# Finding sufficient statistic for Weibull density function

I am given the follow problem and am having trouble finding the sufficient statistic.

Suppose that Y$_1$, Y$_2$, ..., Y$_n$ denote a Weibull density function, given by:

f ( y | $\theta$ ) =

Let $Y_1, Y_2, ... , Y_n$ denote a Weibull density function, given by: $$f (y | \theta ) = \begin{cases} \frac{2y}{\theta}e^\frac{-y^2}{\theta}, & 0 < y \\ 0, & \text{elsewhere} \end{cases}$$ Find the MVUE for $\theta$.

My issue here is again, in regards to finding the sufficient statistic. I begin by taking the likelihood:

L ($\theta$) = $\prod \frac{2y}{\theta}e^\frac{-y^2}{\theta}$

= $(\frac{2}{\theta})^n e^\frac{-\sum{y_i^2}}{\theta}\prod y_i$

How do I know what the sufficient statistic is? Is it:

1. $\prod y_i$
2. $-\sum{y_i^2}$

The answer is supposed to be the second one, but I'm still unclear as to how we know that. Any help will be much appreciated!

• Try taking the log of the likelihood. Which term now appears to be a constant, i.e., one whose value can change without affecting which value of $\theta$ will maximize the log likelihood? Which term's value affects which value of $\theta$ will maximize the log likelihood? – jbowman Nov 1 '17 at 20:37
• You can use the exponential family to get the sufficient stat. In this case 2 is the answer Try to write it in the exponential family and it will be clear to you – Ahmed Nov 1 '17 at 21:45
• @jbowman Ahh, yes, I see it when I take the log likelihood. My only concern is that we technically learned this method before learning the log likelihood, so I am looking for a way to understand the problem assuming I don't yet know that. – agra94 Nov 2 '17 at 0:55
• Well, taking the log is hardly a big deal, but I see what you mean. The logic you can use is that all $\Pi y_i$ does is scale the likelihood up and down, so to speak, it doesn't change its shape - or, more importantly, where the maximum is with respect to $\theta$. Sort of like that $2^n$ term at the front of the expression; you can include it or not, but it won't affect the shape or where the maximum is. This is not true of the other term $-\Sigma y_i^2$. – jbowman Nov 2 '17 at 1:18
• @jbowman Okay I think I get it now. So basically the reason -$\sum{y_i^2}$ is sufficient is because it's behavior is affected by the parameter, which in turn is a function of the likelihood? Sorry that was kind of awkwardly worded. – agra94 Nov 3 '17 at 16:20

By the Neyman factorization theorem it is quite clear the second option is the sufficient statistic.

Just review the theorem,

Let $X_1,X_2,...X_n$ denote a random sample from a distribution that has $\textit{pdf}$ or $\textit{pmf}$ of $f(x;\theta), \theta \in \Omega$. The statistic $Y_1 = u_1(X_1 , ... ,X_n)$ is a sufficient statistic for $\theta$ if and only if we can find two nonnegative functions, $k_1$ and $k_2$ such that $$f(x_1, x_2, ... , x_n;\theta) = k_1[ u_1(x_1, x_2, ... , x_n);\theta ] k_2(x_1, x_2, ... , x_n)$$

where:

$k_1$ is a function that depends on the data $x_1, x_2, ..., x_n$ only through the function $u_1(x_1, x_2,..., x_n)$, and the function $k_2(x_1, x_2, ..., x_n)$ does not depend on the parameter $\theta$

Now look at your $L(\theta)$, the first part is $(\frac{2}{\theta})^n e^\frac{-\sum{y_i^2}}{\theta}$ this is the $k_1$ function, it depends on $y_1, y_2,...,y_n$ only through $-\sum y_i^2$ (note, you should treat $\theta$ as a constant here, since you condition on it here). The second part is $\prod y_i$ which does not depend on $\theta$, it is the $k_2$ function.

So by the factorization theorem, you can directly say $-\sum y_i^2$ is the sufficient statistic for $\theta$.

• Ahh I see. So $\sum{y_i^2}$ is the sufficient statistic because e$^\frac{\sum{y_i^2}}{\theta}$ is a function of both the data and the parameter? ..... Or actually wait. If this is the case, why can't $\frac{2^n}{\theta^n} e^\frac{1}{\theta} \prod y_i$ be k1 and $e^\sum{y_i^2}$ be k2 ?. If so, k1 is still a function of the data and parameter and k2 only includes the data. – agra94 Nov 2 '17 at 0:50
• $e^{\frac{\sum{y_i^2}}{\theta}}\neq e^{\frac{1}{\theta}}e^{\sum{y_i^2}}$ – Deep North Nov 2 '17 at 1:09
• Ahh I've gotten way too far along in math to make that mistake, Thank you – agra94 Nov 2 '17 at 1:14