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

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

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  • $\begingroup$ 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

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