Solutions - 5240-04

(1) Define random variables for partitioning the 30 flips into groups of 10:

Let $A$ be the number of heads in the first 10 flips.

Let $B$ be the number of heads in the middle 10 flips.

Let $C$ be the number of heads in the last 10 flips.

Clearly, $A,B,C\sim \text{Bin}(10,\frac{1}{2})$ and are independent. Note that $X=A+B$ and $Y=B+C$.

---

(2) Compute $E[X]$ and $E[Y]$:

$$\begin{array}{c}E[X]=E[A+B]=E[A]+E[B]=5+5=10\\ \\ E[Y]=E[B+C]=E[B]+E[C]=5+5=10\end{array}$$

---

(3) Compute $E[XY]$:

$$\begin{array}{c}E[XY]=E\left[(A+B)(B+C)\right]\\ \\ \gg \gg E[AB]+E[AC]+E[BC]+E\left[B^2\right]\\ \\ \gg \gg E[A]E[B]+E[A]E[C]+E[B]E[C]+E\left[B^2\right]=75+E\left[B^2\right]\end{array}$$

---

(4) Compute $E\left[B^2\right]$:

Since $E[B]=5$ and $\text{Var}[B]=\frac{5}{2}$, we have $E\left[B^2\right]=5^2+\frac{5}{2}=\frac{55}{2}$.

Thus, $E[XY]=\frac{205}{2}$.

---

(5) Compute $\text{Cov}[X,Y]$:

$$\begin{array}{c}\text{Cov}[X,Y]=E[XY]-E[X]E[Y]=\frac{205}{2}-10\cdot 10=\frac{5}{2}\end{array}$$

---

(6) Compute $\rho [X,Y]$:

$$\begin{array}{c}\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}=\frac{\frac{5}{2}}{\sqrt{5\cdot 5}}\gg \gg \frac{1}{2}\end{array}$$