# Tag Info

7

There is actually no such object as $Y|X$ --- whenever this notation appears, it is an abuse of notation which operates as shorthand for specifying the conditional distribution of a random variable conditional on another random variable.$^\dagger$ Thus, the statement $Y|X = m(X) + \epsilon$ actually doesn't make any sense; the conditionality notation $|X$ ...

4

Let the distinct values be $$X = \{x_1, x_2, \ldots, x_n\}.$$ Let the sample size be $m\, (=3).$ The ordered samples of distinct values are the subset $$\Omega = \{(y_1, y_2, \ldots, y_m) \in X^m\mid y_1\lt y_2\lt \ldots \lt y_m\}.$$ The question seeks a probability distribution on $\Omega$ with probabilities $$p(i_1,i_2,\ldots, i_m) = \Pr((y_1,\ldots, ... 3 The problem with both characterizations is that they ignore the underlying probabilities. Recall that a random variable X is a function that assigns real numbers to elements of the sample space. If a considerable part of the domain of X has no probability, then the range of X may have virtually any property whatsoever but that won't tell you a thing ... 3 Well, even if the range (or support set) of the random variable X is uncountable, X do not necessarily have a density. The answer by @Sebastian mentions measure, and specifically counting measure. But counting measure on an uncountable set isn't very useful, for instance, it is not \sigma-finite. So not very useful in probability. There is an ... 2 Consider the example where your sample space \Omega = \mathbb{R}. This is uncountably infinite. However whether the RV is continuous depends on the used measure. If you would use \mu = \# (i.e. the counting-measure) you could still easily defined a density with respect to \mu that would induce a discrete distribution. In general whether a ... 2 In my opinion, the use of the equality makes dependence on X explicit. The only place I have seen notation like y \vert X is when the distribution of y is being discussed. You see this frequently in Bayesian models like$$ y\vert \mu , \sigma \sim \mathcal{N}(\mu, \sigma)  This notation tells me that there is a prior for $\sigma$ and that the ...

2

Let's clarify some of the issues. Convexity Consider a real-valued function function $f$ defined on an interval $I$ of real numbers. (Everything that follows extends naturally and obviously to functions defined on sets $I\subset \mathbb{R}^d$ provided $I$ is a convex set, but I won't belabor that point. The key property enjoyed by $I$ is that "mixtures" ...

2

Although these statements capture the idea pretty well, we should take some care concerning their logical status: some provide definitions; others are characterizations not unique to discrete random variables; and a couple of them are meaningless (depending on how broadly they are interpreted). It can be useful to bear in mind some examples of discrete ...

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The $n$th central moment is defined as $E[(X-E(X))^n]$, so for $n=1$, we have $E[(X-E(X))]$. By linearity of expectation, this equals $E(X)-E[E(X)]$. Since $E(X)$ already is a constant, so that $E[E(X)]=E(X)$.

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Hi: Just set the seed to some value initially say 123.( outside any loops, before any calculations start ). Then you still have reproducibility and the problem that you describe goes away.

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There may be a good reason for the complications, because the equivalence is not generally true. For it to hold, you have to assume $X$ has nonzero variance and that both $X$ and $Y$ have finite variances, and even then by assuming the correlation of $X$ and $Y$ is $\pm 1$ you can only conclude there exist numbers $a$ and $b$ for which there is a 100% ...

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You can find more detail on this issue in a related question here. Absolute continuity of the CDF is a stronger condition than continuity, and essentially just means that the distribution has a valid density function. For example, if the CDF $F$ of a real scalar random variable is absolutely continuous then there exists a real function $f$ (the density) ...

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On the one hand, it's hard to read the mind of an author of a paper that is 35 years old. On the other hand, the reason may simply be that there are far more tests for a normal than for a uniform distribution. This in turn may be due either to more interest in the normal than in the uniform distribution (the various probabilistic laws imply that we will ...

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