I want to find the Covariance function of a stochastic Process $(X(t))_{t \in I}$ for $X(t) := \mathbb{1}_{[0,t]}(U)$ where $U$ ~Unif$[0,1]$ and $I = [0,1]$.
The covariance function is given by
$$Cov(X(s),X(t)) = \mathbb{E}[(X(s)-\mathbb{E}X(s)) \cdot (X(t)-\mathbb{E}X(t))]$$
The way I understand it (correct me if I'm wrong), because both $X(s)$ and $X(t)$ depend on the same $U$, they are not independent. We have $\mathbb{E}X(s) = \frac{s^2}{2}$ (right?). So we have
$$Cov(X(s),X(t)) =\mathbb{E}[(X(s)-\frac{s^2}{2}) \cdot (X(t)-\frac{t^2}{2})]$$
Now I am not sure how to calculate this. Is this correct (assuming $s \le t$) ?
$$\int_{0}^{1}u\cdot(\mathbb{1}_{[0,s]}(u)-\frac{s^2}{2})\cdot(\mathbb{1}_{[0,t]}(u)-\frac{t^2}{2})du = \int_{0}^{1}u\cdot(\mathbb{1}_{[0,s]}(u)-\frac{s^2}{2})\cdot(\mathbb{1}_{[0,s]}(u)-\frac{t^2}{2})du = \int_{0}^{s}u\cdot(u-\frac{s^2}{2})\cdot(u-\frac{t^2}{2})du = \int_{0}^{s}u\cdot(u^2-u\cdot\frac{s^2+t^2}{2}+\frac{s^2t^2}{2})du = \int_{0}^{s}u^3-u^2\cdot\frac{s^2+t^2}{2}+u\cdot\frac{s^2t^2}{2}du = \frac{s^4}{4}-\frac{s^5+s^3t^2}{6}+\frac{s^4t^2}{4}$$
And how would my result change if instead I had $X(t) := \mathbb{1}_{[0,t]}(U)-t$ with $U$ and $I$ as above?
For every $t\in[0,1]$ random variable $X(t)$ only takes values in $\{0,1\}$ so that: $$\mathbb EX(t)=P(X(t)=1)=P(U\leq t)=t$$
Also $X(t)X(s)$ only takes values on $\{0,1\}$ so that:$$\mathbb EX(t)X(s)=P(X(t)X(s)=1)=P(X(t)=1=X(s))=P(U\leq t,U\leq s)=$$$$P(U\leq\min(t,s))=\min(t,s)$$
Find the covariance by applying the general rule that:$$\mathsf{Cov}(X(t),X(s))=\mathbb EX(t)X(s)-\mathbb EX(t)\mathbb EX(s)$$