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Bumped by Community user
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YCor
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Weighted sum of Standardstandard Brownian Bridgesbridges

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

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

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.

Weighted sum of standard Brownian bridges

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.

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Janak
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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.