Can I bound $P[R > x + \epsilon]$ independently of R?

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I have this probability distribution: $P[\Theta < \varphi] = \frac{\varphi}{\pi}$ for $\phi \in [0,\pi]$.

Now I have $n$ samples of $D = R\Theta$ i.i.d. ($R>0$) and I want to estimate $R$ as $\frac{\max_i\{D_i\}}{\pi}$. I am trying to see if I can find some sort of confidence interval such that I can say:

$R \in [\frac{\max_i\{D_i\}}{\pi}, \frac{\max_i\{D_i\}}{\pi} + \epsilon]$ with probability at least $1 -\delta$. So what I am saying is that I want $\epsilon$ such that

$P[R > \frac{\max_i\{D_i\}}{\pi} + \epsilon] \leq \delta$

But now if I do the calculations I get:

$P[R > \frac{\max_i\{D_i\}}{\pi} + \epsilon] = (1-\frac{\epsilon}{R})^n \leq \delta$

so my $\epsilon$ will depend on $R$ : $\epsilon \geq R(1-\sqrt[n]{\delta})$

Am I doing something wrong? Is there no way to give an absolute error term? Should I use some Markov/Chebyshev inequality?

I should mention, I haven't done statistical inference in a long time so I might be confusing something and the answer might be much more simple.