2
$\begingroup$

I have a math coursework where we are supposed to find the MLEs of the following distribution:

$$ p(x;\alpha) = \frac{1+\alpha x}{2}, x \in [-1,1], \alpha\in[-1,1] $$

I'm just curious as to what it is as I have never seen it before. It is apparently used for modelling something about muon disintegration in physics. Sorry if this should have been posted in the maths or physics sections.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ This is not a pdf : $\int p(x,a)dx\neq 1$ in general. And there's no sense to talk about the MLE of a distribution. $\endgroup$
    – Stéphane Laurent
    Commented Apr 26, 2013 at 17:19
  • 2
    $\begingroup$ @StéphaneLaurent: In which case don't we have $\int_{-1}^1 p(x; a) \, \mathrm{d}x = 1$? $\endgroup$
    – ocram
    Commented Apr 26, 2013 at 17:29
  • 3
    $\begingroup$ It does integrate to 1 for every value of $\alpha$. $\endgroup$
    – random_user
    Commented Apr 26, 2013 at 18:00
  • $\begingroup$ Oops sorry, I thought about $\int_0^1$ ! $\endgroup$
    – Stéphane Laurent
    Commented Apr 26, 2013 at 18:02

1 Answer 1

Reset to default
5
$\begingroup$

It's a two-component-mix of a Uniform(-1,1) and a Triangular(-1,1).

To see this, let f(x) denote our Uniform(-1,1) pdf:

f = 1/2;  domain[f] = {x, -1, 1};      
PlotDensity[f]

enter image description here

and let g(x) denote our Triangular(-1,1) pdf:

g = (1 + x)/2; domain[g] = {x, -1, 1};      
PlotDensity[g]

enter image description here

Then, define the two-component mix pdf h(x):

h = (1 - α) f + α g  // FullSimplify

which returns output:

$(1 + x \alpha)/2$

with domain of support:

domain[h] = {x, -1, 1}  && {-1 <= α <= 1};

Here is a plot of the two-component-mix pdf (using mathStatica/Mathematica):

PlotDensity[h /. α -> Range[-5, 5]/5]

enter image description here

Having said all that, I think it deserves its own name (if it does not have one already) and suggest: Linear distribution

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ +1. It's also a truncated triangular and a truncated linear distribution. (Linear distributions would naturally have a PDF of zero at one endpoint.) $\endgroup$
    – whuber ♦
    Commented Apr 26, 2013 at 19:30
  • $\begingroup$ The term 'truncated' would imply that the domain of support is being constrained. However, that is not the case here ... it remains (-1,1). Linear would not necessarily require that the intercept be 0. But perhaps there are nicer names? Rotated Uniform? Spin the donkey? $\endgroup$
    – wolfies
    Commented Apr 26, 2013 at 20:25
  • $\begingroup$ "Truncated" is a relative term. Here it's clearly the case that the support is constrained to $[-1,1]$. But it doesn't matter... $\endgroup$
    – whuber ♦
    Commented Apr 26, 2013 at 20:26
  • $\begingroup$ I really like this pdf - it's cute and neat and fun - and I think it should be added to the in-built Continuous distribution palette in mathStatica. After further thought, perhaps the name 'Linear' is not quite right, because, it only occupies 90 degrees, from one extreme to the other (as in acute). So, the name I am tinkering with is: "A Cute Linear Distribution", if you will excuse the pun :) $\endgroup$
    – wolfies
    Commented Apr 27, 2013 at 9:36
  • $\begingroup$ The "90 degrees" restriction disappears when this family is extended to include variations of scale. $\endgroup$
    – whuber ♦
    Commented Apr 28, 2013 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.