# Exponential distribution

1. Nov 7, 2007

1. The problem statement, all variables and given/known data

component has lifetime X that is exponentialy distibuted with parameter gamma.

a) if the cost per unit time is a constant, c, what is the expectec cost of its lifetime?
b) if c is not constant, and=c(1-.5e^(a*x) such that a<0. (aka it costs more over time) what is its cost inrespect to its lifetime?

2. Relevant equations

f(x)=gamma*e^(-gamma*x)
E(X)=1/gamma
Var(X)=1/gamma^2

3. The attempt at a solution

i thitnk its just

a) c/gamma
b) (c(1-.5e^(ax)))/gamma

but that seems to easy...

Last edited: Nov 7, 2007
2. Nov 7, 2007

### EnumaElish

To find E[cost] you need to integrate the relevant unit cost (per unit time) weighted by f(x) over the entire domain of X.

3. Nov 8, 2007

what do you mean "weighted by f(x)"?

so integrate c...so cx

and integrate (c(1-.5e^(ax)))/gamma
which is cx-(.5/a)e^(ax)

where x is time

but dunno what to do now

4. Nov 8, 2007

### HallsofIvy

Staff Emeritus
Well, you didn't actually integrate "over the entire domain of X"! You found the anti-derivative which is not quite the same thing. For one thing, the "expected value" does not depend upon x. "Weighted by f(x)" means, here, "multiplied by f(x)" before you integrate which is apparently what you did. What is the entire domain of X?

5. Nov 8, 2007

since x is t then 0 to infinity?

6. Nov 8, 2007

f(x)=gamma*e^(-gamma*x)

so would i go

c*gamma*e^(-gamma*x)
and
(c(1-.5e^(ax)))/gamma* gamma*e^(-gamma*x)

and then integrate form x=0 to infinity?

7. Nov 8, 2007

### EnumaElish

Correct.

8. Nov 8, 2007

c*gamma*e^(-gamma*x)

becomes (-C*gamma^2)*e^(-gamma*x) from 0 to infin
so at x=infin that equals 0
at x=0, it is -c*gamma^2

so c*gamma^2 is my first answer?

and then (c(1-.5e^(ax)))/gamma* gamma*e^(-gamma*x)
becomes c(1-.5e^(ax)))*e^(-gamma*x)
so C(e^(-gamma*x)-e^(-gamma*x+ax))
which becomes Ce^(-gamma*x)-Ce^(-gamma*x+ax)
and again...Ce^(-gamma*x)-Ce^(x(a-gamma))
so integrating that i get

(-C/gamma)e^(-gamma*x)-(C/(a-gamma)e^(x(a-gamma))) from 0 to infin
again, at infin it equals 0
at 0, the first expression is -C/gamma and the 2nd is C/(a-gamma)
so that means 0-(-C/gamma -C/(a-gamma) )
so my answer is C/gamma +C/(a-gamma)?????

9. Nov 8, 2007