Let $\{X(t), t \geqslant 0\}$ be a standard Brownian motion. That is, for every $t \gt 0$, $X(t)$ is normally distributed with mean $0$ and variance $t$.
Then $\{X(t), 0 \leqslant t \leqslant 1 | X(1) = 0\}$, known as the Brownian bridge, is a Gaussian process. That is, for every $0 \lt t \lt 1$, it is multivariate normally distributed.
Thus, I want to find its the marginal mean values and the covariance values.
Remark) There is two true equations. $$\tag{1} E\left[X(s)|X(t)=B\right]=\frac{s}{t}B$$ $$\tag{2} \text{Var}\left[X(s)|X(t)=B\right]=\frac{s(t-s)}{t}$$
- For $s\lt1$,
$$E\left[ X(s) | X(1) = 0 \right] = 0$$
- For $s\lt t\lt 1$,
\begin{align} \text{Cov}\left[ \left(X(s), X(t)\right)|X(1)=0\right] &= E\left[ X(s) X(t) | X(1) = 0 \right] \\ &= E\left[E\left[ X(s) X(t) | X(t), X(1) = 0 \right]|X(1)=0\right] \tag3\\ &= E\left[X(t) E\left[ X(s) | X(t)\right]|X(1)=0\right] \tag4\\ &= E\left[X(t) \left.\frac{s}{t} X(t)\right|X(1)=0\right] \quad \text{by (1)}\\ &= \frac{s}{t}E\left[\left.X^2(t) \right|X(1)=0\right] \\ &= \frac{s}{t}\cdot t(1-t) \quad \text{by (2)}\\ &= s(1-t) \end{align}
I cannot understand equation $(3)$ and equation $(4)$ are derived.
Please give me some explanations with a detailed process.
Thank you for reading my question.
In equation (3), Let $E_1$ denotes the conditional expectation$E(|X(1)=0)$ we use the iterated expectation , that is $E_{1}(X(s)X(t))=E_{1}(E(X(s)X(t)|X(t))$.
In equation (4), since we conditioned to X(t), that means that we know its value , hence X(t) is not random anymore under that assumption: we can remove it from the expectation term as if it was deterministic.
$E_{1}(X(s)X(t))=E_{1}(X(t)E(X(s)|X(t)))$