Variance of a polynomial series from a uniform distribution

Question

I intend to derive the variance of $Z$: $$Z \equiv \alpha_0+\alpha_1X+\alpha_2X^2+\dots + \alpha_MX^M = \sum_{m=0}^{M}\alpha_mX^m $$ for some $0 < M < \infty$ where each $\alpha_m \in \mathbb{R}$ and $X \in [0,1]$ is a standardized random uniform variable.

For $M=2 \mbox{, } Z= \alpha_0+\alpha_1X+\alpha_2X^2$. Therefore, using the relation: $$ \text{Var}(\alpha_1X+\alpha_2Y) =\alpha_1^2\text{Var}(X) +\alpha_2^2\text{Var}(Y) + 2\alpha_1\alpha_2\text{Cov}(X, Y)$$ I found : $$\text{Var(Z)} = \frac{\alpha_1^2}{12}+\frac{4 \alpha_2^2}{45} + \frac{2 \alpha_1\alpha_1}{12}$$

So, I am demanding help in deriving the variance for any $M$ or at least for $M=3,4,5 \text{ and } 6$

Answer 1

We have that $$ \operatorname{Var}\bigg(\sum_{m=1}^Ma_mX^m\biggr) =\sum_{m=1}^Ma_m^2\operatorname{Var}X^m+2\sum_{l=1}^{M-1}\sum_{m=l+1}^Ma_la_m\operatorname{Cov}(X^l,X^m). $$

We need to calculate $$ \operatorname{Var}X^m=\operatorname EX^{2m}-(\operatorname EX^m)^2 $$ and $$ \operatorname{Cov}(X^l,X^m)=\operatorname EX^{l+m}-\operatorname EX^l\operatorname EX^m. $$

If $X\sim U(0,1)$, then $$ \operatorname E X^n=\int_0^1x^n\mathrm dx=\frac{1}{n+1}. $$

I hope this helps.