A random variable Y has a Probability Density Function (PDF) of $f(y) = \frac{1}{4}y^2$ for $ 0 \leq y \leq 2$. Assume Y measures the rate of electrical problems a building gets. To Fix these problems the landlord pays C in dollars, with $C(y)=10y^2-2$. How do I calculate the monthly cost of these problems. Knowing that the mean of $Y$ is $1$ using $\int_{0}^{2}y\cdot \frac{1}{4}\cdot y^2$.
Getting the average cost using probability density function
0
$\begingroup$
probability
average
2 Answers
1
Do you mean the expected monthly cost?
You know that the mean is obtained by integration: $$~\mathsf E(Y) = \int_0^2 y\,f(y)\mathop{\rm d} y~$$
Similarly the expected cost is:
$$\begin{align}\mathsf E(C(Y)) ~&=~ \int_0^2 C(y)\,f(y)\mathop{\rm d} y \\[1ex] &=~ \int_0^2 (10y^2-2)\,(\tfrac 14 y^2)\mathop{\rm d} y \end{align}$$
0
Why not $\int_{0}^{2}(10y^2-2)\frac{1}{4}y^2dy$? When you do the integral, you should get, where $n$ is even digit, $\frac{nn}{n-1}$.