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.
(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)
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