Chapter 3.4, Problem 55E

Suppose that Y is a binomial random variable with n > 2 trials and success probability p. Use the technique presented in Theorem 3.7 and the fact that E{Y(Y − 1)(Y − 2)} = E(Y³) − 3E(Y²) + 2E(Y) to derive E(Y³).

To determine

Derive $E\left(Y^3\right)$.

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then $E\left(Y^2\right)=n\left(n-1\right)p+np$

The random variable Y is binomial random variable having $n>2$ trails and probability os success p. Also, $E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}=E\left(Y^3\right)-3E\left(Y^2\right)+2E\left(Y\right)$.

Consider the term $E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}$.

$\begin{array}{c}E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}={\displaystyle \sum _{y=0}^n\left[y\left(y-1\right)\left(y-2\right)\right]p\left(y\right)}\\ ={\displaystyle \sum _{y=0}^n\left[y\left(y-1\right)\left(y-2\right)\right]\frac{n!}{y!\left(n-y\right)!}{p}^y{\left(1-p\right)}^{n-y}}\\ ={\displaystyle \sum _{y=3}^n\left[y\left(y-1\right)\left(y-2\right)\right]\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{y\left(y-1\right)\left(y-2\right)\left(y-3\right)!\left(n-y\right)!}{p}^y{\left(1-p\right)}^{n-y}}\\ ={\displaystyle \sum _{y=0}^n\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(y-3\right)!\left(n-y\right)!}{p}^y{\left(1-p\right)}^{n-y}}\end{array}$

Add and subtract 3 to the term $\left(n-y\right)!$ in the expression.

$\begin{array}{c}E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}={\displaystyle \sum _{y=0}^n\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(y-3\right)!\left(n-y-3+3\right)!}{p}^y{\left(1-p\right)}^{n-y}}\\ ={\displaystyle \sum _{y=0}^n\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(y-3\right)!\left(n-3-\left(y-3\right)\right)!}{p}^y{\left(1-p\right)}^{n-y}}\end{array}$

Suppose that $z=y-3$ this implies that $y=z+3$.

$\begin{array}{c}E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}={\displaystyle \sum _{y=0}^n\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(y-3\right)!\left(n-3-\left(y-3\right)\right)!}{p}^y{\left(1-p\right)}^{n-y}}\\ =n\left(n-1\right)\left(n-2\right){\displaystyle \sum _{z=0}^{n-3}\frac{\left(n-3\right)!}{\left(z\right)!\left(n-3-\left(z\right)\right)!}{p}^{z+3}{\left(1-p\right)}^{n-\left(z+3\right)}}\\ =n\left(n-1\right)\left(n-2\right){\displaystyle \sum _{z=0}^{n-3}\left(\begin{array}{l}n-3\\ z\end{array}\right)p^z{p}^3{\left(1-p\right)}^{n-3-z}}\\ =n\left(n-1\right)\left(n-2\right)p^3{\displaystyle \sum _{z=0}^{n-3}\left(\begin{array}{l}n-3\\ z\end{array}\right)p^z{\left(1-p\right)}^{n-3-z}}\end{array}$

It is obvious that, $p\left(z\right)=\left(\begin{array}{l}n-3\\ z\end{array}\right)p^z{\left(1-p\right)}^{n-3-z}$ is a binomial function which is based on $\left(n-3\right)$ number of trails. Then, ${\displaystyle \sum _{z}p\left(z\right)}=1$. Thus, the value of function ${\displaystyle \sum _{z=0}^{n-3}\left(\begin{array}{l}n-3\\ z\end{array}\right)p^z{\left(1-p\right)}^{n-3-z}}$ is 1.

That is,

$\begin{array}{c}E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}=n\left(n-1\right)\left(n-2\right)p^3\left(1\right)\\ =n\left(n-1\right)\left(n-2\right)p^3\end{array}$

The value of $E\left(Y^3\right)$ is,

$\begin{array}{c}E\left(Y^3\right)=3E\left(Y^2\right)-E\left\{Y\left(Y-1\right)\left(Y-2\right)\right\}-2E\left(Y\right)\\ =3\left(n\left(n-1\right)p+np\right)-\left(n\left(n-1\right)\left(n-2\right)p^3\right)-2np\left[\text{Using}\text{theorem}3.7\text{and}E\left(Y\right)=np\right]\\ =3n\left(n-1\right)p+3np-n\left(n-1\right)\left(n-2\right)p^3-2np\\ =3n\left(n-1\right)p-n\left(n-1\right)\left(n-2\right)p^3+np\end{array}$

Hence, the expression is $E\left(Y^3\right)=3n\left(n-1\right)p-n\left(n-1\right)\left(n-2\right)p^3+np$.