, which implies that $$f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|J|=\frac{|v|}{2\pi}\exp\left\{\frac{-v^2}{2}(2u^2-2u+1)\right\}$$ Hence \begin{align}f_U(u)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}|v|e^{\frac{-v^2}{2}(2u^2-2u+1)}\text{d}v\\ &=\frac{2}{2}\frac{1}{\pi}\int_{0}^{\infty}ve^{-\overbrace{\frac{v^...


8

The likelihood function is merely proportional to the sampling density, in the sense that you have $L_x(\theta) = k \cdot p(x|\theta)$ for some constant $k > 0$ (though you should note that the likelihood is a function of the parameter, not the data). If you want to use this in your expression for Bayes theorem then you need to include the same scaling ...


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