1. ## Exponential law exercise

Hello ! I am struggling a bit with this probability exercise. Thank you very much for any help with the problem !

X is a continuous random variable distributed following an exponential law 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

I found it by searching on wikipedia, however I don't know how to find it from the density. Was I supposed to derive this expression ?

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

2. ## Re: Exponential law exercise

1) you've done correctly.

2) basically you want to find, given an observation $x$, the parameter $z$ which is the most likely to have caused the observation.

we find the maximum $z$ the way we find any other maximum of a function, setting it's derivative to zero and solving.

$f_z(x) = z e^{-z x}$

$\dfrac{\partial f}{\partial z} = e^{-zx} - z x e^{-z x} = (1-x z) e^{-zx}$

$f_z(x) = 0 \Rightarrow z = \dfrac 1 x$

This isn't surprising since $E[x] = \dfrac 1 z$

3. ## Re: Exponential law exercise

Thank you for the explanation !

When I try to derive fx I obtain -z^2 e^(-zx) though. Is there a différence between df/dz and df/dx ?

4. ## Re: Exponential law exercise

Originally Posted by Jo37
Thank you for the explanation !

When I try to derive fx I obtain -z^2 e^(-zx) though. Is there a différence between df/dz and df/dx ?
you really should know the answer to that one.

of course there is.

you are trying to maximize the density function by adjusting $z$ not $x$. You have a value for $x$, that's your observation.

so you use $\dfrac{df}{dz}=0$