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Probability with Martingales:


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  1. In the definition of $f$, is that really $z$ and not $\lceil |z| \rceil$, $\lfloor |z| \rfloor + 1$ or something?

  2. How exactly do we have the part in the $\color{red}{\text{red}}$ box?

What I tried:

$$\sum_{n=1}^{\infty} \frac{E[Y_n^2]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 ; |X| \le n]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} E\left[\frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right]$$

$$ = E\left[ \sum_{n=1}^{\infty} \frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right] \ \text{by Fubini's theorem}$$

$$ = E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right]$$

$$ = E\left[ |X|^2 \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2}\right]$$

Now observe that for integer $k$

$$\sum_{n=k}^{\infty} \frac{1}{n^2} \le \sum_{n=k}^{\infty} \frac{2}{n(n+1)} = 2 \sum_{n=k}^{\infty} \frac{1}{n(n+1)} = 2 \sum_{n=k}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2 (1/k - 1/(k+1) + 1/(k+1) - 1/(k+2) + ...) = 2(1/k)$$

Hence

$$\sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2} \le \frac{2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2} \le \frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}\right]$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\lceil |X| \rceil}\right] = E\left[2\lceil |X| \rceil\right]$$

$$\le E\left[2 (|X| + 1) \right] < \infty \ \because E[|X|] < \infty$$

Is that right?

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  • $\begingroup$ @Did I edited my question and answer. Please remove the downvotes you made, if any $\endgroup$ – BCLC Jun 21 '16 at 16:16
  • $\begingroup$ Same remark as in your other post. $\endgroup$ – Did Jun 21 '16 at 16:21
  • $\begingroup$ @Did I believe I checked cases. $\endgroup$ – BCLC Jun 21 '16 at 17:10
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$$\sum_{n=1}^{\infty} \frac{E[Y_n^2]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 ; |X| \le n]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} \frac{E[|X|^2 1_{\{|X| \le n]\}}]}{n^2}$$

$$ = \sum_{n=1}^{\infty} E\left[\frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right]$$

$$ = E\left[ \sum_{n=1}^{\infty} \frac{|X|^2 1_{\{|X| \le n]\}}}{n^2}\right] \ \text{by Fubini's theorem}$$

$$ = E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right]$$

$$ = E\left[ |X|^2 \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2}\right]$$

Now observe that for integer $k$

$$\sum_{n=k}^{\infty} \frac{1}{n^2} \le \sum_{n=k}^{\infty} \frac{2}{n(n+1)} = 2 \sum_{n=k}^{\infty} \frac{1}{n(n+1)} = 2 \sum_{n=k}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2 (1/k - 1/(k+1) + 1/(k+1) - 1/(k+2) + ...) = 2(1/k)$$

Hence

$$\sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{1}{n^2} \le \frac{2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2} \le \frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\max\{1,\lceil |X| \rceil\}}\right]$$

$$\to E\left[ \sum_{n=\max\{1,\lceil |X| \rceil\}}^{\infty} \frac{|X|^2 }{n^2}\right] \le E\left[\frac{2|X|^2}{\lceil |X| \rceil}\right] = E\left[2\lceil |X| \rceil\right]$$

$$\le E\left[2 (|X| + 1) \right] < \infty \ \because E[|X|] < \infty$$

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