If $\mathbb{E}|X_n|=O(a_n)$, where $a_n\to 0$ and $X_n$ is a sequence of positive random variables, how large is $Y_n = X_n\ln\left(\frac{1}{X_n}\right)$?

My attempt: by Markov's inequality $\mathbb{E}|X_n|=O(a_n)$ implies $X_n=O_p(a_n)$ and $Y_n = O_p(a_n)\ln\left(\frac{1}{X_n}\right)$. It remains to asses $\ln\left(\frac{1}{X_n}\right)$. For some positive sequence of random variables $Z_n=O_p(1)$

\begin{equation} \begin{aligned} X_n = a_nZ_n& \iff \ln(X_n) = \ln(a_n) + \ln(Z_n) \\ & \iff \frac{\ln\left(\frac{1}{X_n}\right)}{\ln\left(\frac{1}{a_n}\right)} = \frac{\ln(Z_n)}{\ln(a_n)} + 1 \end{aligned} \end{equation} so if we show that the right side is bounded in probability we are done, since $a_n\to 0$.

By definition $Z_n = O_p(1)$ if any $\varepsilon>0$, there exists $M<\infty$ such that $$\sup_{n\in\mathbb{N}}\Pr\left(Z_n>M\right)<\varepsilon.$$

It follows that for any $\varepsilon>0$, there exists $L=\ln(M)$ such that $$\sup_{n\in\mathbb{N}}\Pr\left(\ln Z_n>L\right)<\varepsilon,$$ so $\ln Z_n = O_p(1)$ and $$Y_n = O_p\left(a_n\ln\left(\frac{1}{a_n}\right)\right).$$

Are there flaws in my reasoning? Is there a simpler way to see this result?

My second question is whether we can say something about the order in expectation $$\mathbb{E}\left|X_n\ln\left(\frac{1}{X_n}\right)\right| = O(?)?$$

Since $$\ln(x) = \sum_{j=1}^\infty\frac{(-1)^{j+1}}{j}(x-1)^j,$$ it looks like having only the first moment in expectation is not sufficient. Is this correct?

  • $\begingroup$ I haven't gone through whuber's calculation, but I think you should be using $x log (x) \rightarrow 0$ not looking at each bit of $Y_n$ separately $\endgroup$
    – seanv507
    Commented Jul 25, 2017 at 19:17

2 Answers 2


I believe much might be revealed by contemplating series of random variables like the following:

$$X_n = \left\{\eqalign{\frac{1}{n} & \text{ with probability } 1 - \frac{1}{f(n)^2}e^{-g(n)}\\ f(n)e^{g(n)} & \text{ with probability } \frac{1}{f(n)^2}e^{-g(n)}.}\right.$$

Later we will identify suitable functions $f$ and $g$ after analyzing the roles they play in the asymptotic expectations. For now let's just assume $f(n)$ is nonzero and both diverge as $n$ grows large, with $g(n) \ge n$ for all $n\gt 0$.

By definition of expectation,

$$\eqalign{ \mathbb{E}(X_n) &= \frac{1}{n}\left(1-\frac{1}{f(n)^2}e^{-g(n)}\right) + f(n)e^g(n) \left(\frac{1}{f(n)^2}e^{-g(n)}\right) \\&= \frac{1}{f(n)} + \frac{1}{n} - \frac{1}{nf(n)^2}e^{-g(n)}.}$$


$$\mathbb{E}(X_n) = O\left(n^{-1} + f(n)^{-1}\right),$$

permitting us to take $a_n = n^{-1} + f(n)^{-1}$, which converges to $0$ as required. (Because it does so, and $x \log(1/x)\to 0$ as $x\to 0$, notice that $a_n\log(1/a_n)\to 0$.) Nevertheless the calculation of $\mathbb{E}(Y_n)$ includes a term

$$f(n)e^g(n)\log\left(\frac{1}{f(n)e^{g(n)}}\right) \times \frac{1}{f(n)^2}e^{-g(n)}=-\frac{\log(f(n))}{f(n)} - \frac{g(n)}{f(n)}\tag{1}$$

The other term, equal to

$$\frac{1}{n}\log\left(\frac{1}{1/n}\right) \times \left(1 - \frac{1}{f(n)^2}e^{-g(n)}\right) = \frac{\log{n}}{n}\left(1-\frac{1}{f(n)^2}e^{-g(n)})\right),\tag{2}$$

remains bounded (and converges to zero).

Let's suppose $f$ diverges more slowly than $g$; that is, pick $f$ for which $g(n)/f(n)$ diverges. The sum of $(1)$ and $(2)$ asymptotically is

$$\mathbb{E}(Y_n) = O\left(\frac{g(n)}{f(n)}\right) \to \infty.$$

There do exist such $f$ and $g$ satisfying all the conditions placed on them (positive, divergent, with $g(n)/f(n)$ divergent too): for instance, $g(n)=nh(n)$ (with $h(n) \ge 1$) and $f(n) = n^\epsilon$ works for any $0 \lt \epsilon \lt 1$. Consequently, $\mathbb{E}(Y_n)=O(h(n)n^{1-\epsilon})$ for all $\epsilon\gt 0$ and for all functions $h$ bounded below by $1$.

This shows there is no limit at all on the rate at which $\mathbb{E}(Y_n)$ can diverge.


Since $X_n$ are positive random variables, we do not need the absolute value. We have

$$\{\mathbb{E}X_n\}=O(a_n) \implies \lim_{n\to \infty}\frac{\mathbb{E}X_n}{a_n} < K \in \mathbb R_{++}$$

Then, since also

$$a_n \to 0 \implies \mathbb{E}X_n \to 0 \implies X_n \to 0,\;\;\; n\to \infty$$

since they are positive r.v's. So the sequence of $X$'s converges to the constant zero.

But then

$$Y_n = -X_n \ln X_n \implies \lim_{n \to \infty} Y_n =0 $$

...or maybe plims.

Am I missing something?

  • 1
    $\begingroup$ The original post seems to be asking two related but more delicate questions than the ones you address here. The first concerns the rate at which $Y_n$ converges to zero. The second concerns what happens to its expectation. $\endgroup$
    – whuber
    Commented Aug 1, 2017 at 2:40

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