What is the density of a brownian bridge $X_t=B_t-tB_1$? 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 $\mathcal N(0,t(1-t))$ and so has density
$$f(x)=\frac1{\sqrt{2\pi t(1-t)}}e^{-x^2/2t(1-t)}.$$
