Sum/differences of two normal variables

[t_n_p]

Homework Statement

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

Homework 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)

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:
B_t+s + B_t+s it would be ~N(0,t+s+t+s)
No issues here

But now since I have a negative infront of the B_s, 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 -B_s (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 B_s to a positive?

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

B_t+s + B_t+s - B_s - B_s ~ N (0, (t+s)+(t+s)+(s)+(s))

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

hope it all makes sense

 

[statdad]

If $X$ is $n(\mu, \sigma^2)$

then $aX$ is $n(a\mu, a^2 \sigma^2)$

If $X, Y$ are independent, then

$$\sigma^2_{aX+bY} = a^2 \sigma^2_X + b^2\sigma^2_Y$$

and both of these hold regardless of the sign of $a$ and $b$.