# Finding minimal sufficient statistic and maximum likelihood estimator

Question:

Let $$Y_1, \dots, Y_n$$ be a random sample from a distribution with density function

\begin{align} f_Y(y ; \alpha, \theta) = \begin{cases} \alpha e^{- \alpha ( x - \theta)} & x > \theta,\\ 0 & x \leq \theta, \end{cases} \end{align}

where $$\alpha, \theta > 0$$ are parameters. Find the minimal sufficient statistic and maximum likelihood estimator for $$(\alpha, \theta)$$ when neither is known.

Attempt:

1) Minimal sufficient statistic

$$f_{\boldsymbol{Y}}(\boldsymbol{y}; \alpha, \theta) = \alpha^n e^{-\alpha(\sum y_i - n\theta)} I_{(\theta, \infty)}(y_{(1)})$$ where $$y_{(1)}$$ is the first order statistic and $$I$$ is an indicator function. According to the factorization theorem, $$(\sum y_i, y_{(1)})$$ is sufficient for $$(\alpha, \theta)$$.

To prove minimal sufficiency I need to show that given samples $$\boldsymbol{y}$$ and $$\boldsymbol{x}$$, the ratio of the joint densities $$\frac{f_{\boldsymbol{Y}}(\boldsymbol{y}; \alpha, \theta)}{f_{\boldsymbol{Y}}(\boldsymbol{x}; \alpha, \theta)}$$ involves neither $$\alpha$$ nor $$\theta$$ $$\Leftrightarrow (\sum y_i, y_{(1)}) = (\sum x_i, x_{(1)})$$.

\begin{align} \frac{f_{\boldsymbol{Y}}(\boldsymbol{y}; \alpha, \theta)}{f_{\boldsymbol{Y}}(\boldsymbol{x}; \alpha, \theta)} &= \frac{\alpha^n e^{-\alpha(\sum y_i - n\theta)} I_{(\theta, \infty)}(y_{(1)})}{\alpha^n e^{-\alpha(\sum x_i - n\theta)} I_{(\theta, \infty)}(x_{(1)})}\\ &= e^{-\alpha (\sum x_i - \sum y_i)} \frac{I_{(\theta, \infty)}(y_{(1)})}{I_{(\theta, \infty)}(x_{(1)})}. \end{align}

For this expression to not involve either $$\alpha$$ or $$\theta$$, I need $$\sum x_i = \sum y_i$$. My problem is that I don't see why I need $$y_{(1)} = x_{(1)}$$. As far as I can tell, $$y_{(1)}, x_{(1)} > \theta$$ is good enough.

Where am I going wrong?

2) Maximum likelihood estimator

For $$\alpha, \theta > 0$$,

\begin{align} L(\alpha, \theta ; \boldsymbol{y}) &= \alpha^n e^{-\alpha(\sum y_i - n\theta)}\\ \Rightarrow \ell (\alpha, \theta ; \boldsymbol{y}) &= n \log \alpha - \alpha \big( \sum y_i - n\theta \big)\\ \frac{\partial \ell}{\partial \alpha} &= \frac{n}{\alpha} - \big( \sum y_i - n\theta \big). \end{align}

Setting $$\frac{\partial \ell}{\partial \alpha} = 0$$, I get $$\hat{\alpha} = \frac{n}{\sum y_i - n \theta}$$.

To find $$\hat{\theta}$$, I look at $$L(\alpha, \theta ; \boldsymbol{y}) = \alpha^n e^{-\alpha \sum y_i} e^{\alpha n \theta} I_{(\theta, \infty)}(y_{(1)})$$ and I see that $$L$$ gets larger as $$\theta$$ increases from $$0$$ right until $$\theta = y_{(1)}$$, at which point $$L = 0$$. Since the MLE is defined to be a supremum, I conclude that $$\hat{\theta} = y_{(1)}$$.

So the maximum likelihood estimator for $$(\alpha, \theta)$$ is $$\big( \frac{n}{\sum y_i - n \theta}, y_{(1)}\big)$$.

Is this the correct MLE?

Thank you.

It's more or less correct, but some very minor issues...

1. to verify minimality, your ratio is

$$\frac{e^{-\alpha \sum_ix_i}\times\mathbb{1}_{(0;x_{(1)})}(\theta)}{e^{-\alpha \sum_iy_i}\times\mathbb{1}_{(0;y_{(1)})}(\theta)}$$

This ratio is independent by $$(\alpha;\theta)$$ iff

$$\begin{cases} \sum_ix_i=\sum_iy_i \\ x_{(1)}=y_{(1)} \end{cases}$$

1. Considering the profile likelihood,

$$\hat{\theta}=x_{(1)}$$

Then, substituting $$\hat{\theta}$$ with $$\theta$$ you get

$$\hat{\alpha}=\frac{n}{\sum_i[x_i-x_{(1)}]}$$

Thus

$$(\hat{\alpha};\hat{\theta})=(\frac{n}{\sum_i[x_i-x_{(1)}]};x_{(1)})$$