Weighted sum of standard Brownian bridges

Question

Let $\{B_j\}_{j=1}^k$ be a sequence of Brownian bridges.

Let us consider $$X(t)=\sum_{j=1}^m w_j(t)B_j(t),$$ where $w_j$ are positive weight functions.

Then what can we say about (distribution or may be mean and variance) $X(t)$??

Clearly, $X(t)$ may not be a Brownian bridge as any standardized version of $X(t)$ has has its covariance function of the form $$\mathrm{cov}(X(t),X(t'))=\frac{(\min(t,t')-tt')\sum_{j=1}^m w_j(t)w_j(t')}{\sqrt{\sum_{j=1}^m w^2_j(t)\sum_{j=1}^m w^2_j(t')}}$$ which cannot be written as only as $\min(t,t')-tt'$. This tells us that we cannot standardise $X(t)$ to have a Brownian bridge.

Answer 1

Supposing that they are all independent Brownian bridges with $B_{j}(0)=B_{j}(T)=0$, we have for independent Brownian motions $W_{j}$

$$X(t)=\sum_{j=1}^m w_j(t)B_j(t)=\sum_{j=1}^m w_j(t)W_j(t)-\frac{t}{T}\sum_{j=1}^m w_j(t)W_j(T).$$

For the first term to be a Brownian motion $W_{t}$ we would need

$EW_{t}W_{s}=t\wedge s\sum_{j=1}^m w_j(t)w_j(s)=t\wedge s$ i.e.

$\sum_{j=1}^m w_j(t)w_j(s)=1$ for all $s,t\in [0,T]$.