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I was reading about Stochastic Convergence and I came across a term called Defective Distribution.

Essentially what they refer to as a {Defective Distribution is a distribution that has all the properties of a Cumulative Distribution $F$, i.e.,

$1)$ $F:\mathbb{R}\rightarrow[0,1]$

$2)$ Based on https://en.wikipedia.org/wiki/Cumulative_distribution_function it is also upwards continuous monotonic increasing function

BUT

the Defective Distribution does not have the property of the Cumulative Distribution

$3)$ $\lim_{x \to -\infty}F(x)=0$ and $\lim_{x \to \infty}F(x)=1$

the Defective Distribution has

$3^{'})$ $\lim_{x \to -\infty}F(x)\geq0$ and $\lim_{x \to \infty}F(x)<1$


I would like to ask if there is a know such distribution, i.e. a know Defective Distribution with the property $3^{'})$ ?

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    $\begingroup$ All such "distributions" are mixtures of (a) an ordinary distribution function, (b) a point mass at $-\infty,$ and (c) a point mass at $+\infty.$ This is an immediate consequence of $(3^\prime).$ $\endgroup$
    – whuber
    Commented Oct 11, 2021 at 15:44
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    $\begingroup$ @whuber By placing point masses at $+\infty,-\infty$ we avoiding taking the limit? $\endgroup$
    – Fiodor1234
    Commented Oct 11, 2021 at 16:31
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    $\begingroup$ Actually, no. We first have to extend the real line by adjoining points $\pm\infty$ at either end. (There is a standard construction in topology--a form of "completion" at the "ends" of the space--to do this.) The masses you assign to these points do not help you evaluate limits concerning what happens on the usual real numbers. Using the extended reals like this can be a helpful way to record information about (say) things that have less than a 100% chance of occurring. $\endgroup$
    – whuber
    Commented Oct 11, 2021 at 17:06
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    $\begingroup$ Suppose you are betting, and you start with $\$10$, and at each bet can win $\$1$ with probability $\frac23$ and lose $\$1$ with probability $\frac13$. Let $T$ be the time at which you run out of money. What is $P(T\le t)$ for $t \in \mathbb N$? For example $P(T\le 9)=0$ and $P(T\le 10)=3^{-10}$ Does the distribution of $T$ meet your definition of Defective Distribution since there is a positive probability you never run out of money? $\endgroup$
    – Henry
    Commented Oct 11, 2021 at 20:31
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    $\begingroup$ @Fiodor1234 Yes. $P(T \le t)$ is a weakly increasing function of $t$ (it jumps up for every even integer, starting at $t=10$). It is not continuous (though it is càdlàg like all distributions) but the CDF of a binomial distribution is not continuous either. $\endgroup$
    – Henry
    Commented Oct 11, 2021 at 20:59

1 Answer 1

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Well, certainly.

For instance, the arctangent is monotonically increasing from $\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2}$ to $\lim_{x\to\infty}\arctan x=\frac{\pi}{2}$. So something like

$$ F: x\mapsto \frac{1}{2}+\frac{1-2\epsilon}{\pi}\arctan x $$

will satisfy $\lim_{x\to-\infty}F(x) = \epsilon$ and $\lim_{x\to\infty}F(x) = 1-\epsilon$ for any (small, but) positive $\epsilon$.

plot

R code:

xx <- seq(-10,10,by=.1)
epsilon <- 0.1
plot(xx,1/2+(1-2*epsilon)*atan(xx)/pi,
  type="l",las=1,ylim=c(0,1),xlab="x",ylab="F(x)")
abline(h=c(epsilon,1-epsilon),lty=2)
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    $\begingroup$ I think your formula might be a bit off--it looks like $\frac{1}{2} + \frac{1-2\epsilon}{\pi}\arctan x$ would work. For anyone wondering, the main idea can be described using graph transformations: take the function $f(x) = \arctan x$, and multiply it by $1/\pi$ so its range is $(-1/2, 1/2)$. Then multiply it by some positive constant $k < 1$ so that the graph is slightly more vertically compressed, with a range of $(-k/2, k/2)$. Then add $1/2$ so the range is $((1-k)/2, (1+k)/2)$. In particular if $k = 1-2\epsilon$ then the range is $(\epsilon, 1-\epsilon)$. $\endgroup$
    – mathmandan
    Commented Oct 12, 2021 at 15:24
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    $\begingroup$ @mathmandan: thanks for pointing that out, and of course you are totally right! $\endgroup$ Commented Oct 12, 2021 at 15:51
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    $\begingroup$ +1 We could also do this with any other function not? If $F(x)$ is the CDF of a non-defective distribution, then $G(x) = 0.5 + (1-\epsilon) (F(x)-0.5)$ is the CDF of a defective distribution. $\endgroup$ Commented Oct 12, 2021 at 18:49
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    $\begingroup$ @SextusEmpiricus: yes indeed. Plus, we can of course use $F\circ g(x)$ for any monotonically increasing function $g$. $\endgroup$ Commented Oct 12, 2021 at 19:39

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