'mle' - can we use this command for maximizing log-likelihood estimation of a vector input, which is normally distributed?

I have a vector of particular length, normally (Gaussian) distributed, for which I want to maximize log-likelihood estimation. I have directly given that vector to 'mle' command. But the output I got was not the exact thing I needed. So, how to calculate the log-likelihood estimation? and how to maximise it? It would be a great help if someone could help me in this regard....

[Torsten]

But the output I got was not the exact thing I needed

And what do you need ?

[BIPIN SAMUEL]

@Torsten I have a matrix having the biomedical signal, I want to maximize the log-likelihood of each row using mle. I would be grateful if you could help me with finding the log-likelihood estimation and maximizing it using mle.

[Torsten]

You mean

n = 10000;
mu = 10;
sigma = 2;
X = mu + sigma*randn(n,1) ;
phat = mle(X)

phat = 1×2

10.0047 2.0004

?

Answer 1

[Mayank Sengar]

You can use mle function to calculate the maximum likelihood estimation. Since the logarithmic function is monotonic, maximizing the likelihood is same as maximizing the log of the likelihood. Therefore, taking logarithm of the result will give you the maximum log likelihood estimation.

[BIPIN SAMUEL]

Thank you @Mayank Sengar you mean taking the logarithm of mean and variance got using 'mle' is the log-likelihood estimation.

[Torsten]

@BIPIN SAMUEL

You have data from which you assume that they follow a normal distribution with mean mu and standard deviation sigma. Now you want to estimate mu and sigma for your data. For this, you use "mle" and you get the estimated parameters mu and sigma in return. The method to determine them is to maximize the log likelihood function. In order to get the maximum value of this function, you can use "negloglik" on the result obtained from "fitdist". "fitdist" can be used alternatively to "mle" to estimate your distribution parameters.

If you still don't understand the concept, you should consult Wikipedia before asking permanently the same questions.