Proving sufficiency via likelihood functions

Click For Summary
SUMMARY

The discussion focuses on proving the sufficiency of the product of independent and identically distributed random variables from a power family distribution, specifically when the parameter θ is known. The likelihood function is defined as L(Y1,Y2,...,Yn|θ)=∏i=1nf(yi|θ), and the sufficiency condition requires that L(Y1,Y2,...,Yn|θ)=g(u,θ) x h(y1,y2,...,yn), where g is a function of u and θ, and h does not depend on θ. The participants clarify that the product ∏i=1n Yi serves as the statistic u, and they explore the use of indicator functions to properly define the likelihood function across its valid range.

PREREQUISITES
  • Understanding of likelihood functions in statistics
  • Familiarity with power family distributions
  • Knowledge of sufficient statistics and their properties
  • Basic concepts of indicator functions in probability theory
NEXT STEPS
  • Study the derivation of likelihood functions for power family distributions
  • Learn about the concept of sufficient statistics in depth
  • Explore the application of indicator functions in probability distributions
  • Investigate examples of proving sufficiency using likelihood functions
USEFUL FOR

Statisticians, data scientists, and students studying statistical inference who are looking to deepen their understanding of sufficiency in likelihood functions and power family distributions.

BlueKazoo
Messages
3
Reaction score
0

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?
 
Physics news on Phys.org
BlueKazoo said:
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.
 
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.
 
OK, next. What is the joint distribution of Y1,...,Yn??

I know you said

\prod_{i=1} \frac{\alpha y_i^{\alpha-1}}{\theta^\alpha}

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

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

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.

Try to use this with a suitable A and \Omega.
 
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?:
\prod_{i=1}^{n}\frac{\alpha y_{i}^{\alpha - 1}}{\theta^{\alpha}}
\frac{\alpha^{n}}{\theta^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}
Therefore
g(\prod_{i=1}^{n}y_{i}|\alpha)=\frac{\alpha^{n}}{ \theta ^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}
h(y_{1},y_{2},...,y_{n})=1
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).
 
BlueKazoo said:
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?:
\prod_{i=1}^{n}\frac{\alpha y_{i}^{\alpha - 1}}{\theta^{\alpha}}
\frac{\alpha^{n}}{\theta^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}
Therefore
g(\prod_{i=1}^{n}y_{i}|\alpha)=\frac{\alpha^{n}}{ \theta ^{n\alpha}}(\prod_{i=1}^{n}y_{i})^{\alpha - 1}
h(y_{1},y_{2},...,y_{n})=1
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 f(x)=x^3 for 0≤x≤1 and f(x)=0 otherwise. I would write this as

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

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

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K