What is computed if you maximise likelihood of an exponential distribution without resorting to logs?

23 Views Asked by At

I would like to get some explanation or intuition behind the process of finding the maximum likelihood of an exponential distribution.

Given the following likelihood function:

$L\left( \theta \right) = \theta ^{4}e^{-8.65\theta }$

I would like to find $\theta$ for which the likelihood is maximised. I know I can achieve this by taking the derivative and solving the equation:

$\dfrac{dL\left( \theta \right) }{d\theta }=0$

Having calculated the derivative I obtained:

$\dfrac{dL\left( \theta \right) }{d\theta } = \theta ^{3}\left( 4-8.65\theta \right) \cdot e^{-8.65\theta }$

Solving it for $0$ yields ~ $0.46$.

When one takes a look at the plot of this function, one can see that this function doesn't have the maximum in that place. Since it is $e^{-x}$, the max value goes to the inf.

What am I actually determining, if I say that likelihood is maximised at 0.46?

1

There are 1 best solutions below

0
On

It has a maximum at $\theta\approx 0.4624$ even if it is just $L(0.46)\approx 0.0008$


$$...$$ enter image description here