I want to derive a lower bound for $E\left[\left(\frac{X}{k-X}\right)^{i}\right] $ with $X \sim Bin_{(k-1),p}$ and $ k \in \mathbb{N} $. So far I could prove that \begin{equation} E\left[\frac{X}{k-X}\right] = \frac{p}{1-p}-\frac{p^k}{1-p} \end{equation} by law of unconcious statistician and the identity $\binom{k-1}{l}=\frac{k-l}{l}\binom{k-1}{l-1}$. Now, I use as lower bound the fact that: \begin{equation} \frac{1}{k^{i}}E\left[X^{i}\right]\leq E\left[\left(\frac{X}{k-X}\right)^{i}\right] \leq E\left[X^{i}\right] \end{equation} Then: \begin{equation} E\left[X^{i}\right] = \sum_{l=0}^{k-1}l^{i}\binom{k-1}{l}p^l(1-p)^{k-1-l} \\= \sum_{l=1}^{k-1}l^{i}\binom{k-1}{l}p^l(1-p)^{k-1-l}\geq \sum_{l=1}^{k-1}\binom{k-1}{l}p^l(1-p)^{k-1-l} = 1- (1-p)^{k-1} \end{equation} So this is my lower bound. For the upper bound I use Jensen's inequality on a concave function: \begin{equation} E\left[\left(\frac{X}{k-X}\right)^{i}\right]\leq \left(E\left[\frac{X}{k-X}\right]\right)^{i} = \left(\frac{p}{1-p}-\frac{p^k}{1-p}\right)^{i} \end{equation} Thus, I have: \begin{equation} \frac{1}{k^{i}}\left(1- (1-p)^{k-1}\right) < E\left[\left(\frac{X}{k-X}\right)^{i}\right] < \left(\frac{p}{1-p}-\frac{p^k}{1-p}\right)^{i} \end{equation} Do you have anything better than that?

  • $\begingroup$ I was referring to it with a functional analysis tab as the $\frac{1}{j}$ moment refers to the metric in a $l_p$ space with $p<1$ (the random variable is non-negative). Or am I wrong? $\endgroup$
    – qwert
    Oct 21, 2020 at 13:54
  • $\begingroup$ no, because the function is concave (i < 1) $\endgroup$
    – qwert
    Oct 21, 2020 at 14:28

1 Answer 1


Let $r:=i\in(0,1)$. Your upper bound, based on Jensen's inequality, seems fine.

The lower bound can be greatly improved, at least for large enough $k$. Indeed, the expectation to bound is $$Eg\Big(\frac{1+Y}k\Big),$$ where $g(u):=(1/u-1)^r$ for $u\in(0,1]$ and $Y:=k-1-X\sim\text{Binomial}(k-1,q)$, with $q:=1-p$. For $u\in(0,1)$, $g''(u)$ is of the same sign as $1+r-2u$, so that $g$ is convex on $(0,\frac{1+r}2]$ and concave $[\frac{1+r}2,1]$. Also, $g(1)=h(1)$, $g(r)=h(r)$, and $g'(r)=h'(r)$, where $h(u):=(1-u)/a_r$ and $$a_r:=r^r(1-r)^{1-r}.$$ So, $g\ge h$ on $(0,1]$ and hence $$Eg\Big(\frac{1+Y}k\Big)\ge Eh\Big(\frac{1+Y}k\Big) =\Big(1-\frac{1+EY}k\Big)/a_r=\Big(1-\frac{1+(k-1)q}k\Big)/a_r =\frac1{a_r}\,\frac{k-1}k\,p,$$ so that $\dfrac1{a_r}\,\dfrac{k-1}k\,p$ is a lower bound on the expectation in question. This bound is much greater than your lower bound, $\frac{1}{k^{i}}\left(1- (1-p)^{k-1}\right)$, at least for large enough $k$. (It is easy to see that $1/2<a_r<1$ for $r\in(0,1)$.)

Here are the graphs of $g$ (solid) and $h$ (dotted) for $r=1/4$:

enter image description here

  • $\begingroup$ $h$ works excellent for small $j$, like in the graphical solution $j = 4$. In my case, refers to a sample size, which might be up to a few thousand. Then, $1/k^{i}(1-(1-p)^{k-1})$ performs even better, but it's not usable for me, as I'm particularly interested in the cases $p>5$ (or $p<5$ in your example), when the expectation is greater than 1. $\endgroup$
    – qwert
    Oct 21, 2020 at 20:41
  • $\begingroup$ @qwert : You should specify in advance (rather than after getting an answer) all the conditions that you have, so that people willing to help you would not waste their time and effort. In your case, the parameters are $j,k,p$. What is your full set of conditions on them? What kind of bounds do you want? Why is your (or my?) bound not usable for you? What do you mean by $p>5$? -- $p$ is a probability. $\endgroup$ Oct 21, 2020 at 23:43
  • $\begingroup$ I meant $p>0.5$. Both bounds are really very far away when $j>>1$. I use $\left(\frac{X}{k-X}\right)^{i}$ inside on an estimator and it is likely that most of the cases have $p>0.5$. I specified the condition well, in my opinion. Particularly with the definition $i=\frac{1}{j}, j\in \mathbb{N}$. I was not looking for a lower bound that holds for $j \in \{1,2,3,4\}$ $\endgroup$
    – qwert
    Oct 22, 2020 at 8:47
  • $\begingroup$ The lower hold for all natural $j$. Your conditions $p>0.5$ and $j>>1$ were not (and still are not) mentioned in your post, which asked just for "anything better". Alnyway, what do you mean by "really very far away" and what would be close enough? $\endgroup$ Oct 22, 2020 at 13:47
  • $\begingroup$ Ja. I didn't know the $p > 0.5$ "condition" at the beginning. But I needed a lower bound that produces values greater than 1 - iff $p > 0.5$ as the true value does so. Meanwhile, I have decided to plug in $\frac{l}{k-l}$ into the expectation via unconscious statistician - that's why I didn't update the question. I still think that if I ask for "anything better" then the solution should do so independent of the parameters. In case of $j = 8$ and $p \leq 0.3$, $ \frac{p\left(k-1\right)}{k i^{i}\left(1-i\right)^{1-i}} > \frac{1}{k^{i}}\left(1-(1-p)^{k-1}\right)$ $\endgroup$
    – qwert
    Oct 22, 2020 at 17:30

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