Estimation with a log-likelihood function

[blehbleh]

Homework Statement

Hi, I am trying to solve a time series problem but I am stuck with a function I am not used to work with. I have a basic model AR(1) where e(t) has mean 0 and variance σ^2. The issue is that (e)t follows the density function attached to the post and it's causing me issues because of the gamma function with the (1/L) after. Does anyone know how I need to handle that part of the function to build my log likelihood function for (d,B,σ^2) ? L=Lambda (is a known parameter), G=gamma function, o is sigma

Homework Equations

y(t)=d +B(yt-1)+e(t) model

The Attempt at a Solution

What I have tried so far has let me with the following:

-2/3 ln 2L - ln G(1/L) - Sum(t=2, T) (yt-d-B(yt-1))^L/(2o^2)^(1/2) - (y1-d/(1-B))^L / (2o^2/(1-B^2))^(1/2)Is this way of handling the density function with the Gamma correct (basically ignoring it as it goes away after the derivatives)? If not could anyone shed some light on the way I should approach this to solve it?

Thanks a lot.

 

[mighty2000]

Hi there,

It looks like you are on the right track with your attempt at the solution. The gamma function in the density function is often used to represent the shape of the distribution, so it is important to take it into consideration when building your log likelihood function.

One possible approach would be to use the properties of the gamma function to simplify the expression. For example, you could use the identity: ln(G(1/L)) = -ln(L) + ln(G(L)) to combine the two terms. Additionally, you could use the fact that the gamma function satisfies the property: G(x+1) = xG(x), which may help simplify the expression further.

Another approach could be to use a numerical optimization method, such as maximum likelihood estimation, to find the maximum likelihood estimates of the parameters d, B, and σ^2. This would involve setting up the likelihood function, taking the derivative with respect to each parameter, and solving for the values that maximize the likelihood.

I hope this helps and good luck with your problem!