Thread: Exponential dist expected value help

1. Exponential dist expected value help

Q: The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $\displaystyle C=100+40Y+3Y^{2}$ relates the cost C of completing this operation to the square of the time to completion. Find the mean and variance of C.

My work:

Since $\displaystyle V(C)=E(C^{2})-[E(C)]^{2}$ I need to solve for $\displaystyle E(C^{2})$ and $\displaystyle [E(C)]^{2}$ first.

$\displaystyle E(C^{2})=9E(Y^{4})+240E(Y^{3})+2200E(Y^{2})+8000E( Y)+10000$

So, I started with finding $\displaystyle E(Y^{4})$,

$\displaystyle E(Y^{4})=\int_{0}^{\infty}y^{4}e^{\frac{-y}{10}}dy$, by u-sub we have $\displaystyle u=\frac{y}{10}\Rightarrow\\10du=dy$, so $\displaystyle 10\int_{0}^{-\infty}u^{4}e^{-u}du= 10\int_{0}^{-\infty}u^{3-1}e^{-u}du=10\Gamma(3)$

and so, if that is correct, shouldn't $\displaystyle E(Y^{3})=10\Gamma(2)$ and so, down to $\displaystyle E(Y)=\beta=10$?

Nevertheless, I keep ending up with a negative variance when I use this method, since my $\displaystyle [E(C)]^{2}$ gets so large. Do I just have to crank out every integrand by way of integration by parts instead of using the gamma function?

Some direction would be great,

Thanks

2. Originally Posted by Danneedshelp
Q: The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $\displaystyle C=100+40Y+3Y^{2}$ relates the cost C of completing this operation to the square of the time to completion. Find the mean and variance of C.

My work:

Since $\displaystyle V(C)=E(C^{2})-[E(C)]^{2}$ I need to solve for $\displaystyle E(C^{2})$ and $\displaystyle [E(C)]^{2}$ first.

$\displaystyle E(C^{2})=9E(Y^{4})+240E(Y^{3})+2200E(Y^{2})+8000E( Y)+10000$

So, I started with finding $\displaystyle E(Y^{4})$,

$\displaystyle E(Y^{4})=\int_{0}^{\infty}y^{4}e^{\frac{-y}{10}}dy$, by u-sub we have $\displaystyle u=\frac{y}{10}\Rightarrow\\10du=dy$, so $\displaystyle 10\int_{0}^{-\infty}u^{4}e^{-u}du= 10\int_{0}^{-\infty}u^{3-1}e^{-u}du=10\Gamma(3)$

and so, if that is correct, shouldn't $\displaystyle E(Y^{3})=10\Gamma(2)$ and so, down to $\displaystyle E(Y)=\beta=10$?

Nevertheless, I keep ending up with a negative variance when I use this method, since my $\displaystyle [E(C)]^{2}$ gets so large. Do I just have to crank out every integrand by way of integration by parts instead of using the gamma function?

Some direction would be great,

Thanks
$\displaystyle E(Y^n) = \int_0^{+\infty} y^n e^{-y/10} \, dy = 10^{n+1} \Gamma(n+1)$.