Exercise : exponential law and maximum likelihood estimation

Question

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

Answer 1

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 $\overline x=\frac1n\sum_{i=1}^n x_i$.