MLE for X ~ Exp(λ)

StatsBeginner

New Member
Let (X1, ... Xn) be a random sample from a random variable. Find the maximum likelihood estimator for λ.

I know the answer to this but do you have a link to a step by step solution? I found sites which show all stages for finding the estimator for other distributions only.

Dason

Wikipedia has a walkthrough on the exponential distribution page.

StatsBeginner

New Member
Thanks but I'm not sure if this is the way I'm expected to be doing this. Is it possible to find the estimator using the fact that fx(k) for Poisson is (λ^k/k!)e^-λ?

Dason

What? Why would that... Why would . What?

Words. None.

Dason

Why would you bring the poisson into this? You want to estimate the MLE of the parameter for an exponential distribution. To do that you find the value of the parameter that maximizes the likelihood of the exponential distribution. Unless X ~ Exp(lambda) is your code for a poisson distribution...

StatsBeginner

New Member
I'm sorry, I think that lambda should indicate a Poisson here.

StatsBeginner

New Member
Never mind, I'm too confused! I think it's an exponential then. But are you sure there aren't any other simpler ways of finding the estimator than the one on Wikipedia?

Dason

The method on wikipedia is pretty much the universal method for finding an MLE. You get the likelihood (which they do), typically convert to using the log-likelihood (because it simplifies things), take the derivative and set equal to 0 and then solve for the parameter.

Have you found any MLEs before? Is this for a class?

StatsBeginner

New Member
I think I'm just tired, everything's clear now. It's the first time I see the equation which sets the derivative not only equal to 0 but also considers it <0 and >0. What's the point of setting this derivative <0 and >0? Can't I just omit it?

Dason

Using that additional information is one way to show that the value is a maximum and not a minimum.

StatsBeginner

New Member
Wikipedia uses λexp(-λx1). Is it the same as λe^(−λx)? If so, why do they write λexp(-λx1)?

GretaGarbo

Human
The expression "exp(x)" means the same thing as "e^x". It is just two different ways of writing the same thing. The number "e" is about 2.718.

So "lambda*exp(-lambda*x)" is the same thing as "lambda*e^(-lambda*x)".

By the way, the exponential distribution is a distribution for a continuous variable, like the time between two random events.

The exponential distribution is related to the Poisson distribution. But the Poisson distribution is for a discrete variable, like number of events within a specified time period. So the Poisson variable could take values like 0, 1, 2, 3, 4,....