Recent content by Alex_Doge

Thanks a lot for the help. I solved it now. The red graph is the probability. The other two are the bounds of the floor function. The level is then 0.05. The plots helped me understand it and gave the hint sense.

That means I have this: \sum_{k=1}^{\lfloor3/q\rfloor} q (1-q)^{k-1} =1-(1-q)^{\lfloor3/q\rfloor}\geq \alpha and need to find alpha. Can I look at 1-(1-q)^{\lfloor3/q\rfloor} and see where it's minimum is, for q \in (0,1) , and then define alpha to be just lower? What do you mean by "large"...

I calculated that Probability P(X < 3/q) = \sum_{k=1}^{\lfloor 3/q \rfloor} q (1-q)^{k-1} =q\sum_{k=1}^{\lfloor3/q\rfloor}(1-q)^{k-1}=(1-‌q)^{\lfloor3/q\rfloor} How do I calculate this lower bound? Like this: (1-‌q)^{\lfloor3/q\rfloor}=1 and then solve for q? Or do I minimize and maximize to...

So because q lies in C with the probability 1, C is a confidence interval? For k \geq 4 the probability is not 1 for large k. Is that a problem?

Yea it looks better now: But how do I continue from here?

Oh sorry, I'm getting tired, been stuck on this problem all day now. That's a strange plot. What does this mean then? :) Is it something similar to a delta function, or have I made a mistake plotting it?

First of all thanks for the detailed answer. Sorry I didn't know it would turn out to be a calculus problem. The integral gives P(0 < q < 3/k) = \int_0^{3/k} k(k+1) q (1-q)^{k-1} \, dq=\frac{12\cdot3^k k^{-k} (-1)^k(1-1/3k)^{k+1}+k-3}{k-3} It converges towards 0.8 if I'm not mistaken What...

Hello Physicsforum Homework Statement I have a problem proving this: Given C(x)=[0, 3/x] for all x\in\chi, with \chi=\Omega being the sample space and P_q=Geom(q) being the geometric distribution. I have to show that C(x) is a confidence Interval for q but I don't know how to get started...