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Proving sufficiency via likelihood functions

  1. Aug 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Let Y1,Y2,...,Yn denote independent and identically distributed random variables from a power family distribution with parameters α and θ. Then, if α, θ > 0,

    f(y|α, θ)={αy(α-1)α, 0≤y≤θ; 0, otherwise.

    If θ is known, show that ∏i=1n Yi is sufficient for α.

    2. Relevant equations
    Well, I know that the likelihood function is L(Y1,Y2,...,Yn|θ)=∏i=1nf(yi|θ).

    I also know that a statistic is sufficient if L(Y1,Y2,...,Yn|θ)=g(u,θ) x h(y1,y2,...,yn) Where g(u, θ) is a function of only u and θ, and h(y1,y2,...,yn) doesn't have θ. I am not 100% clear on what exactly this entails in practice since there are very few examples in my book which deal with proving sufficiency this way.

    3. The attempt at a solution
    I simply plugged the first formula into the the second, so I got:
    i=1nαyiα-1α
    (α/θα)(∏i=1nyi)α-1

    I have no idea how to handle the remaining product function. I have some more problems in my homework where I reach an impasse at this point. Can I stop here, or am I missing some steps here?
     
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  3. Aug 27, 2012 #2

    micromass

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    What is u in the first place? Is it your statistic?? What is your statistic in this case??

    Secondly, θ is the unknown parameter. What is the unknown parameter in this case?? From this sentence

    I already know that θ won't be the unkown parameter.
     
  4. Aug 27, 2012 #3
    Ah, whoops, I should have used different variables in the problem than in the set of given equations.

    I believe "u" in g(u|θ) would be the ∏i=1nYi i this case. In g(u|θ) and L(...|θ), θ is just whatever variable I need to prove the first one is sufficient for.

    So I guess I'm trying to find g(∏i=1nYi|α) and h(Y1...Yn), the latter of which would not contain α). I apologize for the confusion, though I believe the presentation of the book is a bit wonky to begin with and I'm not entirely sure how to use these equations.
     
  5. Aug 27, 2012 #4

    micromass

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    OK, next. What is the joint distribution of Y1,...,Yn??

    I know you said

    [tex]\prod_{i=1} \frac{\alpha y_i^{\alpha-1}}{\theta^\alpha}[/tex]

    but that isn't correct. The reason that it's not correct is because this form only holds for [itex]0\leq y_i\leq \theta[/itex]. But we want the formula to hold for all [itex]y_i[/itex].

    To make it work, you'll need to use characteristic functions. These are defined for a set A as

    [tex]I_A:\Omega\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l}0 ~\text{if} ~x\notin A\\ 1~\text{if}~x\in A\end{array}\right.[/tex]

    Try to use this with a suitable A and [itex]\Omega[/itex].
     
  6. Aug 27, 2012 #5
    Oh, so you basically mean that it's incorrect because I didn't take into account where the function is 0 otherwise.

    If any y is not between 0 and θ, the value of the function is only dependent on that specific y and therefore dependent on the product of all the y's. So the function is sufficient for α, since all other values besides that product y are inconsequential to the final value, right?

    Buuut, let's assume all y's fall between 0 and θ. Then would I solve it like this?:
    [itex]\prod_{i=1}^{n}\frac{\alpha y_{i}^{\alpha - 1}}{\theta^{\alpha}}[/itex]
    [itex]\frac{\alpha^{n}}{\theta^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}[/itex]
    Therefore
    [itex]g(\prod_{i=1}^{n}y_{i}|\alpha)=\frac{\alpha^{n}}{ \theta ^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}[/itex]
    [itex]h(y_{1},y_{2},...,y_{n})=1[/itex]
    Since n and θ are constants.

    I should have probably set up the indicator function earlier, and I'm not too familiar with them, but I'll try anyways. I guess A could be the range of numbers between 0 and θ? And Ω would be some form of my function? Honestly, I don't understand how to set them up in this case, and none of my resources have examples of this (in this context anyways).
     
  7. Aug 27, 2012 #6

    micromass

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    OK, that seems good if all y are between 0 and θ. But now you need to work with indicator functions.

    OK. Let's try an example. Let's say that I have [itex]f(x)=x^3[/itex] for 0≤x≤1 and f(x)=0 otherwise. I would write this as

    [tex]f(x)=x^3 I_{[0,1]}(x)[/tex]

    So in this case, A=[0,1] and [itex]\Omega=\mathbb{R}[/itex]. Do you see why this works??
    What is it going to be in your case?
     
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