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Math Help - convergence weakly/in-distribution

  1. #1
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    convergence weakly/in-distribution

    Do "weak convergence" and "convergence in distribution" mean the same thing?

    Also, what does it mean for a distribution to not have a density? I can't think of a distribution that doesn't have a density (probably because I'm used to being given a density and integrating to the distribution).

    I'm sorry if these seem like stupid questions, but I feel stupid when it comes to probability theory.
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  2. #2
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    Hello,

    Weak convergence and convergence in distribution are the same (in probability theory at least)

    A discrete distribution is not a distribution which has a density ! (Poisson, binomial, ...)

    A characteristic of a distribution having a density (with respect to the Lebesgue measure) is that the probability of being at a given point is 0. We always talk about the probability of being in an interval.
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    Were only my text and instructor one-hundredth as clear and concise as you...

    So might it be put that a distribution has a density if and only if it is the distribution of a continuous random variable? (I say these things, but am unsure even of what they mean...I'm working on it.)
    Last edited by cribby; March 4th 2010 at 01:07 PM.
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    So might it be put that a distribution has a density if and only if it is the distribution of a continuous random variable?
    Unfortunately no !

    It can just be said that a random variable which has a density is continuous.
    That's kind of explained in this Wikipedia article :
    However, sometimes a continuous random variable can be "mixed", having part of its probability spread out over an interval like a typical continuous variable, and part of it concentrated on particular values, like a discrete variable.

    A random variable has an associated probability distribution and frequently also a probability density function. Probability density functions are commonly used for continuous variables.
    But in most cases - at least for now - you can assume that there's an equivalence between continuous rv and its distribution having a density.


    Were only my text and instructor one-hundredth as clear and concise as you...
    Haha, thanks... But I don't think I'm as skilled as they are. I'm just a student lol!
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    I'm trying to make sense of the statement of an example:

    Let X_n = \frac{1}{n},\, n = 1,2,3,\dotsand let X = 0. Let f_n,\, f be the corresponding density functions and let F_n,\, F be the corresponding distribution functions. Show that...

    The X_n are discrete uniform random variables, right? Why does the problem statement refer to their "density functions"? Is it just verbal abuse?
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    Capital letters F usually denote the cumulative distribution function (I suggest you have a google search on it). Sometimes called cdf.
    Contrary to the (probability) density (function) (sometimes called pdf), it's defined for any distribution.

    The cumulative distribution function is defined as F(t)=\mathbb{P}(X\leq t).

    Is it somehow clear to you ?

    The X_n are discrete uniform random variables, right?
    No, not the way you defined it
    Are you sure about what you wrote ?
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    I apologize--I'm using a mixture of examples I get from the internet and a text (in which the author, Billingsley, apparently has a stick in a very uncomfortable place) to fabricate some measure of understanding for myself.

    Billingsley doesn't call a lot of things by what the rest of the mathematical world seems to call them. The phrase "cumulative distribution function" never appears, just "distribution function". Similarly, "probability density function" never appears, I only see "density (-ies)". Not a huge deal, but I've come across some situations where Billingsley omitting a qualifier turned me upside-down when trying to cross-reference with other material.

    The example comes from a webpage, and I posted it nearly exactly as it appears. Apparently I'm just not understanding what the example is giving me. The example asks to show three things: (a) f_n(x)\to 0, \forall x \in \mathbb{R}, (b) F_n(x)\to 0 if  x \leq 0,\, F_n(x) \to 1 if  x>0, and (c) F_n(x) \to F(x),\, \forall x \neq 0.

    With that in mind, what were the random variables in the example statement defined as, then? Thanks again for your help.
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    I guess what might be throwing me off is that in Billingsley I'm reading that a random variable, its distribution, and its distribution function are discrete if the probabililty measure has a countable support. The text says that a support is a Borel set for which the probability measure on this set is 1. For each of the random variables X_n as defined in that webpage example, isn't the set \{\frac{k}{n} \vert k=1,2,\dots ,n\} a countable support?
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