Let $W(t)$ be a standard Brownian motion (BM), in particular $W(t) \sim \mathcal{N}(0,t)$. Then it is easily shown that $\int_0^T W(t) dt \sim \mathcal{N}(0, T^3/3)$.
Question: What is the distribution of $X := W(T) + \int_0^T W(t) dt$? Are these two random variables not dependent? Since they are, how do we know $X$ is normal? I've always thought that the sum of two normals is not necessarily normal, especially when they are correlated. Thanks!
As a linear combination of a centered gaussian process, the random variable $X$ is centered normal and its distribution is fully characterized by its variance $\sigma^2_T=E(X^2)$, where $$ \sigma^2_T=E(W(T)^2)+2\int_0^TE(W(T)W(t))\mathrm dt+2\int_0^T\!\!\!\int_0^t E(W(t)W(s))\mathrm dt\mathrm ds, $$ which should be evaluated easily. Alternatively, $$ X=\int_0^T W(t)\mathrm d\mu_T(t), $$ for some measure $\mu_T$ one can write down explicitely, hence $$ \sigma^2_T=\iint_{[0,T]^2}\min\{t,s\}\,\mathrm d\mu_T(t)\mathrm d\mu_T(s). $$