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!

  • $\begingroup$ 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? $\endgroup$
    – jbowman
    Nov 1, 2017 at 20:37
  • $\begingroup$ 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 $\endgroup$
    – Ahmed
    Nov 1, 2017 at 21:45
  • $\begingroup$ @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. $\endgroup$
    – jmoore00
    Nov 2, 2017 at 0:55
  • $\begingroup$ 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 $-\Sigma y_i^2$. $\endgroup$
    – jbowman
    Nov 2, 2017 at 1:18
  • $\begingroup$ @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. $\endgroup$
    – jmoore00
    Nov 3, 2017 at 16:20

1 Answer 1


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 $$f(x_1, x_2, ... , x_n;\theta) = k_1[ u_1(x_1, x_2, ... , x_n);\theta ] k_2(x_1, x_2, ... , x_n)$$


$k_1$ 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 $\theta$.

  • $\begingroup$ 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. $\endgroup$
    – jmoore00
    Nov 2, 2017 at 0:50
  • $\begingroup$ $e^{\frac{\sum{y_i^2}}{\theta}}\neq e^{\frac{1}{\theta}}e^{\sum{y_i^2}}$ $\endgroup$
    – Deep North
    Nov 2, 2017 at 1:09
  • $\begingroup$ Ahh I've gotten way too far along in math to make that mistake, Thank you $\endgroup$
    – jmoore00
    Nov 2, 2017 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.