EX 3.22 vehicle diagnostic test cost

• Vehicle Diagnostic Test Cost
  • Cost Function:
    • The cost $h(X)$ of the diagnostic test is given by:
    • $h(X)=20+3X+0.5X^2$
  • Variables:
    • $X$: The number of cylinders in the vehicle’s engine (random variable)
    • $Y=h(X)$: The cost of the diagnostic test (random variable)
• Probability Distributions:
  • pmf of $X$: - The probability mass function (pmf) for $X$
  • Derived pmf of $Y$: - The pmf of $Y$ derived from the cost function $h(X)$

x	4	6	8
p(x)	.5	.3	.2

implies

y	40	56	76
p(y)	.5	.3	.2

With $D^{\ast }$ denoting possible values of $Y$ ,

$$\begin{array}{rl}E(Y)& =E\left[h(X)\right]=\sum _{D^{\ast }}y\cdot p(y)\\ & =(40)(0.5)+(56)(0.3)+(76)(0.2)\\ & =h(4)\cdot (0.5)+h(6)\cdot (0.3)+h(8)\cdot (0.2)\\ & =\sum _{D}h(x)\cdot p(x)\end{array}\text{\hspace{1em}(3.11)}$$

• Expected Value of $Y$ Without Determining pmf
  • According to Equation (3.11):
    • It is unnecessary to determine the pmf of $Y$ to calculate $E(Y)$
  • Reasoning:
    • The desired expected value $E(Y)$ is a weighted average of the possible $h(x)$ values (instead of the $x$ values)
    • This approach simplifies the calculation by focusing on $h(x)$, the cost function, rather than $Y$‘s distribution