Skip to main content
1 of 2
Janak
  • 213
  • 1
  • 7

Weighted sum of Standard Brownian Bridges

Suppose, $\{B_j\}_{j=1}^k$ 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 $$cov(X(t),X(t'))=\frac{(\min(t,t')-tt')\sum_{j=1}^m w_j(t)w_j(t')}{\sqrt{\sum_{j=1}^m w^2_j(t)\sum_{j=1}^m w^2_j(t')}}$$ which cannot be written as only as $\min(t,t')-tt'$. This tells us that we cannot standardised $X(t)$ to a have a Brownian Bridge.

Janak
  • 213
  • 1
  • 7