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Let $\lambda \in \mathbb{R}, \lambda > 0$ and let $X, Y, Z \sim P(\lambda)$ (they have Poissons distribution) independent random variables. Calculate $Var (XYZ) $.

I tried by calculating $ \mathbb{E} (XYZ) ^2 ( = \lambda ^6)$ because $X,Y,Z$ are independent and $(\mathbb{E} (XYZ) )^2 ( = \lambda ^6)$ (let $g$ be function so $g(X) = X^2$ and then because $X,Y,Z$ are independent so are $g(X), g(Y), g(Z))$ which means $Var(XYZ) =0$. Is that correct?

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I can't follow your argument ! It's really unclear. Since $X,Y,Z$ are independent, indeed $$\mathbb E[XYZ]=\mathbb E[X]\mathbb E[Y]\mathbb E[Z]$$ and $$\mathbb E[(XYZ)^2]=\mathbb E\left[X^2\right]\mathbb E\left[Y^2\right]\mathbb E\left[Z^2\right].$$ Now, $$\mathbb E[X^2]=\mathbb E[Y^2]=\mathbb E[Z^2]=\lambda +\lambda ^2.$$ This because $$Var(X)=\lambda =\mathbb E[X^2]-\mathbb E[X]^2=\mathbb E[X^2]-\lambda ^2.$$ (same with $Y,Z$). I let you conclude.

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For a Poisson variable, $E(X^2)=\lambda(\lambda+1)$, and not $\lambda$.

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  • $\begingroup$ Your comment brings it to the point. $\endgroup$
    – callculus
    May 11 '19 at 16:09

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