# Lévy measure and compensation

In the Lévy-Itô decomposition it's necessary to compensate small jumps. That's clear. The small jumps are perhaps non-summable. But why are the jumps quared summalbe?

In the "ordinary" proofs of the Lévy Khintchine Formula or the Lévy-Itô Decomposition i can't get the point where we exactly use THIS and where the proof fails if we would not use this. I think there is a connection in both proofs, because the integrand in the L-K formula is near the origin something to the power of 2.

Are there basic results where it's proven that the x^2 integrated with respect to a Lévy measure is finite without using the Lévy-Khintchine Formula or the L-I decomposition?

-

I'm not sure what last part of the question means, but there are martingale ways of proving that the jumps must be square summable.  If you only have small jumps, you can localize to get an $\mathbb L ^2$ martingale, and large jumps can be put in the bounded variation term, in which case you start with an $\mathbb L^2$ semi martingale with bounded jumps. – mike Jun 25 '12 at 12:07
But using this martingale ways it turns out that, substracting the big jumps, the rest is a $\mathbb{L}^p$ martingale for every $p \geq 1$. So also for $p=1$ and hence it should be also simple summable, but this don't work... – kuemmelsche Jun 25 '12 at 12:52
I should add that it's not alone needed to substract the big jumps $W_t=X_t-\sum_{s\leq t}\Delta X_t \mathbb{I}_{\left\{ \left|\Delta X_t \right| \geq 1 \right\}}$ but also compensate by $W_t - \mathbb{E}W_t$ to get a $\mathbb{L}^p$ martingale. But why this implys that $X$ has summable squared jumps? – kuemmelsche Jun 25 '12 at 13:04
I had in mind looking at the quadratic variation, which should have a component which is the sum of the squares of the jumps, but maybe it is not as simple as I thought. – mike Jun 26 '12 at 19:03
Oh okay, while looking at the quadratic variation there also appers the sum of small jumps. But how i can proof that the quadratic variation process of a Lévy process is always finite? – kuemmelsche Jun 28 '12 at 9:15
show 2 more comments
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9457756280899048, "perplexity_flag": "middle"}