Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!).

share|cite|improve this question
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 $$L = \exp\left(x\frac{(\mu_1-\mu_0) }{\sigma^2}\right)\exp\left(\frac{\mu_0^2-\mu_1^2}{2\sigma^2}\right).$$

share|cite|improve this answer
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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.