Is ${Y(t) : t ≥ 0}$ second order stationary? Let $ {X(t) : t ≥ 0} $ be a stochastic process that satisfies the following conditions:
$-{X(t) : t ∈ T} ∈ M2$$
-It has independent increments
-It has stationary increments
Define a new stochastic process ${Y (t) : t ≥ 0}$ as:
$$Y (t) = X(t + 5) − X(t + 2)$$
Is ${Y(t) : t ≥ 0}$ second order stationary?
I have really tried but I get confused in the properties, I appreciate the help
 A: I think I had misunderstood the question, the OP wants to show that $Y$ is weakly stationary, not that $Y$ is M2 stationary.
So we have a starting process which is M2, that is the joint distributions up to two variables are shift invariant, i.e. $X(t) \sim X(t+\tau)$ and $(X(t),X(t')) \sim (X(t+\tau),X(t'+\tau))$.
We want to know that the process $Y(t)=X(t+5)-X(t+2)$ is second order stationary, meaning that the mean, variance and covariance are shift invariant (also called weakly stationary), so:
1- $E[Y(t)]=E[X(t+5)-X(t+2)]=E[X(t+5)]-E[X(t+2)]=0$
2- $E[Y(t)Y(t')]=E[(X(t+5)-X(t+2))(X(t'+5)-X(t'+2))]=$
$=E[X(t+5)X(t'+5)]-E[X(t+5)X(t'+2)]-E[X(t+2)X(t'+5)]+E[X(t+2)X(t'+2)]=$
Now since the process is $M2$ stationary all joint distributions of two variables (and in particular the means) are invariant after the transformation:
$t \rightarrow t+\tau$, $t' \rightarrow t'+\tau$
showing that:
$E[Y(t)Y(t')]=E[Y(t+\tau)Y(t'+\tau)]$
Therefore since the mean of the process is zero we have shown that the covariances are stationary.
Comment: I was not able to show that the process $Y(t)$ is $M2$ (i.e. to show that the distribution of $(Y(t),Y(t'))$ is shift invariant, but probably this is not a part of the question ? I suspect it is false but maybe somebody can also comment on this ?
