Sum of normally distributed variables is a normally distributed variable?

Question

Consider two Wiener processes: $$ \begin{aligned} X_a &\sim\mathcal N(0,a) \\ X_{a-b} &\sim\mathcal N(0,a-b) \end{aligned} $$ How do I show that: $$ X_a - X_{a-b} \sim\mathcal N(.,.) $$ That is, how do I show that $X_a - X_{a-b}$ is Gaussian?

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I know that $E(X_a - X_{a-b}) = 0$

and $$ \begin{aligned} \text{Var}(X_a - X_{a-b}) &= E(X_a^2) - E(X_{a-b}^2) \\ &= \text{min}(a,a) -\text{min}(a-b,a-b) \\ &= a - a - b \\ &= b \end{aligned} $$

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Note that both $X_a$ and $X_{a-b}$ have the following properties:

• They are normally distributed with mean $0$.
• Their sample paths are continuous.
• Its covariance function is $\rho(s,t) := \text{Cov}(X_s,X_t)=E(X_sX_t)=\text{min}(s,t)$.

Answer 1

$X_a = X_{a-b} + \int_{a-b}^a dX_s$

$\Rightarrow X_a - X_{a-b} = X_{a-b} + \int_{a-b}^a dX_s - X_{a-b}$

$ = \int_{a-b}^a dX_s$

$ = X_{a - (a-b)} = X_b$.

Since $\{X_t\}_{t \in [0,T]}$ is a Wiener process and $b \in [0,T]$, then $X_b$ is Gaussian by the fact that $X_b \sim N(0,b)$ (from the definition of a Wiener process).