# Proving $E\left[\left|\frac1n\sum_{m=1}^n Y_m-Y_m1_{[0,R)}(|Y_m|)\right|\right]\leq E\left[|Y_1|, \,|Y_1|\geq R\right]$

I'm doing an exercise from a book about Markov Processes, where I need to prove that

$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E\left[|Y_1|, \,|Y_1|\geq R\right],$$

where $C_m=Y_m-Y_m1_{[0,R)}(|Y_m|),\; R>0$ and all the $Y_i, i=1,...,m$ are i.i.d random variables.

I'm not 100% sure if my approach is correct but this is what I tried:

$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\underset{tri.eq.}{\leq}\frac1n\sum_{m=1}^n E[|C_m|] \underset{i.i.d.}{=}\frac1n\cdot n E[|C_1|]=E[|C_1|].$$

Now, I was thinking could it be proved like this. Let me denote:

$$\mu(|C_1|)_a^b=\frac{1}{b-a}\int_{a}^b|C_1|\;dy=\frac{1}{b-a}\left[\int_{a}^{-R}|C_1| \;dy+\underbrace{\int_{-R}^R|C_1|\;dy}_{=0}+\int_{R}^b|C_1|\;dy\right]$$

$$=\frac{1}{b-a}\left[\int_{a}^{-R}|Y_1|\;dy+\int_{R}^b|Y_1|\;dy\right].$$

Next, I denote:

$$\mu(|Y_1|, \,|Y_1|\geq R)_a^b=\frac{1}{b-a-2R}\left[\int_{a}^{-R}|Y_1| \;dy+\int_{R}^b|Y_1| \;dy\right].$$

Now, since $\frac{1}{b-a}<\frac{1}{b-a-2R}\;$ I have $\;\mu(|C_1|)_a^b \leq \mu(|Y_1|, \,|Y_1|\geq R)_a^b$. By setting the integration domain to $a=-\infty$ and $b=\infty$ I get:

$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E[|C_1|]=\mu(|C_1|)_{-\infty}^{\infty} \leq \mu(|Y_1|, \,|Y_1|\geq R)_{-\infty}^{\infty}=E\left[|Y_1|, \,|Y_1|\geq R\right].$$

I'm a bit uncertain about this if I made something illegal in my reasoning. I'm very unsure about the last part, when I equate the expectations and my $\mu$-functions. Is this valid? I'd appreciate if I could get some comments on this and maybe even an alternative better solution. Thank you!

P.S. sorry for posting this question two times. In the first post I noticed a flaw and decided to delete, edit and re-post.

UPDATE:

To make my question more clear I will restate it:

A) Is my solution to the problem correct, and B) if not, how should I solve this problem?

Problem in reference book (page 16, problem 1.3.2 part b))

• Wait, $$C_1=Y_1\mathbf 1_{|Y_1|\geqslant R}$$ hence $$E(|C_1|)=E(|Y_1|;|Y_1|\geqslant R)$$ no?
– Did
Aug 25, 2017 at 16:47
• What is $E\left[|Y_1|, \,|Y_1|\geq R\right]$? Conditional expectation?
– Dap
Aug 25, 2017 at 18:05
• Hi @Did I also thought at first it could be what you suggested, but then it seemed to me that in $E(|C_1|)$ we are taking the mean of positive and zero values, whereas with $E(|Y_1|, |Y_1|\geq R)$ we are only taking the mean of positive numbers. Thank you! Aug 25, 2017 at 22:15
• @Dap I checked the notation from my book. Author states: $E(X,A)$ notation means: "Expected value of $X$ on event $A$". Aug 25, 2017 at 22:34
• Not sure where you are going... The very first formula in my first comment seems to settle the matter, no?
– Did
Aug 26, 2017 at 6:39

I will post my own answer here, thank you everybody for your help!

I need to prove that:

$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\underset{tri.eq.}{\leq}\frac1n\sum_{m=1}^n E[|C_m|] \underset{i.i.d.}{=}\frac1n\cdot n E[|C_1|]=E[|C_1|].$$

Now, I was thinking could it be proved like this. Let me denote:

$$\mu(|C_1|)_a^b=\frac{1}{b-a}\int_{a}^b|C_1|\;dy=\frac{1}{b-a}\left[\int_{a}^{-R}|C_1| \;dy+\underbrace{\int_{-R}^R|C_1|\;dy}_{=0}+\int_{R}^b|C_1|\;dy\right]$$

$$=\frac{1}{b-a}\left[\int_{a}^{-R}|Y_1|\;dy+\int_{R}^b|Y_1|\;dy\right].$$

Next, I denote:

$$\mu(|Y_1|, \,|Y_1|\geq R)_a^b=\frac{1}{b-a-2R}\left[\int_{a}^{-R}|Y_1| \;dy+\int_{R}^b|Y_1| \;dy\right].$$

Now, since $\frac{1}{b-a}<\frac{1}{b-a-2R}\;$ I have $\;\mu(|C_1|)_a^b \leq \mu(|Y_1|, \,|Y_1|\geq R)_a^b$. By setting the integration domain to $a=-\infty$ and $b=\infty$ I get:

$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E[|C_1|]=\mu(|C_1|)_{-\infty}^{\infty} \leq \mu(|Y_1|, \,|Y_1|\geq R)_{-\infty}^{\infty}=E\left[|Y_1|, \,|Y_1|\geq R\right].$$