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I am a student, and I disagree with the solutions our TA has prepared. I am seeking verification that I am correct or explanation as to why I am wrong. It seems to be a disagreement or misunderstanding of the definition of uniform boundedness.

The question boils down to this: For what choices of $a, b \in \mathbb{R}$, is the following risk function uniformly bounded as a function of $\mu$ (here, $\sigma$ is a constant)?

$$ R(\mu ; a, b) = a^2 + 2a(b-1)\mu + (b-1)^2\mu^2 + b^2 \frac{\sigma^2}{n}$$

Clearly, we must have $b=1$ to knock out the polynomial and linear terms in $\mu$. Once $b=1$, we have:

$$ R(\mu ; a, b=1) = a^2 + \frac{\sigma^2}{n} $$

Our TA says this is uniformly bounded for all choices of $a$. By my understanding, UNIFORM boundedness means there must exist one single bound $M > 0$ such that $ \mid R(\mu ; a, b=1) \mid \le M, \forall a, \forall \mu$. That is, our choice of the bound $M$ must be good for any choice of the parameter $a$.

Clearly there is no one bound to contain this function for all $a \in \mathbb{R} $: If we choose $M = 100$, choose $a=101$ (the other term is strictly positive), and our function is not bounded by M. I say we need $a \in [-1,1]$, so that $a^2 \le 1$, and then we can bound the family of functions with a constant $1+ \frac{\sigma^2}{n}$. It seems like he is thinking only of boundedness, not uniform boundedness.

Apologies for the trivial matter; but where else to go for a dispute over definitions than to the community? Thank you very much!

PS: the actual context is decision theoretic statistics and risk.

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  • $\begingroup$ Usually, we speak about uniformly bounded set of functions, not about a u.b. function "with parameters". Also, you should specify with respect to what you want the "uniform boundedness" - a,b, or both of them? "Our TA says this is uniformly bounded for all choices of a." -- seems ok: that means, if you choose any $a$ (and now it's a constant) and $b=1$, then there are no more parameters. If the question reads "for which set $A\subseteq\mathbb{R}^2$ is $\{f(a,b,\mu):\, (a,b)\in A\}$ uniformly bounded?", then the answer can be, as you say, anything like $I\times \{1\}$ where $I$ is bounded. $\endgroup$ – Peter Franek Feb 22 '15 at 21:53
  • $\begingroup$ @PeterFranek I edited the original post. We are asked to pick $a$ and $b$ so the function is u.b. w.r.t. $\mu$. I was thinking of $a$ and $b$ as particular values from the index set $\mathbb{R}^2$. Given that interpretation, am I correct to specify $a \in [-1,1]$? Next question: is my interpretation the correct one given the question statement? If my TA is right, then what is the difference between "uniform" boundedness and just boundedness? $\endgroup$ – RMurphy Feb 22 '15 at 22:40
  • $\begingroup$ What is a "uniformly bounded function"? It doesn't make much sense for "one function". You can simply say that the function is bounded; and that is iff $b=1$ and $a$ is arbitrary. Or you can concider the set of functions $\{f(a,b,\mu): (a,b)\in A\}$ for some fixed $A\subseteq\mathbb{R}^2$. This seems to be closer to your ideas: then it works for $A=[-1,1]\times \{1\}$ as you propose. Maybe look here: en.wikipedia.org/wiki/Uniform_boundedness $\endgroup$ – Peter Franek Feb 22 '15 at 22:46
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    $\begingroup$ The sense I'm getting is that our question was poorly worded, or that statisticians understand "uniform bounded" in a different way. The link you included shows the definition I am familiar with from Real Analysis, and is the one I operated under. I agree: I don't understand what a "uniformly bounded function" is. I may need to see the professor for clarification, but at least I will have a better guess of what they think "uniform bounded" means on all our questions for the current homework assignment. Thanks. $\endgroup$ – RMurphy Feb 22 '15 at 22:50

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