Example 8-6 Section
Suppose the p.m.f. of the discrete random variable \(X\) is:
| x | 0 | 1 | 2 | 3 |
| f(x) | 0.2 | 0.1 | 0.4 | 0.3 |
What is \(E(2)\)? What is \(E(X)\)? And, what is \(E(2X)\)?
This example leads us to a very helpful theorem.
- If \(c\) is a constant, then \(E(c)=c\)
- If \(c\) is a constant and \(u\) is a function, then:
\(E[cu(X)]=cE[u(X)]\)
Proof
Example 8-7 Section
Let's return to the same discrete random variable \(X\). That is, suppose the p.m.f. of the random variable \(X\) is:
| x | 0 | 1 | 2 | 3 |
| f(x) | 0.2 | 0.1 | 0.4 | 0.3 |
It can be easily shown that \(E(X^2)=4.4\). What is \(E(2X+3X^2)\)?
This example again leads us to a very helpful theorem.
Let \(c_1\) and \(c_2\) be constants and \(u_1\) and \(u_2\) be functions. Then, when the mathematical expectation \(E\) exists, it satisfies the following property:
\(E[c_1 u_1(X)+c_2 u_2(X)]=c_1E[u_1(X)]+c_2E[u_2(X)]\)
Before we look at the proof, it should be noted that the above property can be extended to more than two terms. That is:
\(E\left[\sum\limits_{i=1}^k c_i u_i(X)\right]=\sum\limits_{i=1}^k c_i E[u_i(X)]\)
Proof
Example 8-8 Section
Suppose the p.m.f. of the discrete random variable \(X\) is:
| x | 0 | 1 | 2 | 3 |
| f(x) | 0.2 | 0.1 | 0.4 | 0.3 |
In the previous examples, we determined that \(E(X)=1.8\) and \(E(X^2)=4.4\). Knowing that, what is \(E(4X^2)\) and \(E(3X+2X^2)\)?
Using part (b) of the first theorem, we can determine that:
\(E(4X^2)=4E(X^2)=4(4.4)=17.6\)
And using the second theorem, we can determine that:
\(E(3X+2X^2)=3E(X)+2E(X^2)=3(1.8)+2(4.4)=14.2\)
Example 8-9 Section
Let \(u(X)=(X-c)^2\) where \(c\) is a constant. Suppose \(E[(X-c)^2]\) exists. Find the value of \(c\) that minimizes \(E[(X-c)^2]\).
Note that the expectations \(E(X)\) and \(E[(X-E(X))^2]\) are so important that they deserve special attention.