The original question is:

For $t\in [0,1]$, we define $X_t=B_t-tB_1$, where $\{B_t:t\geq 0\}$ is a standard Brownian motion. Find the density of $X_t$ .

After reading several resources, I think $X_t$ is a normal distribution. However, since most of the books concerning on the process itself other than the distribution of $X_t$, can someone confirm this?


Yes, it is normal. Brownian motion is a Guassian process, which means that its finite dimensional distributions are all multivariate normal, and in particular any linear combination of $B_{t_1},\ldots,B_{t_n}$ is normal for any indices $t_1,\ldots,t_n$.

To find the density of $X_t$, we need only calculate its mean and variance. Its mean can be found as $$E(X_t)=E(B_t)-tE(B_1)=0$$ and its variance can be found by calculating its second moment using $E(B_sB_t)=s\wedge t$: $$\operatorname{Var}(X_t)=E(X_t^2)=E(B_t^2)-2tE(B_tB_1)+t^2E(B_1^2)=t-2t^2+t^2=t(1-t).$$ Hence for $t\in(0,1)$, $X_t$ is $\mathcal N(0,1-t)$ and so has density $$f(x)=\frac1{\sqrt{2\pi t(1-t)}}e^{-x^2/2t(1-t)}.$$

  • $\begingroup$ You missed a $t^2$ while calculating the variance… $\endgroup$ – Gono Dec 29 '17 at 18:44
  • $\begingroup$ Books said $X_t$ is a Brownian motion conditional on $X_1 = 0$. Does this affect anything about the distribution of $X_t$? $\endgroup$ – Jango Dec 29 '17 at 19:02
  • $\begingroup$ @Gono Thanks for the correction! $\endgroup$ – Jason Dec 29 '17 at 19:07
  • 2
    $\begingroup$ @Jango I'm not totally sure what you mean by affecting the distribution - what you mention is an alternative characterization (or definition) of a Brownian bridge, but this fact requires some proof. $\endgroup$ – Jason Dec 29 '17 at 19:13


  • $X_t=(1-t)B_t - t(B_1 - B_t)$.
  • $B_1 - B_t \sim N(0,1-t)$ and $B_t \sim N(0,t)$ are independent.
  • $\begingroup$ Actually it's $$B_t-B_1 \sim N(0,1-t)$$ because $0 \le t \le 1$ $\endgroup$ – Gono Dec 29 '17 at 18:36
  • 1
    $\begingroup$ @Gono Thanks for the correction! $\endgroup$ – angryavian Dec 29 '17 at 18:38

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