This is the situation. I have an estimation of the position $(x_t,y_t)$ of an object with its covariance $\Sigma_p$ and an estimation of its speed $(v_x, v_y)$ with its covariance $\Sigma_v$. Actually, the two estimations should be correlated since the speed is computed from the position estimation, but for simplicity, we can assume they are independent.
Now, I want to estimate the covariance of the future position of the object $(x_{t+1}, y_{t+1}) = (x_t,y_t) + (v_x, v_y) \cdot dt$, where $(v_x, v_y) \cdot dt$ can be represented with a RVE with covariance $\Sigma_1 = dt^2 \cdot \Sigma_v$. So, I have to compute the covariance of the sum of two RVE (namely $X$ and $Y$), the first with covariance $\Sigma_p$, the second with covariance $\Sigma_1$.
Ok, let's try to do it:
$cov(X+Y) = E((X+Y)(X+Y)^T) - E(X+Y)E(X+Y)^T$ $= E((X+Y)(X^T+Y^T)) - E(X+Y)E(X^T+Y^T)$ $= E(XX^T + YY^T + XY^T + YX^T) - (E(X)E(Y))(E(X^T)E(Y^T))$ $= E(XX^T)-E(X)E(X^T) + E(YY^T)-E(Y)E(Y^T) + E(XY^T)-E(X)E(Y^T) + E(YX^T)-E(Y)E(X^T)$ $= cov(X) + cov(Y) + E(XY^T)-E(X)E(Y^T) + E(YX^T)-E(Y)E(X^T)$
Is it correct? Now, can I rewrite the formulation in order to compute $cov(X+Y)$ only in function of $cov(X)$ and $cov(Y)$ since I do not have the PDF of $X$ and $Y$?
Thanks in advance!
You've mentioned that you can assume that $X$ and $Y$ are independent, in which case all the terms in your expression other than $\operatorname{Var}(X)$ and $\operatorname{Var}(Y)$ cancel.
Are you sure independence is a good assumption for your use case? Why do you want to estimate the variance of $X+Y$? is it to estimate the error of your prediction? In that case I wonder if you could estimate a confidence interval in a better way (here it seems you will make a gaussian assumption to give a CI).