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May 8 at 17:43 comment added Ziv @MathRevenge Is the sum the same as something like $P(I_1) + \sum_{n = 1} P(I_n)$? (Write $n = \sum_{k=1}^n 1$, then use Tonelli's theorem to swap the sums.) That would imply you only need to bound $P(I_n)$ from below.
Apr 28 at 16:18 comment added Iosif Pinelis This goal should be, and should have been, stated explicitly in your post.
Apr 28 at 16:09 history edited MathRevenge CC BY-SA 4.0
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Apr 28 at 16:04 comment added MathRevenge Unfortunately my goal is to show that, in my case, the series $\sum_{n=2}^{\infty}n(P(I_{n-1})-P(I_{n}))$ is divergent.. and in order to prove it I thought to bound that expression with a function $g(n)$ such that the series of $ng(n)$ is divergent. (If not, I otherwise have to show that the first series is convergent instead)
Apr 28 at 14:37 comment added Iosif Pinelis Doesn't then your question reduce to this: show that $P(I_{n-1})-P(I_n)>0$ for all large enough $n$?
Apr 28 at 14:22 comment added MathRevenge In the sense that in my case $X_i = \cos{\theta_i}$ where $\theta_i$ is uniformly distributed in $[0,2\pi]$. Maybe it could be useful..
Apr 28 at 13:55 comment added Iosif Pinelis You can provide the distribution of $X_1$? In what sense?
Apr 28 at 13:48 comment added MathRevenge I mean, the distribution of $X_1$. I was not precise, sorry.
Apr 28 at 13:31 comment added Iosif Pinelis You can provide what?
Apr 28 at 13:23 comment added MathRevenge Not necessarily! If needed I can provide it. @IosifPinelis
Apr 28 at 13:20 comment added Iosif Pinelis Do you want $g$ to be independent of the distribution of $X_1$?
Apr 28 at 13:08 history edited MathRevenge CC BY-SA 4.0
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Apr 28 at 13:01 history asked MathRevenge CC BY-SA 4.0