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Prove $E[(\Delta B_j)^4]=3(\Delta t_j)^2$ where the Delta stands for the change of something i.e $B_j-B_{j-1}=\Delta B_j$ and the $B_j$ stand for the standard Brownian motion

I won't show my step here since i know it is a few step computation, i think my mistake on conceptual got me wrong from start

Here is the second degree example with answer: https://math.stackexchange.com/questions/1308705/show-a-lemma-of-brownian-motion-is-valid

The difficulty is that i can't just use the second degree example to convert it into four degree solution...

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  • $\begingroup$ Why it is off topic? $\endgroup$ – Victor Jun 2 '15 at 18:49
  • $\begingroup$ @avid19 - This is a few step problem, however it seems to me that the internet only provide the second power example and i don't think i am close to the real concept yet. $\endgroup$ – Victor Jun 2 '15 at 18:52
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    $\begingroup$ "Why it is off topic?" Because This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Did Jun 2 '15 at 19:16
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    $\begingroup$ @MichaelHardy On the other hand, one could expect that a user active on the site for nearly 4 years with more than 250 questions under their belt could have become even moderately interested in the basics of the functioning of the site, don't you think? Or, all this is just a bad play, where some users deliberately mimic surprise, with the goal to continue to knowingly post questions whose format explicitely contradicts (good sense and) the rules of the site. In the end, who ought what, one wonders. $\endgroup$ – Did Jun 2 '15 at 21:18
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    $\begingroup$ Like what? Please be specific. $\endgroup$ – Did Jun 2 '15 at 21:56
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This is merely the fact that if $X\sim N(0,\sigma^2)$ then $\operatorname{E}(X^4)=3\sigma^4$. If $Z=X/\sigma\sim N(0,1)$ then clearly $\operatorname{E}(Z^4)=\operatorname{E}(X^4)/\sigma^4$, so it is enough to prove $\operatorname{E}(Z^4)=3$. $$ \operatorname{E}(Z^4)=\frac 1 {\sqrt{2\pi}}\int_{-\infty}^\infty z^4 e^{-z^2/2}\,dz. $$ Let $u=z^2/2$ so that $du = z\ dz$. Then we have $$ 2\cdot\frac 1{\sqrt{2\pi}}\int_0^\infty (2u)^{3/2} e^{-u}\,du=2\cdot\frac 1 {\sqrt{2\pi}} \cdot2\sqrt2\cdot \Gamma(5/2) = 2\cdot\frac 1 {\sqrt{2\pi}} \cdot\frac32\cdot\frac12\cdot\Gamma(1/2) $$ $$ =2\cdot\frac 1 {\sqrt{2\pi}}\cdot 2\sqrt2 \cdot\frac32\cdot\frac22\cdot\sqrt\pi = 3. $$

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