2
$\begingroup$

Let $X_1,X_2,\dots ,X_n$ be a random sample from a distribution with pdf $f(x,\theta)$. Find a level $\alpha$ most powerful test of $H:\theta=\theta_0$ against $K:\theta = \theta_1$ when $$f(x,\theta)=\theta x^2I_{(\theta, \infty)}(x),\theta_0 \neq \theta_1$$

I consider $$\frac{p_1(x)}{p_0(x)}=\dfrac{\theta_1^n(\Pi x_i)^2 \Pi(I_{(\theta_1, \infty)}(x_i))}{\theta_0^n(\Pi x_i)^2 \Pi(I_{(\theta_0, \infty)}(x_i))}\\=(\theta_1/\theta_0)^n\dfrac{\Pi(I_{(\theta_1, \infty)}(x_i))}{ \Pi(I_{(\theta_0, \infty)}(x_i))}$$ Then I need to consider $\frac{p_1(x)}{p_0(x)}>k$ for a Neyman-Pearson test but I am stuck. Please help.

$\endgroup$
1
  • 2
    $\begingroup$ Fix your density. Did you try drawing the likelihood? It might clarify some issues for you $\endgroup$
    – Glen_b
    Commented Mar 31, 2014 at 23:20

1 Answer 1

2
$\begingroup$

First of all, $f(x;\theta) = \theta x^2 I_{(\theta, \infty)}(x)$ is not a valid probability density function (it doesn't integrate to unity). I suspect that you actually meant $f(x;\theta) = \theta x^{-2} I_{(\theta, \infty)}(x)$ which does integrate to unity for any $\theta \in \mathbb{R}$. So I will assume this from now on (and assuming $\theta_0 < \theta_1$).

To perform Neymann-Pearson, as you stated, you need to consider the how the ratio $\frac{p_1(X)}{p_0(X)}$ varies as the random variable $X$ varies, where $X$ follows the null hypothesis, in this case, we have that the $X_i \sim f(x, \theta_0)$. The intuition is that we reject $H_0$ if a realisation of the random ratio is large (it's kind of like a probabilistic 'proof' by contradiction - we know $H_0$ to be true, but find $H_1$ appears more likely than even $H_0$, which can't possibly have occurred because $H_0$ is true - this is a contradiction, therefore, $H_1$ must be true...).

Note that we only care how the ratio behaves as a function of the $x$'s, and note that $\theta_0$ and $\theta_1$ are fixed constants. Let me take off from where you left off:

$\begin{align} \frac{p_1(x)}{p_0(x)} &= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{\prod_i I_{(\theta_1, \infty)}(x)}{\prod_i I_{(\theta_0, \infty)}(x)}\\ &= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{ \mathbb{1}_{x_i \geq \theta_1, \forall i}}{\mathbb{1}_{x_i \geq \theta_0, \forall i}}\\ &= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{\mathbb{1}_{\text{min}(x_i) \geq \theta_1}}{\mathbb{1}_{\text{min}(x_i) \geq \theta_0}}\\ \end{align}$

Note that the indicator function in the denominator is always unity since the $x_i$'s obey $H_0$. Therefore, looking at the ratio $\frac{p_1(x)}{p_0(x)}$, we see that it is an increasing function of $\text{min}(x_i)$. Thus, our rejection criterion is equivalent to rejecting $H_0$ if $\text{min}(X_i)$ is large. I will leave it at that for you to complete...

$\endgroup$
4
  • 1
    $\begingroup$ Please check the guidelines for answering self-study questions (scroll down to "Answering...". If possible, could you try to guide/hint more on future questions (at least until the student has clearly figured it out -- or at least several days have passed -- at which point a fuller answer is good). You've left a little to do, which is commendable, but I think I'd normally lean a little harder toward leaving more for the student. $\endgroup$
    – Glen_b
    Commented Mar 31, 2014 at 23:22
  • 1
    $\begingroup$ To clarify, I think the first 50-70% of what you wrote is great, but I think the point about the minimum is the central insight that (to me at least) appeared to be the aim for the student to discover for themselves; it would probably be better to guide the student toward that insight if possible rather than state outright first up. Anyway, +1 for what's a very good answer. $\endgroup$
    – Glen_b
    Commented Mar 31, 2014 at 23:31
  • $\begingroup$ I see - sorry for not adhering to the guidelines - i'm quite new on the scene. You are definitely right about leaving more for the student :) $\endgroup$
    – queenbee
    Commented Mar 31, 2014 at 23:43
  • $\begingroup$ queenbee - if you write great answers like that, I hope I get to upvote many more of them. I wish mine were as clear and helpful. $\endgroup$
    – Glen_b
    Commented Apr 1, 2014 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.