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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!).

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2 Answers 2

up vote 1 down vote accepted

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 $$L = \exp\left(x\frac{(\mu_1-\mu_0) }{\sigma^2}\right)\exp\left(\frac{\mu_0^2-\mu_1^2}{2\sigma^2}\right).$$

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Yes I am interested in those many others. I only know Gausian. –  Seyhmus Güngören Oct 21 '12 at 12:30

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?

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No( it is for $\Omega=\mathbb{R^+}$ –  Seyhmus Güngören Oct 21 '12 at 17:30
    
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. –  Seyhmus Güngören Oct 21 '12 at 18:07
    
Maybe you need to edit the question to include these restrictions. –  Dilip Sarwate Oct 21 '12 at 18:10
    
It is already there. $\Omega=\mathbb{R}$. –  Seyhmus Güngören Oct 21 '12 at 18:20
    
At any case thank you very much for posting an answer, –  Seyhmus Güngören Oct 21 '12 at 18:23

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