# What is the variance of a Brownian Motion?

In my attempt to digest the answers to my previous question about stochastic integrals, I have stumbled upon yet another question that I need some clarification on...

Simply, what is the variance of a Brownian Motion $W_t$? Is it $t$ or $1$?

I ask this because, my notes said that the Brownian motion $(W_t)_{t \geq 0}$ has variance $t$. But, I was looking at Ito's formula(Ito theorem, it seems to be interchangeably called) and it introduces what is called an "Ito process", as follows,

$X_t$ is an Ito process given by $dX_t = U_tdt + V_tdt$ with $X_0=x_0$.

Which is okay... but in an example worked, it tries to solve the integral in my previous question $\int W_sdW_s$ this time using the Ito formula.

It starts off saying "the stochastic equation for a Brownian motion is when, $U_t-0$ and $V_t=1$ ..."

The reason this made me think "is the variance $1$ for a Brownian motion"? Is, in the following tutorial video,

at around 12:00 the lecturer writes down the same formula ($dX_t = U_tdt + V_tdt$ this one) but with $U_t$ and $V_t$ replaced with notations $\mu$ and $\sigma$ respectively. To me, all along, I thought $\mu$ and $\sigma$ denotes the "mean" and "variance" especially in a context like this(statistics, stochastic modelling etc). So $U_t=0$ makes sense to me, IF this is the mean $\mu$ matching the notation in the video.
But variance? If $\sigma$ is the variance(of a Brownian Motion), is it not $t$?
• It is $t$. What you are seeing in Ito's formula is that the derivative of the variance is $1$. – Ian Dec 2 '15 at 1:25