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Proving sufficiency via likelihood functions

  • Thread starter BlueKazoo
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Homework Statement


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.

Homework 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.

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?
 

Answers and Replies

  • #2
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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.
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

If θ is known,
I already know that θ won't be the unkown parameter.
 
  • #3
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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.
 
  • #4
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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].
 
  • #5
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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).
 
  • #6
22,097
3,277
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.
OK, that seems good if all y are between 0 and θ. But now you need to work with indicator functions.

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).
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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