# exponential distribution question

Printable View

• Sep 22nd 2008, 06:26 PM
lllll
exponential distribution question
I'm having trouble setting up the following equation:
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 $C=100+4Y+3Y^2$ relates to the cost of $C$. Find the mean of $C$.

initially would it be:

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

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

then just integrate out?
• Sep 22nd 2008, 07:54 PM
mr fantastic
Quote:

Originally Posted by lllll
I'm having trouble setting up the following equation:
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 $C=100+4Y+3Y^2$ relates to the cost of $C$. Find the mean of $C$.

initially would it be:

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

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

then just integrate out?

No:

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

2. There's an easier way:

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

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

If you want you can also get $E[Y^2]$ by integration.
• Sep 22nd 2008, 09:51 PM
lllll
Quote:

Originally Posted by mr fantastic
No:

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

2. There's an easier way:

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

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

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

Where did the 2 come from?
• Sep 22nd 2008, 11:11 PM
lllll
Nevermind, I figured it out. But on a side note would finding the Variance of the cost follow a similar approach?
• Sep 22nd 2008, 11:22 PM
mr fantastic
Quote:

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:

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

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

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

To get $E[Y^3]$ and $E[Y^4]$ you'll have to integrate. (Alternatively you could use the moment generating function).
• Sep 23rd 2008, 12:03 AM
lllll
if you were to integrate it out would it be:

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

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

$=\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000(10)+660(220)+24y^3+9y^4) \ dy$ ?
• Sep 23rd 2008, 12:15 AM
mr fantastic
Quote:

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

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

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

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

No. $E \left[C^2\right] = \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy$. 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:

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

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