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# quadratic variation of Brownian motion

###### Theorem.

Let $(W_{t})_{{t\in\mathbb{R}_{+}}}$ be a standard Brownian motion. Then, its quadratic variation exists and is given by

$[W]_{t}=t.$ |

As Brownian motion is a martingale and, in particular, is a semimartingale then its quadratic variation must exist. We just need to compute its value along a sequence of partitions.

If $P=\{0=t_{0}\leq t_{1}\leq\cdots\leq t_{m}=t\}$ is a partition of the interval $[0,t]$, then the quadratic variation on $P$ is

$[W]^{P}=\sum_{{k=1}}^{m}(W_{{t_{k}}}-W_{{t_{{k-1}}}})^{2}.$ |

Using the property that the increments $W_{{t_{k}}}-W_{{t_{{k-1}}}}$ are independent normal random variables with mean zero and variance $t_{k}-t_{{k-1}}$, the mean and variance of $[W]^{P}$ are

$\displaystyle\mathbb{E}\left[[W]^{P}\right]$ | $\displaystyle=\sum_{{k=1}}^{m}\mathbb{E}\left[(W_{{t_{k}}}-W_{{t_{{k-1}}}})^{2% }\right]=\sum_{{k=1}}^{m}(t_{k}-t_{{k-1}})=t,$ | ||

$\displaystyle\operatorname{Var}\left[[W]^{P}\right]$ | $\displaystyle=\sum_{{k=1}}^{m}\operatorname{Var}\left[(W_{{t_{k}}}-W_{{t_{{k-1% }}}})^{2}\right]=\sum_{{k=1}}^{m}2(t_{k}-t_{{k-1}})^{2}$ | ||

$\displaystyle\leq 2|P|\sum_{{k=1}}^{m}(t_{k}-t_{{k-1}})=2|P|t.$ |

Here, $|P|=\max_{k}(t_{k}-t_{{k-1}})$ is the mesh of the partition. If $(P_{n})_{{n=1,2,\ldots}}$ is a sequence of partitions of $[0,t]$ with mesh going to zero as $n\rightarrow\infty$ then,

$\mathbb{E}\left[([W]^{{P_{n}}}-t)^{2}\right]\leq 2|P_{n}|t\rightarrow 0$ |

as $n\rightarrow\infty$. This shows that $[W]^{{P_{n}}}\rightarrow t$ in the $L^{2}$ norm and, in particular, converges in probability. So, $[W]_{t}=t$.

## Mathematics Subject Classification

60H10*no label found*60J65

*no label found*

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