Expected Value and Variance Calculations using PCE

Expected Value

$$\boxed{{\mathbb{E}}_{\xi }[X]=q_0}$$

Derivation

$${\mathbb{E}}_{\xi }[X]={\mathbb{E}}_{\xi }\left[\sum _{i=0}^{\infty }{q}_i{\Phi }_i(\xi )\right]=\sum _{i=0}^{\infty }{q}_i{\mathbb{E}}_{\xi }\left[{\Phi }_i(\xi )\right]$$

Since ${\Phi }_0=1$ and all higher-order polynomials are constructed to have zero mean (i.e., $\mathbb{E}\left[{\Phi }_i(\xi )\right]=0$ for $i>0$ ), we get:

$${\mathbb{E}}_{\xi }[X]=q_0\cdot 1+\sum _{i=1}^{\infty }{q}_i\cdot 0=q_0$$

Variance

$$\boxed{{\mathbb{V}}_{\xi }[X]=\sum _{i=1}^{\infty }{q}_i^2}$$

Derivation

$${\mathbb{V}}_{\xi }[X]={\mathbb{E}}_{\xi }\left[{\left(X-{\mathbb{E}}_{\xi }[X]\right)}^2\right]={\mathbb{E}}_{\xi }\left[{\left(\sum _{i=0}^{\infty }{q}_i{\Phi }_i(\xi )-q_0\right)}^2\right]$$

Simplifying:

$${\mathbb{V}}_{\xi }[X]={\mathbb{E}}_{\xi }\left[{\left(\sum _{i=1}^{\infty }{q}_i{\Phi }_i(\xi )\right)}^2\right]={\mathbb{E}}_{\xi }\left[\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }{q}_i{q}_j{\Phi }_i(\xi ){\Phi }_j(\xi )\right]$$

Using orthonormality:

$${\mathbb{V}}_{\xi }[X]=\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }{q}_i{q}_j{\mathbb{E}}_{\xi }\left[{\Phi }_i(\xi ){\Phi }_j(\xi )\right]=\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }{q}_i{q}_j{\delta }_{ij}=\sum _{i=1}^{\infty }{q}_i^2$$

Assuming $\xi$ is orthonormal.