Yes, sure!
First off, we need to recognise that B(t) can be written as B(t) - B(0), since B(0) = 0.
Now B(t) = B(t) - B(0) is a standard Brownian Motion increment. Standard Brownian Motion increments:
a) are normally distributed
b) have a mean of 0
c) have a variance equal to the time lag, which is t - 0 = t.
So B(t) has a N(0,t) distribution.
Now we need to recall properties of normally distributed random variables. Say X is N(a, b). Then:
- cX is N(a, c^2 b)
- X + c is N(a + c, b)
- exp(X) is logN(a, b)
To see why the first bullet holds, work out the mean and the variance.
E[cX] = c E[X] = ca
var[cX] = c^2 var[X] = c^2 b
To see why the second bullet holds, work out the mean and the variance.
E[X + c] = E[X] + c = a + c
var[X + c] = var[X] = b (note that additive constants do not alter the variance).
To see why the third bullet holds, we use the rule that:
If log Y is N(a, b) then Y is logN(a,b).
Equivalently, if X is N(a, b) then exp(X) is logN(a, b).
This follows by setting X = log Y so that Y = exp(X).
Your question is a case of the first bullet.
sigma B(t) is N(0, sigma^2 t)
Re the other query, you are right to say that usually:
var(a + b) = var(a) + var(b) + 2cov(a, b).
When a and b are independent, as is the case in the picture, the covariance term disappears as a and b do not relate to each other, and we are left with:
var(a + b) = var(a) + var(b)
Hope this all helps
Anna
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