1. ## exponential distribution question

I'm having trouble setting up the following equation:
The length of time $\displaystyle Y$ necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $\displaystyle C=100+4Y+3Y^2$ relates to the cost of $\displaystyle C$. Find the mean of $\displaystyle C$.

initially would it be:

$\displaystyle \frac{1}{\beta}e^{-\frac{y}{\beta}} = \frac{1}{10}e^{-\frac{y}{10}}$

$\displaystyle \int^{\infty}_0 y \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy$

then just integrate out?

2. Originally Posted by lllll
I'm having trouble setting up the following equation:
The length of time $\displaystyle Y$ necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $\displaystyle C=100+4Y+3Y^2$ relates to the cost of $\displaystyle C$. Find the mean of $\displaystyle C$.

initially would it be:

$\displaystyle \frac{1}{\beta}e^{-\frac{y}{\beta}} = \frac{1}{10}e^{-\frac{y}{10}}$

$\displaystyle \int^{\infty}_0 y \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy$

then just integrate out?
No:

1. It's wrong. It should be $\displaystyle \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy$ (note the missing y at the front).

2. There's an easier way:

$\displaystyle E[C] = E[100+4Y+3Y^2] = E[100] + 4 E[Y] + 3 E[Y^2] = 100 + 4 (10) + E[Y^2]$.

Now recall that $\displaystyle Var[Y] = E[Y^2] - (E[Y])^2$. And you should know that for an exponential distribution, $\displaystyle Var[Y] = (E[Y])^2$. Therefore $\displaystyle E[Y^2] = 2 (E[Y])^2 = 200$.

If you want you can also get $\displaystyle E[Y^2]$ by integration.

3. Originally Posted by mr fantastic
No:

1. It's wrong. It should be $\displaystyle \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy$ (note the missing y at the front).

2. There's an easier way:

$\displaystyle E[C] = E[100+4Y+3Y^2] = E[100] + 4 E[Y] + 3 E[Y^2] = 100 + 4 (10) + E[Y^2]$.

Now recall that $\displaystyle Var[Y] = E[Y^2] - (E[Y])^2$. And you should know that for an exponential distribution, $\displaystyle Var[Y] = (E[Y])^2$. Therefore $\displaystyle E[Y^2] = {\color{red}2} (E[Y])^2 = 200$.

If you want you can also get $\displaystyle E[Y^2]$ by integration.
Where did the 2 come from?

4. Nevermind, I figured it out. But on a side note would finding the Variance of the cost follow a similar approach?

5. Originally Posted by lllll
Nevermind, I figured it out. But on a side note would finding the Variance of the cost follow a similar approach?
Yes, but it'll be more tedious and there's no avoiding some integration:

$\displaystyle Var[C] = E[C^2] - (E[C])^2$.

You already know E[C] so you have half the answer. The easy half, unfortunately.

$\displaystyle C^2 = (100 + 4Y + 3Y^2)^2 = \, .......$.

To get $\displaystyle E[Y^3]$ and $\displaystyle E[Y^4]$ you'll have to integrate. (Alternatively you could use the moment generating function).

6. if you were to integrate it out would it be:

$\displaystyle \int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy$

$\displaystyle =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000y+660y^2+24y^3+9y^4) \ dy$

$\displaystyle =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000(10)+660(220)+24y^3+9y^4) \ dy$ ?

7. Originally Posted by lllll
if you were to integrate it out would it be:

$\displaystyle \int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy$

$\displaystyle =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000y+660y^2+24y^3+9y^4) \ dy$

$\displaystyle =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000(10)+660(220)+24y^3+9y^4) \ dy$ ?
No. $\displaystyle E \left[C^2\right] = \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy$. This time you have a y^2 at the front that shouldn't be there.

But why do all this when all you really need to do is use integration to find these two things:

$\displaystyle E[Y^3] = \int^{\infty}_0 y^3 \, \frac{1}{10}e^{-\frac{y}{10}} ~ dy$

$\displaystyle E[Y^4] = \int^{\infty}_0 y^4 \, \frac{1}{10}e^{-\frac{y}{10}} ~ dy$