# Log-likelihood and sufficient statistic of exponential pdf

Let $$X_1, . . . , X_n$$ be a random sample from $$f(x,θ)=exp \{−(x−θ)\}exp\{−exp\{−(x−θ) \} \}$$ with$$−∞< θ <∞, −∞< x <∞$$. I have to find a sufficient and complete statistic and a MLE for $$θ$$.

I'm not sure if my approach is correct or there's a way to simplify the calculations.

For the sufficient statistic I tried this: $$p(X,θ) = \prod _{i=1}^n e^{-\left(x_i-\theta \right)} e^{-e^{-\left(x_i-\theta \right)}}$$ $$=e^{\left(n\theta \right)}e^{\left(-\sum_{i=1}^{n}x_i\right)}e^{-\sum_{i=1}^{n}e^{-\left(x_i-\theta \right)}}$$ $$=e^{\left(\theta-\bar{x} \right)n}e^{-\sum_{i=1}^{n}e^{-\left(x_i-\theta \right)}}$$

And defined $$T(X)=\bar{x}$$ and $$h(x)=1$$. But I'm stuck in proving this statistic is complete.

For the MLE aplied the Log-likelihood

$$l(\theta,x)= n\theta-\sum_{i=1}^{n}x_i-\sum_{i=1}^{n}e^{-\left(x_i-\theta \right)}$$

$$\frac {\partial [l(\theta,x)] } {\partial\theta}=n-\sum_{i=1}^{n}e^{-\left(x_i-\theta \right)}=0$$

I would like to know if there is a way to simplify this, or a better approach to obtain the MLE.

1. First of all your sufficient estimator is wrong.

The density can be written in the following way

$$f_X(x|\theta)=e^{\theta-x-e^{\theta-x}}$$

This can be viewed in the following way

$$f_X(x|\theta)=e^{-x}e^{\theta-e^{\theta}e^{-x}}$$

This shows that $$f_X(x|\theta)$$ belows to the Exponential family thus

$$S=\Sigma_x e^{-x}$$

is Sufficient and Complete.

1. MLE. Without doing any calculation, just at this point you know that the MLE is a function of the sufficient estimator (it's a property of MLE)

The likelihood is

$$L(\theta) \propto e^{n\theta-e^{\theta}\Sigma_xe^{-x}}$$

let's take the log

$$l(\theta)=n\theta-e^{\theta}\Sigma_xe^{-x}$$

let's derivating $$l(\theta)$$

$$l^*(\theta)=n-e^{\theta}\Sigma_xe^{-x}$$

$$\hat{\theta}_{ML}=log\frac{n}{\Sigma_xe^{-x}}=log \frac{n}{S}$$
...a function of $$S$$, as already known.