is there any closed answer for parameter of this distribution:f(x|teta) = 1/teta such that : 0<x< teta by mle method?

Yes, there is a closed-form solution for the parameter of this distribution using the Maximum Likelihood Estimation (MLE) method. Here are the steps:

1. First, we need to set up the likelihood function. Given a sample X1, X2, ..., Xn from this distribution, the likelihood function is:

   L(teta|X) = (1/teta^n) * I(X_(n) < teta)

   where X_(n) is the maximum of X1, X2, ..., Xn, and I is the indicator function that is 1 if X_(n) < teta and 0 otherwise.

2. To maximize this function, we take the derivative with respect to teta and set it equal to zero. However, because of the indicator function, the likelihood function is not differentiable at teta = X_(n).

3. Therefore, we need to consider the likelihood function for teta > X_(n) and teta < X_(n) separately.

   For teta > X_(n), the likelihood function is (1/teta^n), which is decreasing in teta.

   For teta < X_(n), the likelihood function is 0.

4. Therefore, the maximum likelihood estimate of teta is X_(n), the maximum observed sample.

So, the MLE for teta is max(X1, X2, ..., Xn).