# Thread: mixture of normals inequality

1. ## mixture of normals inequality

Let C be a convex and compact set in R^n, and let T = {theta_1,...,theta_k} be a set of points not in C such that T and C are hyperplane separated. Let {c_1,...,c_k} be, respectively, the closest points in C to {theta_1,...,theta_k}. WLOG, suppose ||c_1 - theta_1|| >= ||c_i - theta_i|| for all i <= k.

Suppose we are interested in testing

H_1: X comes from an equally-weighted mixture of normals with covariance = I_n and means = T

against

H_0: X comes from an equally-weighted mixture of normals with covariance = I_n and means = {c_1,...,c_k}

Let the power of this level-alpha likelihood ratio test be P

Let the power of the level-alpha likelihood ratio test of H_1: X ~ N(theta_1,I_n) against H_0: X ~ N(c_1,I_n) be Q.

Is it possible to show that P <= Q?

2. ## Re: mixture of normals inequality

Originally Posted by student?
Let C be a convex ...

...Is it possible to show that P <= Q?
Yes, I think so. In fact, I suspect you can also show it if C is quasi-convex...because your're using a weak inequality.

3. ## Re: mixture of normals inequality

Originally Posted by Dragan
Yes, I think so. In fact, I suspect you can also show it if C is quasi-convex...because your're using a weak inequality.
ah..yes, actually just notice that the test has less power than the one in which the mean is known to be one of {c_1,...,c_k}

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