# Proving sufficiency via likelihood functions

BlueKazoo

## 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?

Staff Emeritus
Homework Helper
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.

BlueKazoo
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.

Staff Emeritus
Homework Helper
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 $\Omega$.

BlueKazoo
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).

Staff Emeritus
Homework Helper
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?