If M(x) where x is greater or equal to 0 is Poisson with rate lambda, what is E[M(x)*M(x+y)]?
I believe the problem can be solved as follows. Please confirm if my invoking of the ind. increments and stationary properties are correct:
$M(x+y) = M(x) + M(x+y) - M(x)$, due to the independent increments property. This is basically breaking up $M(x+y)$ into two intervals, [0,x] and (x,x+y].
$E[M(x)*M(x+y)]$
$= E[M(x)*(M(x)+M(x+y)-M(x))]$
$= E[M(x)^2] + E[M(x)*(M(x+y)-M(x))]$
$= E[M(x)^2] + E[M(x)*M(y)]$ due to stationary property can subtract M(x) from M(x+y)
Due to the stationarity property both M(x) and M(y) are Poisson with rate lambda per unit time.
I am looking at my textboox and $E[M(x)^2]$ is written as:
$lambda*x + (lambda*x)^2$ and I don't see why that is.
The full answer given is: $lambda*x + (lambda*x)^2 + lambda*x*lambda*y$
I don't see why there are three terms, and in particular I don't see why we have the first two (I am clear about the third one).
Most importantly, is the following correct about the stationary and independent increments property:
*Are the independent increments and stationary properties (or just one or the other) the ones that allow you to do the following?:
$M(x+y) = M(x)+M(x+y)-M(x))$ and then further $M(x+y)-M(x)=M(y)$?