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Edit: *facepalm*


I'm working through old qualifying exams, and I got the first two parts of this problem, but am stuck on the last part. Here's the set up:

Let $X$ be a random variable with pdf $f(x)=c(3+x), \text{ if } -3<x<2$ (zero elsewhere).

The part I haven't figured out is finding the almost sure limit of $\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{\{0\leq X_i \leq 1\}} X_i^3$, where $\{X_1,\ldots,X_n\}$ is a random sample from $X$, and $\mathbf{1}$ is the indicator function.

I'm trying to make use of $$\sum_{n=1}^{\infty}P\Big(|Y_n-Y|>\epsilon \Big)<\infty \implies Y_n \xrightarrow{\text{a.s.}}Y.$$

So (hoping that it converges to zero), by Markov's inequality, I wrote $$P\Bigg(\Bigg|\frac{1}{n}\sum_{i=1}^{n}\mathbf{1}_{\{0\leq X_i \leq 1\}} X_i^3\Bigg|>\epsilon \Bigg) \leq \frac{\mathbb{E}\Big[\Big|\frac{1}{n}\sum_{i=1}^{n}\mathbf{1}_{\{0\leq X_i \leq 1\}} X_i^3\Big|\Big]}{\epsilon}.$$ Now, I know that $\mathbb{E}[\mathbf{1}_A]=P(A)$, and that if $U$ and $V$ are independent, then $\mathbb{E}[UV]=\mathbb{E}[U]\mathbb{E}[V]$, but I don't know if we have independence here (between indicators and $X_i$'s). Can I split up the expectation term into something easily computable? Or, maybe it is easily computable and I'm just not seeing it?

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$Y_i = 1_{\{0\leq X_i \leq 1\}} X_i^3$ are iid. Use the SLLN. –  cardinal Jan 6 '13 at 4:31
    
Got it. Thanks! –  pedrosuavo Jan 6 '13 at 4:39
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From the viewpoint of keeping the Q&A database in order, it would be better if either @cardinal or the OP added a short answer to the question. –  user53153 Jan 6 '13 at 6:18
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1 Answer

up vote 3 down vote accepted

Thanks to cardinal for pointing this out.

Since $Y_i:= \mathbb{1}_{\{0\leq X_i \leq 1\}}X_i^3$ are iid, by the strong law of large numbers, we have that $$\frac{1}{n}\sum_{i=1}^{n}Y_i \xrightarrow{\text{a.s.}} \mathbb{E}[Y_i]=\frac{19}{250}.$$

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