Find the variance of a quadratic cost function

Question

Given a beta distribution $f (y)=2 (1-y)$ and $C=10+20Y+4Y^2$, find $\mbox{Var} [C]$.

I already know $\alpha=1$ and $\beta=2$, so $\mu =\frac13$ and $\sigma^2=\frac{1}{18}$.

I also found that $\mathbb{E} (C)=\frac{52}{3}$ using the fact that $\mathbb{E} (aY+b)=a\mathbb{E} (Y)+b$. I tried calculating $\mathbb{E} (C^2)-[\mathbb{E} (C)]^2$ but got stuck because didn't know how to do $E [aY^3]$ or $E [aY^4]$.

Can anyone help? This is problem 4.97 in Wackerly 6th ed, btw.

Answer 1

\begin{align} \mathbb{E}[Y^n]&= \int_0^1 y^nf(y) dy \\ &= 2\int_0^1y^n (1-y) dy \\ &= 2\int_0^1y^n -y^{n+1} dy \\ &= 2 \left[ \frac{1}{n+1}-\frac{1}{n+2}\right]\\ &=\frac{2}{(n+1)(n+2)} \end{align}

Also use the properties that $\mathbb{E}[aY^n] = a\mathbb{E}[Y^n]$, you should be able to solve the problem.

Answer 2

In matrix form,

$$C = \begin{bmatrix} 10\\ 20\\ 4\end{bmatrix}^{\top} \begin{bmatrix} 1\\ Y\\ Y^2\end{bmatrix}$$

It's not hard to show that

$$\begin{array}{rl} \mbox{Var} (C) &= \begin{bmatrix} 10\\ 20\\ 4\end{bmatrix}^{\top} \left( \begin{bmatrix} \mathbb E (1) & \mathbb E (Y) & \mathbb E (Y^2)\\ \mathbb E (Y) & \mathbb E (Y^2) & \mathbb E (Y^3)\\ \mathbb E (Y^2) & \mathbb E (Y^3) & \mathbb E (Y^4)\end{bmatrix} - \begin{bmatrix} \mathbb E (1) & \mathbb E (Y) & \mathbb E (Y^2)\\ \mathbb E (Y) & \mathbb E (Y) \, \mathbb E (Y) & \mathbb E (Y) \, \mathbb E (Y^2)\\ \mathbb E (Y^2) & \mathbb E (Y^2) \, \mathbb E (Y) & \mathbb E (Y^2) \, \mathbb E (Y^2)\end{bmatrix} \right) \begin{bmatrix} 10\\ 20\\ 4\end{bmatrix}\\ &= \begin{bmatrix} 10\\ 20\\ 4\end{bmatrix}^{\top} \begin{bmatrix} 0 & 0 & 0\\ 0 & \mathbb E (Y^2) - \mathbb E (Y) \, \mathbb E (Y) & \mathbb E (Y^3) - \mathbb E (Y) \, \mathbb E (Y^2)\\ 0 & \mathbb E (Y^3) - \mathbb E (Y^2) \, \mathbb E (Y) & \mathbb E (Y^4) - \mathbb E (Y^2) \, \mathbb E (Y^2)\end{bmatrix} \begin{bmatrix} 10\\ 20\\ 4\end{bmatrix}\\ &= \begin{bmatrix} 20\\ 4\end{bmatrix}^{\top} \begin{bmatrix} \mbox{Var} (Y) & \mathbb E (Y^3) - \mathbb E (Y) \, \mathbb E (Y^2)\\ \mathbb E (Y^3) - \mathbb E (Y^2) \, \mathbb E (Y) & \mathbb E (Y^4) - \mathbb E (Y^2) \, \mathbb E (Y^2)\end{bmatrix} \begin{bmatrix} 20\\ 4\end{bmatrix}\\ &= 400 \, \mbox{Var} (Y) + 160 \left( \mathbb E (Y^3) - \mathbb E (Y) \, \mathbb E (Y^2) \right) + 16 \left( \mathbb E (Y^4) - \mathbb E (Y^2) \, \mathbb E (Y^2) \right)\end{array}$$

One can now use Siong's formula for $\mathbb E (Y^n)$ to compute the variance.