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I need to prove equivalence of definitions of standard Brownian bridge:

  1. $B_t-tB_1$, $t \in [0,1]$
  2. Process whose law is defined as $\mathcal{L}((B_{t_1},\dots,B_{t_k})|B_1=0)$, $0 <t_1<\dots<t_k<1$, where the conditioning is defined as:

$\mathbb{E}[f(B_{t_1},\dots,B_{t_k})|B_1=0]=\lim_\limits{\varepsilon\rightarrow 0}\mathbb{E}[f(B_{t_1},\dots,B_{t_k})| |B_1| \leq \varepsilon]$.

I was able to prove that both processes have the same expectation and covariance, but I'm missing the fact that the second process is a Gaussian process to complete the proof. I would be thankful for any help!

mikasa
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    I guess noting the following fact is enough: A multivariate normal distribution, conditioned to lie in an affine subspace, is again normal. – Sangchul Lee Nov 01 '25 at 13:52
  • @SangchulLee is that clear from the definition of a Gaussian process? If not, how could I prove it? Thank you in advance! – mikasa Nov 01 '25 at 14:02
  • Addition of random variables results in convolution of their distriubtions i.e. product of their Fourier transforms.

    Products of exponentials are exponentials of the sum of exponents. Its follows,that the mix of two Gaussians is a Gaussian with sums of means and sum of variances.

     – Roland F Nov 02 '25 at 09:29

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