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Let $\left(X_n\right)_{n\geq1}$ be a sequence of random variables defined on a given fixed probability space $\left(\Omega\text{, }\mathcal{F}\text{, }\mathbb{P}\right)$.
Consider that $\mathbb{P}\left(\vert X_n-X\vert>\varepsilon\right)=\mathbb{E}\{1_{\{\vert X_n-X\vert >\varepsilon\}}\}$. Then, consider the event $\{\omega:\vert X_n-X\vert>\varepsilon\}$. One is allowed to state that $$\{\omega:\vert X_n-X\vert>\varepsilon\}\Leftrightarrow\Bigg\{\omega:\frac{\vert X_n-X\vert^p}{\varepsilon^p}>1\Bigg\}$$ Henceforth $\dfrac{\vert X_n-X\vert^p}{\varepsilon^p}>1$ on the event $\{\omega:\vert X_n-X\vert>\varepsilon\}= \{\vert X_n-X\vert>\varepsilon\}$. So, I think I could state that the below inequality holds true $$\mathbb{P}\left(\vert X_n-X\vert>\varepsilon\right)=\mathbb{E}\{1_{\{\vert X_n-X\vert >\varepsilon\}}\}<\mathbb{E}\Bigg\{\frac{\vert X_n-X\vert^p}{\varepsilon^p}1_{\{\vert X_n-X\vert >\varepsilon\}}\Bigg\}$$ However, on my book I read $$\mathbb{P}\left(\vert X_n-X\vert>\varepsilon\right)=\mathbb{E}\{1_{\{\vert X_n-X\vert >\varepsilon\}}\}\leq\mathbb{E}\Bigg\{\frac{\vert X_n-X\vert^p}{\varepsilon^p}1_{\{\vert X_n-X\vert >\varepsilon\}}\Bigg\}$$ So, my question is: why is equal sign $(=)$ included in the inequality on my book? I cannot figure out why one could be allowed to include it, given the above reasoning.

Strictly_increasing
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  • If $a < b$ then $a \le b$ so the book only looses a tiny bit of information. Also there seem to be cases where $<$ doesn't hold. What if $|X_n-X|>\epsilon$ has measure 0 – Gribouillis Jun 06 '20 at 08:55
  • Yeah, but in the end this turns out to be essential, since the conclusion is that $\lim\limits_{n\to\infty}\mathbb{P}\left(|X_n-X|>\varepsilon\right)\leq0$ which implies that $\lim\limits_{n\to\infty}\mathbb{P}\left(|X_n-X|>\varepsilon\right)=0$. If instead one ends up with $\lim\limits_{n\to\infty}\mathbb{P}\left(|X_n-X|>\varepsilon\right)<0$, this is clearly not possible since probability is non-negative. So, I think it makes a big difference, doesn'it? @Gribouillis – Strictly_increasing Jun 06 '20 at 08:58
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    When you take a limit, you will lose the strict inequality, it will have to become non-strict. e.g. $-1/n < 0$. Then take limit, you have to conclude that $\lim_{n \to \infty} (-1/n) \leq 0$ – SBK Jun 06 '20 at 09:00
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    If you take a limit a strict inequality becomes large. No the question here is does a strictly positive function have a positive integral. This is not true on a set of measure $0$ – Gribouillis Jun 06 '20 at 09:00
  • I understand your explanation "When you take a limit, you will lose the strict inequality" intuitively, especially thanks to your example. However, could I find some rigorous result telling me so? Some reference? @T_M – Strictly_increasing Jun 06 '20 at 09:09
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    As pointed out in the other answer, it seems like the limit was not your real issue. The issue is that actually you really need the non-strict inequality to be correct. If $f$ is integrable and $f > 1$ on the set $S$ then $\int f \mathbf{1}_S d\mu \geq \mu(S)$. I cannot put $>$ because we might have $\mu(S) = 0$, in which case both sides are zero. – SBK Jun 06 '20 at 09:12
  • Perfect, thank you again @T_M – Strictly_increasing Jun 06 '20 at 09:14

3 Answers3

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I was surprised how often I had this question when teaching analysis.

The notation $x < y$ means ''$x$ is less than $y$''.

The notation $x \leq y$ means ''$x$ is less than or equal to $y$'',

i.e. $x$ is less than $y$ or $x$ is equal to $y$.

SBK
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  • Yes, of course, I know this. The doubt arose since the conclusion on the book is that $\lim\limits_{n\to\infty}P(|X_n−X|>\varepsilon)\leq0$ which implies that $\lim\limits_{n\to\infty}P(|X_n−X|>\varepsilon)=0$. If instead one ends up with $\lim\limits_{n\to\infty}P(|X_n−X|>\varepsilon)<0$, this is clearly not possible since probability is non-negative. However, above I have been told by other users that when taking limits, strict inequality 'gets larger' – Strictly_increasing Jun 06 '20 at 09:06
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    OK sure. But you didn't ask about the limit in the question, you asked why one is allowed to include a non strict inequality. – SBK Jun 06 '20 at 09:09
  • Answer you're looking for has been provided. – SBK Jun 06 '20 at 09:10
  • Yes, of course, you are right, sorry. And thank you a lot – Strictly_increasing Jun 06 '20 at 09:11
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The strict inequality doesn't hold when $\mathbb{P}(|X_n - X|>\varepsilon) = 0$ because the integral of a strictly positive function on a set of measure $0$ is $0$.

Gribouillis
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Your question seems to be when does the expectation preserve strict inequality. Here is a sufficient condition: When does the integral preserve strict inequalities? This might be useful for future reference.

As you can see, this condition is not satisfied here. One of the possible problems is that $$\mathbb{P}(|X_n-X| > \epsilon) = 0$$ is possible (for example, when $X_n = X$ a.s.) and then both sides are $0$, so we have equality. Thus strict inequality cannot hold in general

J. De Ro
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