2

Let $W$ be defined as $W=(\sum_{i=1}^{n}V_{i})^{2}$ and $\left|V_{i}\right|\leq 1$. Determine that $\mathbb{E}[\left|W\right|]<\infty$.

What I have so far updated version:

$$\mathbb{E}\left[\left|W\right|\right]=\mathbb{E}\left[\left|\left(\sum_{i=1}^{n} V_{i}\right)^{2}\right|\right]=\mathbb{E}\left[\left|\sum_{i=1}^{n} V_{i}\cdot\sum_{i=1}^{n} V_{i}\right|\right]\leq \mathbb{E}\left[\sum_{i=1}^{n}\left| V_{i}\right|\cdot\sum_{i=1}^{n}\left| V_{i}\right|\right] = \\ \sum_{i,j}^{n}\mathbb{E}\left[\left| V_{i}\right|\cdot\left| V_{j}\right|\right] $$

Note in the 3rd step I used the Cauchy-Schwarz and in the 4th step linearity of expectation. But now I am stuck, what step do I take next?

Since I believe you can also write

$$\mathbb{E}\left[\left|\left(\sum_{i=1}^{n} V_{i}\right)^{2}\right|\right]=\mathbb{E}\left[\left|\sum_{i=1}^{n} V_{i}^{2}+2\sum_{i<j}^{n} V_{i}V_{j}\right|\right].$$

This confuses me.

Lech121
  • 333
  • 1
  • 8
  • Well, $\Bbb E$ is indeed linear, but not multiplicative, so you can only conclude that $$\Bbb E\left[\sum_i|V_i|\cdot\sum_j|V_j|\right] = \sum_{i,j}\Bbb E\left[|V_i|\cdot|V_j|\right],.$$ – Berci Apr 07 '13 at 22:00
  • Thank you I had a feeling I was doing something that wasn't allowed but I didn't see it. However I don't see how I can conclude $\sum_{i,j}\mathbb{E}\left[\left|V_{i}\right|\cdot\left|V_{j}\right|\right]< \infty$. Am I on a wrong path and should take a different approach? – Lech121 Apr 07 '13 at 22:06
  • If $|V_i| \leq 1$ for all $i$, it seems to me that $-n \leq \sum_{i=1}^n V_i \leq n$, no? – Dilip Sarwate Apr 07 '13 at 22:10
  • Is it really that simple than it would follow immediately right? I would simply have $\mathbb{E}[\left|(\sum_{i=1}^{n}V_{i})^{2}\right|]\leq \mathbb{E}[\left| n^{2}\right|]=n^{2}<\infty$ – Lech121 Apr 07 '13 at 22:18
  • Well, if you're certain that the random variable $|V_i|\cdot|V_j|$ has expected value, then, it's $\le 1$, and there you have again $n^2$. – Berci Apr 07 '13 at 22:19
  • If it is $\leq 1$, wouldn't it be $n$ and not $n^{2}$? – Lech121 Apr 07 '13 at 22:26
  • The argument is strikingly similar to the one used here, no? – Did Apr 07 '13 at 23:16

1 Answers1

0
  • A bounded random variable has a finite expectation.
  • We have $|W|\leqslant n^2$.
Davide Giraudo
  • 172,925