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Say you have:

$f(x_i,y_i)=\lambda \mu \exp\{-\lambda x_i-\mu y_i\}$

However you only observe:

$Z_i=\min (X_i,Y_i)~$ & $~U_i=\begin{cases}1~\text{if}~Z_i=X_i \\ 0~\text{if}~Z_i=Y_i\end{cases} $

We are asked to find a minimally sufficient statistic for this model.

What I have done is to find the marginal distributions $Z_i\sim\exp(\lambda+\mu)$ and $U_i\sim\text{Bernoulli}(\frac{\lambda}{\lambda+\mu})$. Then using the ratio theorem I found the minimally sufficient statistic for each which are essentially just the sums. So I was wondering if together the two statistics that I found would be minimally sufficient for $(\lambda,\mu)$ or do I really have to use the joint distribution to find the statistic.

Thanks

ActuariallyImpaired
  • 321

1 Answers1

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Figured it out, the trick is to write the joint distribution in the following way

$ f(z_i,u_i;\lambda,\mu) = \dfrac{ \exp \left [ \left ( \left ( \log (\lambda^2)-\lambda \right ) - \left ( \log (\mu^2)-\mu \right ) \right ) u_i z_i + \left ( \log (\mu^2)-\mu \right ) z_i \right ] }{\lambda+\mu} $

And then use the ratio theorem to show that

$\dfrac{l(\lambda,\mu|\mathbf{z},\mathbf{u})}{l(\lambda,\mu|\mathbf{s},\mathbf{v})} =\exp \left[ \left ( \left ( \log(\lambda^2)-\lambda \right ) - \left ( \log(\mu^2)-\mu \right ) \right ) \left ( \sum_{i=1}^n u_i z_i - \sum_{i=1}^n v_i s_i \right ) + \left ( \log(\mu^2)-\mu \right ) \left ( \sum_{i=1}^n z_i - \sum_{i=1}^n s_i \right ) \right ] $

The above is constant in terms of $\lambda$ and $\mu$ if and only if $\sum_{i=1}^n u_i z_i = \sum_{i=1}^n v_i s_i$ and $\sum_{i= 1}^n z_i = \sum_{i=1}^n s_i$ therefore $T(\mathbf{z},\mathbf{u})=(\sum_{i=1}^n u_i z_i,\sum_{i= 1}^nz_i)$ is the minimal sufficient statistic for $(\lambda,\mu)$

ActuariallyImpaired
  • 321
  • We can actually show that $Z_i$ and $U_i$ are independent (see https://math.stackexchange.com/questions/1049403/what-is-the-joint-distribution-of-z-minx-y-and-i-z-x?noredirect=1&lq=1). Don't know how you wrote the joint distribution, but looks like a minimal sufficient statistic is simply $(\sum u_i,\sum z_i)$. – StubbornAtom Jan 19 '20 at 18:24