Log-likelihood and sufficient statistic of exponential pdf

Question

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.

Answer 1

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.

2. 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 $l(\theta)$

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

Which immediately leads to

$$\hat{\theta}_{ML}=log\frac{n}{\Sigma_xe^{-x}}=log \frac{n}{S}$$

...a function of $S$, as already known.