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Consider $X_t=B_t-tB_1$ for $0\leq t\leq 1$, where $B_t$ represents the standard Brownian motion. I derived that $E(X_t)=0$ and $\operatorname{Cov}(X_t,X_s)=\min(t,s)-st$. Now, I am asked the joint distribution of $X_t$ and $X_s$ for some fixed $s,t\in[0,1]$. My first thought was that the joint distribution is the bivariate normal with mean $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ and covariance matrix $\Sigma = \begin{pmatrix} t(1-t) & \min(t,s)-st \\ \min(t,s)-st & s(1-s) \end{pmatrix}$. However, I do not know how to justify this choice. Is it the bivariate normal?

Davide Giraudo
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david_rios
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    It is! Indeed, we know that $(B_t,B_s,B_1)$ is a Gaussian vector, therefore, for any $(a_t,a_s,a_1)$, we have that $a_tB_t+a_sB_s+a_1B_1$ is Gaussian. You are if $c_t(B_t-tB_1)+c_s(B_s-sB_1)$ is Gaussian. – LucaMac Apr 15 '21 at 09:31
  • Excellent! Thank you – david_rios Apr 15 '21 at 13:46

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