2
$\begingroup$

Currently, I have a likelihood function in the following form:

$$L(\lambda|\mathbf{x})=\prod_{i=1}^{n}(x_i+(-1)^{x_i}e^{-\lambda})$$

with $x_i$ taking on values of $0$ or $1$, $\lambda>0$.

I have tried taking $log$ and differentiating the likelihood function, but have been unable of finding the maximum likelihood estimate of $\lambda$, as the result becomes

$$\frac\partial{\partial\lambda}log(L(\lambda|\mathbf{x}))=\sum_{i=1}^n\frac{-1}{1-x_ie^\lambda}$$

Can anyone give me some hints on how to do that?

$\endgroup$
5
$\begingroup$

Assume that the number of $x_i=0$ in your dataset is $n_0$ and the number of $x_i=1$ is $n_1$. Your likelihood function is: $$L(\lambda)=(e^{-\lambda})^{n_0}(1-e^{-\lambda})^{n_1}$$ and the log-likelihood is: $$l(\lambda)=-\lambda n_0+n_1 \ln (1-e^{-\lambda})$$ It is easy to differentiate this function to obtain the following MLE: $$\lambda = \ln \frac{n_0+n_1}{n_0}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy