# Thread: Monotone likelihood ratio

1. ## Monotone likelihood ratio

I am trying to show that the monotone likelihood ratio for a uniform distribution [0,theta] is the maximum value (X1,...Xn). I know how to compute the monotone likelihood ratio by using fn(x given theta1)/fn(x given theta2), but then how do i show this is the maximum (X1,...Xn)?? Thanks

2. do you know that the likelihood function is

$\displaystyle {1\over \theta^n} I_{X_{(1)}>0} I_{X_{(n)}<\theta}$

3. Yes so the MLR becomes (theta2^n)/(theta1^n). But what is the step that shows that that is the maximum (X1,...Xn)??

4. I'm not quite sure what your two theta's are
are you testing theta1 vs theta2?

5. Monotone likelihood ratio - Wikipedia, the free encyclopedia

It is like the formula on this page only our class uses theta1 as f(x) and theta2 as g(x)

6. Suppose $\displaystyle \theta_2>\theta_1$. Then the ratio becomes $\displaystyle \frac{\theta_1^n}{\theta_2^n}\frac{I(X_{(n)},\thet a_2)}{I(X_{(n)},\theta_1)}$ [where $\displaystyle I(a,b)=1 ,$ if $\displaystyle a\leq b$ and 0 otherwise.]

$\displaystyle =\frac{\theta_1^n}{\theta_2^n}\frac{1}{1}$ , if $\displaystyle 0<X_{(n)}<\theta_1$
and $\displaystyle \frac{\theta_1^n}{\theta_2^n}\frac{1}{0}$ if $\displaystyle \theta_1<X_{(n)}<\infty$

that is $\displaystyle = \frac{\theta_1^n}{\theta_2^n}$ ,if $\displaystyle 0<X_{(n)}<\theta_1$
and $\displaystyle = \infty$ if $\displaystyle \theta_1<X_{(n)}<\infty$

So as $\displaystyle X_{(n)}$ increases, the ratio also increases. Hence MLR in $\displaystyle X_{(n)}$.

7. Thank you, but how do i show the statistic is the maximum(X1,...Xn)

8. Originally Posted by andyaddition
Thank you, but how do i show the statistic is the maximum(X1,...Xn)
How do you write the pdf (or the joint pdf) of a $\displaystyle U(0,\theta)$ distribution?

9. 1/theta^n ??

10. You are missing something. The joint pdf is given by $\displaystyle f(x_1,x_2,....,x_n)=\frac{1}{\theta^n},$ if $\displaystyle 0<X_i<\theta$ $\displaystyle \forall i=1(1)n$. Since all $\displaystyle x_i$ are less than $\displaystyle \theta$, it is enough to ensure that max (X1,X2,....,Xn), which is denoted by $\displaystyle X_{(n)}$, is always less than $\displaystyle \theta$. Hence the joint pdf becomes $\displaystyle f(x_1,x_2,....,x_n)=\frac{1}{\theta^n},$ if $\displaystyle 0<X_{(n)}<\theta$. Now you move to the ratio.