Finding covariance function

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

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,

$$ \text{Cov}(X(t),X(v))= 1 + tv+ t²v²$$

Answer 2

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 $X(t)$.