# 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 (http://planetmath.org/QuadraticVariationOfASemimartingale). 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 (http://planetmath.org/SubintervalPartition) 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, $[W]_{t}=t$.

Title quadratic variation of Brownian motion QuadraticVariationOfBrownianMotion 2013-03-22 18:41:25 2013-03-22 18:41:25 gel (22282) gel (22282) 6 gel (22282) Theorem msc 60H10 msc 60J65 QuadraticVariation