Suppose that $E[X]$ is known where $X$ is a random variable. Is it possible to compute $E[X^3]$ with only this information? I am thinking of resorting to moment generating functions but I am not sure if I have sufficient information.
This problem came about while I was considering this:
Let $G(t)$ be a stationary Gaussian process with mean 0, variance 1 and correlation function $\rho(\tau) = E[G(t+\tau)G(t)]$. Suppose that $X(t) = G(t)^3.$ I then need to compute $E[X(t+\tau)X(t)]$ which is just the same as the problem I posed above.
Although I am not sure if MGF is the way to go since on the second example in the book I'm reading, the above quantity $E[X(t+\tau)X(t)]$ has also been computed for $X(t)=e^{G(t)}$ which I do not think could be achieved using MGF...
any suggestions? Help appreciated!
Of course not. Fortunately, this is not the problem you are interested in... To solve the problem you are interested in, let us first recall a crucial fact:
(This specific aspects of Gaussian processes is one of the many reasons why they are so useful (and so used) for modeling purposes).
Here, for every fixed $(t,\tau)$, $$(G(t),G(t+\tau))=(U,rU+sV),$$ where $(U,V)$ is i.i.d. standard normal, $r=\rho(\tau)$ and $s=\sqrt{1-r^2}$, hence the process $X$ defined by $$X(t)=G(t)^3,$$ for every $t$, satisfies $E(X(t))=E(U^3)$ and $E(X(t)^2)=E(U^6)$, that is, $$ E(X(t))=0,\qquad E(X(t)^2)=15,$$
and $$ X(t)X(t+\tau)=U^3(rU+sV)^3=r^3U^6+3r^2sU^5V+3rs^2U^4V^2+s^3U^3V^3, $$ with $E(U^6)=15$, $E(U^5V)=E(U^5)E(V)=0$, $E(U^4V^2)=E(U^4)E(V^2)=3\cdot1$ and $E(U^3V^3)=0$, hence $$ E(X(t)X(t+\tau))=15\cdot r^3+3\cdot3rs^2=3r(3+2r^2). $$ Since $E(X(t)^2)=15$, one can summarize this as the fact that, for every $\tau$, $$ \rho_X(\tau)=\rho_G(\tau)\,\left(\frac35+\frac25\rho_G(\tau)^2\right). $$ This uses the first moments of a standard normal random variable, which are $E(U^{2n+1})=0$ for every $n$ and $E(U^2)=1$, $E(U^4)=3$, $E(U^6)=15$.