What is the name of a distribution such that it looks like uniform distribution on $[a, b]$, but the density is not a horizontal line, but a line with a certain slope.
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$\begingroup$ Why not call it a trapezoidal distribution? That's the shape we see. $\endgroup$– Russ LenthCommented Sep 6, 2014 at 2:13
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2 Answers
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A distribution with a linear PDF can be considered a special case of a (truncated) Pareto distribution, Beta distribution, or power distribution. Only particular values of the parameters in these distribution families will give a linear PDF, of course.
Among other things, such a distribution is a truncated Generalized Pareto Distribution. The Wikipedia parameterization of the PDF (for the case where the endpoints $a$ and $b$ are finite, as they must be for a linear graph) can be expressed in terms of the basic function
$$f(x; \eta) = (1 - \eta x)^{1/\eta - 1}$$
for $0\le x \le 1/\eta$. (This can then be rescaled by $\sigma\ne 0$ and shifted by $\mu$ to obtain the most general form. Here I have set $\eta = -\xi$ which will be a positive number.)
Evidently this function is linear in $x$ if and only if $1/\eta - 1=1$; that is, $\eta = 1/2$ (and so $\xi=-1/2$). Its graph equals zero at the endpoint $x=1/\eta = 2$. By truncating it, though, we would obtain the general form stated in the question.
The same thing can be obtained by truncating a generalized Beta Distribution. Its PDF is proportional to
$$x^{\alpha-1}(1-x)^{\beta-1}$$
for $0\le x \le 1$, whence the Beta$(2,1)$ and Beta$(1,2)$ distributions are linear. As with the Pareto$(\xi=-1/2)$ distribution, the graphs of these PDFs are zero at one endpoint. The general linear PDF described in question is obtained in the same way via truncation, rescaling, and shifting.
Finally, Mathematica defines a "power distribution" as one having a PDF proportional to
$$f(x; k, a) = x^{a-1}$$
for $0 \le x \le 1/k$. The case $a=2$ gives a linear PDF, identical to Beta$(2,1)$. Rescaling it by $k$ and recentering it with a parameter $\mu$, and (once again) truncating it will yield the general PDF described in the question.
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$\begingroup$ The power/Beta$(\alpha, 1)$ distributions can also be characterized as "log exponential" distributions: see this answer. For instance, when $X$ has an Exponential (aka Gamma$(1,1)$) distribution, then $(1/\sqrt{e})^X=\exp(-X/2)$ has a Beta$(2,1)$ distribution with linear density rising from $0$ at $x=0$ to $2$ at $x=1$. Adding a constant to $X$ rescales it and adding a constant to the entire result shifts it. Truncating that on the left gives the general form requested in the question. $\endgroup$– whuber ♦Commented Sep 16, 2014 at 20:50
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It might be the triangular distribution. The general triangular distribution with support $[a,b]$ has 1 parameter $c\in[a,b]$ corresponding to the mode. When $c=a$ or $c=b$, the p.d.f. is a straight line segment.
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$\begingroup$ This distribution bears its name because the density has a triangular shape, not that of a line segment of fixed slope! $\endgroup$– whuber ♦Commented Sep 5, 2014 at 22:17
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$\begingroup$ @whuber That is correct. As I mentioned, the pdf is shaped as a straight line when c=a or c=b (the interval boundaries). It is triangular when c is not equal to a or b. Is this not correct? $\endgroup$– a.arfeCommented Sep 5, 2014 at 23:07
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2$\begingroup$ Thanks for the answer. I think your statement is correct. On the other hand, the limitation is that the triangular distribution have 0 density at least one of the ends (i.e., a or b). That is, it cannot be used to describe a density function where the density on the edges are non-zero. Let me know if I missed anything $\endgroup$ Commented Sep 6, 2014 at 0:45
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2$\begingroup$ It is true that any triangular distribution with its mode at one end of the range has a linear density. However, the general linear density doesn't necessarily have $f(x)=0$ as either end. Think instead of a mixture of a mode-at-one-end triangular density with a uniform (making, in effect, a trapezoid with the two parallel sides parallel to the y-axis). There's some discussion of the issue here $\endgroup$– Glen_bCommented Sep 6, 2014 at 0:46