sum/difference of two normal variables

Sep 2009
1. The problem statement, all variables and given/known data

I have Bt+s ~ N (0, t+s) and Bs ~ N(0,s).
I want to find the mean and variance of Bt+s + Bt+s - Bs - Bs.

2. Relevant equations

(i) I know that the sum of independent normals is again normal with mean = sum of means and variance = sum of variances.

(ii) I also know that cX ~ N(cμ, c2σ2)

3. The attempt at a solution

Obviously the mean is easy to calculate since it is 0 all round, but I'm having problems with the variance.

If I wanted to find distribution of:
Bt+s + Bt+s it would be ~N(0,t+s+t+s)
No issues here

But now since I have a negative infront of the Bs, how will that come into effect? Notice how property (i) above is regarding the sum of independent normals, what about the difference between two independent normals?

Do I change -Bs (by property (ii) above) so that it has a distribution ~ N ((-1)0, (-1)2(s))

Does this change the negative out the front of Bs to a positive?

i.e, -Bs ~ N (0, s) and then sum the variances so that:

Bt+s + Bt+s - Bs - Bs ~ N (0, (t+s)+(t+s)+(s)+(s))

i.e. Bt+s + Bt+s - Bs - Bs ~ N (0, 2t+4s)?

hope it all makes sense


MHF Hall of Honor
Mar 2008
P(I'm here)=1/3, P(I'm there)=t+1/3

Well, by the sight of the names, I guess you're talking about brownian motions, right ?

So you know that B(t+s)-B(s) follows a normal distribution N(0,t)

So B(t+s)+B(t+s)-B(s)-B(s)=2*(B(t+s)-B(s))=2*N(0,t)=N(0,4t), isn't it ?

It should answer your question