# Proving sufficiency via likelihood functions

1. Aug 27, 2012

### BlueKazoo

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?

2. Aug 27, 2012

### micromass

Staff Emeritus
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.

3. Aug 27, 2012

### 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.

4. Aug 27, 2012

### micromass

Staff Emeritus
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 $\Omega$.

5. Aug 27, 2012

### 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).

6. Aug 27, 2012

### micromass

Staff Emeritus
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 $f(x)=x^3$ 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?