Exercise : exponential law and maximum likelihood estimation I'm stuck on this problem, especially question n°2, I have no idea how to proceed at all. Thank you for your help !
X is a continuous random variable distributed following an exponential distribution of parameter z with a density
Fx(x) = z* exp(-z*x) when x>= 0
Fx(x) = 0 when x<0
1) Write the expression of the distribution function of X and give the probability that X goes above the value x= 1 when z = 1
The distribution function is the derivative of Fx(x), so :
F(x) = 1 - exp(-z*x) when x>=0
F(x) = 0 when x<0
F(x) = P(X <= 1)
F(x) = 1 - exp(-1*1) = 1 - exp(-1)
So for X’s value to go above 1 : 1 – (1 - exp(-1))
2) Using the maximum likelihood method, give the expression of the estimator of z
I do not know at all how to proceed for this question 

The likelihood function is
$L(\lambda)=\prod_{i=1}^n f(x_i,\lambda)=\prod_{i=1}^n \lambda e^{-\lambda x_i}$
$L(\lambda)=\lambda ^ne^{-\lambda \sum_{i=1}^n x_i}$
Taking logs on both sides.
$\ln L=n\cdot \ln(\lambda)- \lambda\cdot \sum_{i=1}^n x_i$
Now differentiate and set the derivative equal to zero. Then solve the equation for $\lambda$. For this purpose it is good to know that $\overline x=\frac1n\sum_{i=1}^n x_i$.
