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 $cov(X,X^2) = E[X^3] - E[X]*E[X^2] = \frac{3}{\lambda} - \frac{1}{\lambda} * \frac{2}{\lambda} = \frac{3\lambda -2}{\lambda^2}$ ?


1 Answer 1


I don't think you need to use the Erlang distribution for that.

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: $$cov(X, X^2) = \frac{6}{\lambda^3} - \frac{1}{\lambda}\frac{2}{\lambda^2} = \frac{4}{\lambda ^3}$$

  • $\begingroup$ Thank you, you are right, Erlang can be interpreted as the sum of INDEPENDENT exponential random variables and X^2 are surely not two independent X.. $\endgroup$
    – Antonello
    Oct 1, 2021 at 11:57
  • $\begingroup$ 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. $\endgroup$ Oct 1, 2021 at 15:11
  • $\begingroup$ Thanks, I made a mistake, the formula should be $E(X^k) = \frac{k!}{\lambda^k}$. $\endgroup$
    – Pohoua
    Oct 1, 2021 at 17:07
  • $\begingroup$ OK. +1 for promptly correcting your answer. $\endgroup$ Oct 1, 2021 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.