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

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?

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