Question

If E (X + C) = 8 and E (X – C) = 12, then the value of C is
(a) – 2
(b) 4
(c) – 4
(d) 2

Answer 1

Hint: Here, we have to calculate the value of the constant C. We are given the expectation of X + C and X – C as 8 and 12 respectively. We will expand their expectations so that we can easily eliminate E(X) and calculate C. We will expand the two by opening the brackets as \[E\left( X+C \right)=E\left( X \right)+C\] and calculate C by using the elimination method.

Complete step-by-step answer:
We are given that,
\[E\left( X+C \right)=8\]
This means that we have the expectation of X + C as 8.
We know that,
\[E\left( X+C \right)=E\left( X \right)+C\]
So, from this, we can write
\[E\left( X+C \right)=E\left( X \right)+C=8\]
\[\Rightarrow E\left( X \right)+C=8......\left( i \right)\]
Now, we also have that
\[E\left( X-C \right)=12\]
This means that we have the expectation of X – C as 12.
We know that,
\[E\left( X-C \right)=E\left( X \right)-C\]
By using this formula for E(X – C) = 12, we get,
\[E\left( X-C \right)=E\left( X \right)-C=12\]
\[\Rightarrow E\left( X \right)-C=12......\left( ii \right)\]
Now, we will use the equation (i) and (ii) to solve for the value of C. We will use the elimination method to eliminate E(X) and find C.
Equation (i) – Equation (ii)
\[\begin{align}
  & E\left( X \right)+C=8 \\
 & E\left( X \right)-C=12 \\
 & \underline{-\text{ }+\text{ }-} \\
 & \text{ }2C=-4 \\
\end{align}\]
\[\Rightarrow 2C=-4\]
\[\Rightarrow C=\dfrac{-4}{2}\]
\[\Rightarrow C=-2\]
So, we get the value of C as – 2.
So, the correct answer is “Option A”.

Note: While eliminating, keep the correct sign, else students can often get the wrong answer, i.e. \[E\left( X+C \right)\ne E\left( X \right)+E\left( C \right).\] Remember always the expectation of constant is always constant only. While expanding, keep track of negative sign, i.e. \[E\left( X-C \right)=E\left( X \right)-C.\]