Covariance of $X^2$ and $X^3$ when $X$ is exponentially distributed

Question

Here is my work....

$\begin{align*} Cov(Y,Z) &= E(YZ) - E(Y)E(Z)\\ &= E(X^2\cdot X^3) - E(X^2)E(X^3)\\ &= E(X^5) - E(X^2)E(X^3) \end{align*}$

And we know $E(X^n) = \frac{n!}{\lambda^n}$ so,

$\begin{align*} Cov(Y,Z) &= \frac{5!}{\lambda^5} - \frac{2!}{\lambda^2}\cdot \frac{3!}{\lambda^3}\\ &= \frac{120}{\lambda^5} - \frac{12}{\lambda^5}\\ &= \frac{108}{\lambda^5} \end{align*}$

Does this seem like it is correct? Thanks!

Answer 1

Your answer looks right, though maybe you want to prove the identity $E(x^n) = {n! \over \lambda^n}$. If you want a funky proof (assuming x ~ $\lambda e^{-\lambda x}$):

$$ E(x^n) = \lambda \int^\infty_0 x^n e^{- \lambda x} dx \\ = \lambda \int^\infty_0 \left((-1)^n {d^n \over d \lambda^n} e^{-\lambda x} \right) dx\\ = (-1)^n \lambda {d^n \over d \lambda^n} \int^\infty_0 e^{-\lambda x} dx \\ = (-1)^n \lambda {d^n \over d \lambda^n} {1 \over \lambda} \\ = {n! \over \lambda^n} $$

The third line follows from the second because integration over x and differentiation WRT $\lambda$ is commutative.

Answer 2

It is correct.

If you try the R code

set.seed(2014)
cases  <- 1000000
lambda <- 10
X      <- rexp(cases, lambda)
cov(X^2, X^3)

you get $0.001079931$ which is closer to $\dfrac{108}{10^5}$ than you have any reasonable right to expect