0
$\begingroup$

One addition to the title: For $(\Omega,\cal{F},P)$, $\Omega=\mathbb{R}$.

Thanks in advance.

I hope there are some others than only Gaussian (with same variance!).

2 Answers 2

1

There are many examples, but for example take two normal distributions with the same variance $\sigma^2$ and means $\mu_0 \lt \mu_1$ where $$L = \exp\left(x\frac{(\mu_1-\mu_0) }{\sigma^2}\right)\exp\left(\frac{\mu_0^2-\mu_1^2}{2\sigma^2}\right).$$

  • 0
    Yes I am interested in those many others. I only know Gausian.2012-10-21
1

If the conditional densities $f_1$ and $f_0$ are exponential densities with parameters $\lambda_1$ and $\lambda_0$ respectively where $\lambda_0 > \lambda_1$, then $$L(x) = \frac{f_1(x)}{f_0(x)} = \frac{\lambda_1\exp(-\lambda_1x)}{\lambda_0\exp(-\lambda_0x} = \frac{\lambda_1}{\lambda_0}\exp((\lambda_0-\lambda_1)x)$$ is an increasing function of $x$ on $[0,\infty)$ that increases from $\frac{\lambda_1}{\lambda_0} < 1$ to $\infty$. Does that work for you?

  • 0
    No( it is for $\Omega=\mathbb{R^+}$2012-10-21
  • 0
    I think it is of no use. The densities should be on the real axis not on the positive side. $L(x)$ is always positive definite I agree. I need to have $f_0$ and $f_1$ defined on real axis and the likelihood should be invertible.2012-10-21
  • 0
    Maybe you need to edit the question to include these restrictions.2012-10-21
  • 0
    It is already there. $\Omega=\mathbb{R}$.2012-10-21
  • 0
    At any case thank you very much for posting an answer,2012-10-21