Which is the covariance matrix between X ~ exponential (lambda) and X squared?

I need to find the covariance matrix between $$X \sim exp(\lambda)$$ and its square $$X^2$$.

Can I interpret $$X^n$$ as an Erlang distribution (for which I can have the expectation/variance on Wikipedia :-) ) with parameters $$\lambda$$ and $$n$$ and hence $$var(X^2) = 2/\lambda^2$$ and $$cov(X,X^2) = E[X^3] - E[X]*E[X^2] = \frac{3}{\lambda} - \frac{1}{\lambda} * \frac{2}{\lambda} = \frac{3\lambda -2}{\lambda^2}$$ ?

As you said: $$cov(X, X^2) = E(X\times X^2) - E(X)E(X^2) .$$
Now, if $$X \sim \mathcal{E}xp(\lambda)$$, you have that: $$E(X^k) = \frac{k!}{\lambda^{k}}$$ (this is easily shown using the moment generating function of $$X$$).
So in the end: $$cov(X, X^2) = \frac{6}{\lambda^3} - \frac{1}{\lambda}\frac{2}{\lambda^2} = \frac{4}{\lambda ^3}$$
• Why are you using $\dfrac{1}{\lambda}$ as the value of $E[X]$? According to your formula, $$E[X] = E[X^1] = \frac{k!}{\lambda^{k+1}}\bigg|_{k=1} = \frac{1}{\lambda^{1+1}}=\frac{1}{\lambda^2}$$ isn't it? and similarly for the other terms. Oct 1, 2021 at 15:11
• Thanks, I made a mistake, the formula should be $E(X^k) = \frac{k!}{\lambda^k}$. Oct 1, 2021 at 17:07