Question

What is the expected value of a constant variable?

Answer 1

To solve our problem, we will first understand the difference between the terms ‘constant’ and variables. Constants are the terms that have a fixed value on the number line, that is, they cannot take any other value other than the value already assigned to them. Whereas a variable can take any arbitrary value from the number line system.

Complete step-by-step solution:
Now, by the term “a constant variable”, the question mean that there is a variable, say ‘x’. The property of this variable ‘x’ is that it can take only a certain constant value, say ‘p’. So, we need to find the expected value of this value under this condition.
Mathematically, our above statement could be written as:
$\Rightarrow E\left( x \right)=p$
Where, the function in ‘E’ represents the expected value of our variable ‘x’. This statement is
correct and makes sense because the variable ‘x’ can take only one value and so on “average”,
it assumes that value as well. Let us understand this with the help of an example. Let us say for the time being that ‘x’ equals constant 10. Then, we can create a sample of say size 20 in ‘x’. But in that sample all the twenty terms, that is,
${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}..................{{x}_{20}}$ will be equal to 10, as ‘x’ cannot take any other value. Then, the average or expected value of ‘x’ will be equal to:
$\begin{aligned} & \Rightarrow E\left( x \right)=\dfrac{\sum\limits_{i=1}^{20}{{{x}_{i}}}}{\sum\limits_{i=1}^{20}{1}} \\\ & \Rightarrow E\left( x \right)=\dfrac{10+10+10+...........+10}{20} \\\ & \therefore E\left( x \right)=10 \\\ \end{aligned}$
Hence, from our above observation, the expected value of a constant variable is the constant itself.

Note: The term used in our question, a “constant variable” has no significance as such. But, it has a certain degree of ambiguity to itself. Before finding the values or using these terms in a proof, one should first of all completely and properly understand the meaning of such terms and then proceed ahead in the problem.