Given are 3 independent normal variables $X_i=\mathcal{N}(\mu_i,\sigma^2)_{i=1,2,3}$ with expectations $\mu_i$ and equal variance $\sigma^2$.
What is the expectation of their product $\mathbb{E}[X_1 \cdot X_2 \cdot X_3]$ ?
Given are 3 independent normal variables $X_i=\mathcal{N}(\mu_i,\sigma^2)_{i=1,2,3}$ with expectations $\mu_i$ and equal variance $\sigma^2$.
What is the expectation of their product $\mathbb{E}[X_1 \cdot X_2 \cdot X_3]$ ?
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Do the separate integrals to find:
$${\cal E}[X_1 X_2 X_3]$$
$$= {\cal E}[X_1]{\cal E}[X_2]{\cal E}[X_3]$$
$$= \underbrace{\left( \int\limits_{x=-\infty}^\infty \frac{x}{\sqrt{2 \pi} \sigma} e^{-(x - \mu_1)^2/(2 \sigma^2)}\ dx \right)}_{\mu_1} \underbrace{\left( \int\limits_{y=-\infty}^\infty \frac{y}{\sqrt{2 \pi} \sigma} e^{-(y - \mu_2)^2/(2 \sigma^2)}\ dx \right)}_{\mu_2} \underbrace{\left( \int\limits_{x=-\infty}^\infty \frac{z}{\sqrt{2 \pi} \sigma} e^{-(z - \mu_3)^2/(2 \sigma^2)}\ dx \right)}_{\mu_3}$$
$$= \mu_1 \mu_2 \mu_3,$$
independent of the variances. Note that the three variances could differ from each other and you'd get the same result.