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Hello Guys!

How do you interpret the log likelihood of -189.95?

is this good or bad?

NOTE: I am using probit model.

Thanks guys!.

The log likelihood is used for comparing two models. A single value has no meaning.

Thanks Startz for the reply,

I only have one probit regression, but just seeing the value itself. how would you interpret it?. I need for my thesis.

Thanks.

The log likelihood is used for comparing two models. A single value has no meaning.

Thanks Gareth,

sorry for the misunderstanding there. I know now what to do.

Thanks Gareth and Startz!

hello Eviws users:
I am PhD student in agricultural Economics. I want to analyse the yield risk of crops. I am using Normal, beta, Johanson SU and SB for the yield risk analysis. To analyse the distribution function, I am using maximum likelihood estimation or say log likelihood. For this I am using Eviews 6.0 version and I start facing problem from the begining; I am giving equation besed on Eview guide.
The equation I gave is
@logl logl1
res = yield - c(1) - c(2)*t- c(3)*t^2
Sigma= c(4) + C(5)*t)
var = sigma^2
logl1 = log(@dnorm(res/@sqrt(var))) - log(var)/2

't' is time, it is time series data. yield is crop yield. The mean and variance equations are drawn from one published paper. When I give run command, the error message ups up as "missing value in @logl series at current coefficients at observation 1". I have know Idea where problem is and how to solve it. Thank you for ur suggestion.