Question: Let $X(t) = A\_0 +A\_1t+A\_2t^2$, where $A\_i's$ are uncorrelated random variables with mean $0$ and variance $1$. Find the mean function and covariance function of $X(t)$.

I do not have any clue on how to even begin this. Any help would be appreciated!

Answer: 
1) Mean Function:
Remember that expectation $\mathbb{E}$ is linear.
$$
\mathbb{E}[X(t)] = \mathbb{E}[A\_0 + A\_1t+ A\_2t^2] = \mathbb{E}[A\_0] + \mathbb{E}[A\_1]t + \mathbb{E}[A\_2]t^2 = 0
$$

2) Covariance Function:
Remember the definition $\text{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\cdot\mathbb{E}[Y]$.
$$
\text{Cov}(X(t),X(v)) = \mathbb{E}[X(t)X(v)] - \mathbb{E}[X(t)]\cdot\mathbb{E}[X(v)] =
\mathbb{E}[X(t)X(v)]
$$
Because you already know that the mean function is zero.
$$\mathbb{E}[X(t)X(v)] = \mathbb{E}[(A\_0 + A\_1t+ A\_2t^2)\cdot(A\_0 + A\_1v+ A\_2v^2)]$$
As the functions are uncorrelated and their means are zero $\mathbb{E}[A\_iA\_j] = \mathbb{E}[A\_i]\mathbb{E}[A\_j] = 0$, only the square terms matters.
$$\mathbb{E}[X(t)X(v)] = \mathbb{E}[A\_0^2] + \mathbb{E}[A\_1^2]tv + \mathbb{E}[A\_2^2]t^2v^2 = 1 + tv+ t²v²$$
So,

$$
\text{Cov}(X(t),X(v))= 1 + tv+ t²v²$$
The question is, what's a formula for $m(t)=E(X(t))$ and for $$C(t,u)=\mathrm{Cov}(X(t),X(u))=E(X(t)X(u))-(E(X(t))(E(X(u))?$$ Which you can work out because you know the first two moments of the $A\_i$ used in the formula for $X(t)$.