Mode of a mixed distribution

[yaukwankiu]

Does the "mode of a mixed continous-discrete distribution" make any sense at all?

I have seen an example like that somewhere in the ASM Exam-P manual (9th edition), and I am puzzled.
I mean, the continuous part of the pdf is not invariant under scaling. Its magnitude will shrink if you make your scale finer (e.g. for a uniform distribution, 1000 hits in the range 1-100 -> 100 hits in the range 1-10).

Naively I would think the discrete part is modelled by the dirac delta function, which has a magnitude of infinity.

That would mean that any point on the discrete part, however small, would trump any continuous part, however large.

Question: Would something like this appear in the actual exam?
Thank you very much.

[Gandalf]

No, worrying about the mode of a mixed continuous-discrete distribution is far beyond the scope of Course P.

By the way, Wikipedia says this about continuous distributions:

As noted above, the mode is not necessarily unique, since the probability mass function or probability density function may achieve its maximum value at several points x1, x2, etc.. When a probability density function has multiple local maxima, it is common to refer to all of the local maxima as modes of the distribution (even though the above definition implies that only global maxima are modes). Such a continuous distribution is called multimodal (as opposed to unimodal).

Candidates: don't try that on your exam (I think). Go for the overall maximum on a continuous distribution.

However, if you were to believe that local maxima of a continuous distribution are modes, then by extension local maxima of a mixed discrete/continuous distribtution should be modes.

But don't worry about encountering that on Course P.

[krzysio]

Originally Posted by Gandalf

No, worrying about the mode of a mixed continuous-discrete distribution is far beyond the scope of Course P.

By the way, Wikipedia says this about continuous distributions: Candidates: don't try that on your exam (I think). Go for the overall maximum on a continuous distribution.

However, if you were to believe that local maxima of a continuous distribution are modes, then by extension local maxima of a mixed discrete/continuous distribtution should be modes.

But don't worry about encountering that on Course P.

I most certainly disagree. In my opinion this is a possible exam problem.
Yours sincerely,
Krzys' Ostaszewski

[Gandalf]

Originally Posted by krzysio

I most certainly disagree. In my opinion this is a possible exam problem.
Yours sincerely,
Krzys' Ostaszewski

You are free to disagree. Some variations could appear. For example, if X had a discrete distribution with unique mode x, Y had a continuous distribution with unique mode y, and Z was a mixture of X and Y, then
1. If x=y (both had the same mode), they could probably ask the mode of Z.
2. If y was not among the answer choices, maybe they ask "which of the following is a mode of Z?"
Beyond that, it would be pretty absurd to ask, IMO, and even those two are questionable.

Here's something an author of Course P study material posted

Originally Posted by daaaave

There is no standard definition of the mode for mixed distributions, so something like that shouldn't come up on the exam. If you are asked for the mode, it should either be a purely continuous distribution, in which case you find x to maximize f(x), or a purely discrete distribution, in which case you find x to maximize P[X=x].

The only time I have ever seen someone rigorously define the mode for a mixed distribution, he did so in such a way that you would find the largest point mass, even if there is some point x such that f(x) is greater than the largest point mass. He felt it was a sufficiently non-standard definition that he spent quite a bit of his paper discussing it.

Maybe ASM contains a rigorous definition of the mode for a mixed distribution, but even if it does ASM is not on the SOA syllabus, nor is any text definitive for the SOA syllabus. The two questions I said might be possible problems satisfy properties I think everyone would agree the mode of a mixed distribution ought to satisfy in those specific cases, if the mode of a mixed distribution exists at all. (They would satisfy the rigorous definition daaaave said he saw proposed once.)

If (hard as it is to imagine) they don't take my advice and do ask it, your best choice by far is to go with the mode of the discrete distribution. You may not be respected among those who really understand probability theory, but you'll probably match the answer the SOA considers correct.

[yaukwankiu]

A continuous distribution can be compared with a discrete one
only only when the continuous and uncountable probability amplitude is collected into discrete bins.

It may make sense, because, at the end of the day, as are physical objects, money isn't really infinitely divisible.

Then you are left to argue, in which currency should we measure it and whether we should use dollars or cents as basic units. Of course the answer depends on your taste, political affiliation or personal wealth.

Regards,
Y.

[Gandalf]

Originally Posted by yaukwankiu

A continuous distribution can be compared with a discrete one
only only when the continuous and uncountable probability amplitude is collected into discrete bins.

It may make sense, because, at the end of the day, as are physical objects, money isn't really infinitely divisible.

Then you are left to argue, in which currency should we measure it and whether we should use dollars or cents as basic units. Of course the answer depends on your taste, political affiliation or personal wealth.

Regards,
Y.

That's an interesting way of looking at it, and inspires the following hypothetical question:

Suppose X has density function f(x) = 1 - |x| for [-1,-.5) and (-.5,1];
f(-.5) = 3 (that's a density, not a probability mass)
f(x) = 0 elsewhere

What is the mode of X?

Don't worry; you'll never see that one on Course P. I'm not sure if there is a correct answer. I observe that the cdf here is exactly the save as the cdf would be if f(-.5) = .5, so it might be hard to argue that the mode is not 0. Perhaps some would argue that my "density" is not a valid density function.

[yaukwankiu]

Originally Posted by Gandalf

That's an interesting way of looking at it, and inspires the following hypothetical question:

Suppose X has density function f(x) = 1 - |x| for [-1,-.5) and (-.5,1];
f(-.5) = 3 (that's a density, not a probability mass)
f(x) = 0 elsewhere

What is the mode of X?

Don't worry; you'll never see that one on Course P. I'm not sure if there is a correct answer. I observe that the cdf here is exactly the save as the cdf would be if f(-.5) = .5, so it might be hard to argue that the mode is not 0. Perhaps some would argue that my "density" is not a valid density function.

A point has measure zero so it makes no difference on your probability distribution.
(http://en.wikipedia.org/wiki/Probabi...mal_definition) See also this discussion (http://en.wikipedia.org/wiki/Probabi..._distributions) involving the Dirac delta function.