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Brownian motion is said to follow a path where each value is normally distributed with mean $\mu t$ and variance $\sigma^2 t$.

What is the basis for the relation that variance varies directly proportional to the 1st power of $t$?

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    $\begingroup$ It follows directly from the definition of the process (specifically, the variance of the increments and the fact that they are independent). Are you asking for a formal proof of this fact? Or an intuitive explanation or illustration? $\endgroup$
    – Chris Haug
    Aug 31, 2018 at 12:09
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    $\begingroup$ Hi: Hamilton's Time Series Analysis has a decent discussion of this question without the use of stochastic differential equation type math. What you want to look at depends heavily on your math background and also as chris haug said, on what you want. $\endgroup$
    – mlofton
    Aug 31, 2018 at 12:43
  • $\begingroup$ @ChrisHaug How about both? I don't have any info on the proof or the intuition. $\endgroup$
    – Dom Jo
    Aug 31, 2018 at 13:09
  • $\begingroup$ @mlofton I graduated in engineering. But i work in finance. I was trying to study Black Scholes equation from scratch. $\endgroup$
    – Dom Jo
    Aug 31, 2018 at 13:10
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    $\begingroup$ In a random walk, independent random "impulses" are imparted to a particle at regular intervals. Since the variance of a sum of independent variables is the sum of their variances, the variance at the end must be directly proportional to the time elapsed. Thus, this linear relationship becomes a basic requirement of any probability model of a such a phenomenon. $\endgroup$
    – whuber
    Aug 31, 2018 at 15:56

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