# Finding mean, variance of a stochastic process given covariances

We are given two uncorrelated stochastic processes: $X, Y$.

For $s, t \ge 0$ we have: $$\mathbb{E}X_t = t^3+5, \ \mathbb{E}Y_t = t+3, \ cov(X_t, X_s) = \sin t \cdot \sin s, \ cov(Y_t, Y_s) = \cos t \cdot \cos s$$

Define $Z_t = X_t + Y_t$.

I am trying to find $\mathbb{E}Z_t, \ \mathbb{D}^2 Z_t$ and $cov(Z_t, Z_s)$.

$\mathbb{E}Z_t = \mathbb{E}X_t + \mathbb{E}Y_t = t^3 + t + 8$.

But when it comes to variance, I get:

$\mathbb{D}^2 Z_t = \mathbb{D}^2 (X_t + Y_t) = \mathbb{E}(X_t^2 + 2X_tY_t + Y_t^2) - (\mathbb{E}X_t + \mathbb{E}Y_t)^2 = \mathbb{E}(X_t^2) + 2\mathbb{E}(X_tY_t) + \mathbb{E}(Y_t^2) - \mathbb{E}(X_t)^2 - 2\mathbb{E}X_t \mathbb{E}Y_t - (\mathbb{E}Y_t)^2 = \mathbb{D}^2 X_t + \mathbb{D}^2Y_t + 2 cov(X_t, Y_t) = \mathbb{D}^2 X_t + \mathbb{D}^2Y_t$ and we can compute that because we know the values of $cov (X_t, X_s)$ and expected values. Is that correct?

For $cov(Z_t, Z_s)$ I get $\mathbb{E}(Z_tZ_s) - \mathbb{E}Z_t \mathbb{E}Z_s = \mathbb{E}(X_tX_s + X_tY_s + X_sY_t + Y_sY_t) - \mathbb{E}X_t\mathbb{E}X_s - \mathbb{E}X_t\mathbb{E}Y_s - \mathbb{E}Y_t\mathbb{E}X_s - \mathbb{E}Y_t \mathbb{E}Y_s$.

And here again we get $cov (X_t, Y_t) = 0$ and use mean and covariance of $X_t, X_s, Y_t, Y_s$. Is that right?

I will be grateful for any insight, because I'm not sure if I understand the meaning of uncorrelated processes correctly.

• Yeah, looks good. – saz Oct 8 '15 at 19:08