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I am reading the book of Feller called "An Introduction to Probability Theory and Its Applications, Vol I" (third edition, page 227) and am stuck at the moment he explains the notion of variance of a random variable.

In particular, Feller briefly states that the expectation and the variance are just the moments of first and second order, respectively. Additionally, Feller states that if the $r$th moment exists, then as the in equality $|X|^{r-1} \leq |X|^{r}+1$ is true, the preceding $(r-1)$st moment exists as well.

The actual text below:

The except from Feller's book

Could you help me to figure out why the inequality has this form and where it came from? Why there is exactly "$+ 1$" on the right side?

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  • $\begingroup$ Hi, welcome. You may be overthinking this. So here's a hint: why don't you draw $|X|^{r-1}$ and $|X|^r$, say for $r=2,3$. Pay particular attention to the regions $[0,1]$ and $(1, \infty)$ (or $(-\infty, -1)$ and $[-1,0]$). $\endgroup$ – Jim Jun 11 '19 at 11:10
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$$E|X|^{r-1} = E\left[|X|^{r-1} \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r-1} \mathbb{1}_{(|X|> 1)} \right]$$

$$\leq E\left[1 \times \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r-1} \mathbb{1}_{(|X|> 1)} \right]$$

$$\leq E\left[1 \times \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r} \mathbb{1}_{(|X|> 1)} \right]$$

$$\leq 1 + E|X|^r$$

Where the inequalities are justified as follows:

  1. The events are disjoint and partition the sample space

  2. On the set $\{ |X|\leq 1 \}$, clearly $|X|^{r-1}\leq 1$

  3. On the set $\{ |X|>1 \}$, clearly $|X|^{r-1} \leq |X|^r$

  4. $E[|X| \times \mathbb{1}_A ] \leq E|X|$

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  • $\begingroup$ Thank you a lot! I think I have got it. As I understand, $E|x|^{r-1}$ is less then $max(1, E|X|^r)$ taking in account that expectation is monotonic. So the preceding moment is always bound and can never be infinite. Could you point out how the notation inside expectations you have used is called (those products with ones)? My bad I've never seen it before.. $\endgroup$ – mtmrv Jun 11 '19 at 6:52
  • $\begingroup$ They're indicator functions. The function $\mathbb{1}_A$ is $1$ if the event $A$ occurs, and $0$ if $A$ does not occur. They are used to decompose the situation into different events so that they may be treated seperately. Since $A \cup A^c$ always occurs, and exactly one of $A$ or $A^c$ occurs, we have $ 1_A + 1_{A^c}=1$. Hence we can write $X = X\times 1_A + X\times 1_{A^c}$. We can then use the properties of each event $A$ and $A^c$ to prove various inequalities (eg. $|X|^{r-1}<|X|^r$ on the set $A=\{|X|>1\}$) $\endgroup$ – Xiaomi Jun 11 '19 at 7:07
  • $\begingroup$ Awesome tool, thank you very much again! $\endgroup$ – mtmrv Jun 11 '19 at 7:28

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