1. ## Exponential decay question

What is the memoryless property?

2. Originally Posted by bluebiro
Let X represent the life time of a washing machine with c.d.f. F(X) = 1 - exp (-X^ p) for values of X greater than 0, where p > 0 is a parameter. If I have been using the washing machine for X years without failured, is it now better, worse or the same quality as a new washing machine?

I've tried to apply the exponential model to this but i find the question quite confusing Any help/explanation would be greatly appreciated
Read 3.2 of this: Exponential distribution - Wikipedia, the free encyclopedia

3. let X be the lifetime of the washing machine

$f(x) = pe^{-px}$

$P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)}$

EDIT: $= \frac{P(X>x+t)}{P(X>x)} = \frac{e^{-p(x+t)}}{e^{-px}}$ $= \frac{e^{-px}e^{-pt}}{e^{-px}} = e^{-pt}$

notice that $P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt}$

so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new

4. Originally Posted by Random Variable
let X be the lifetime of the washing machine

$f(x) = pe^{-px}$

$P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)}$

$= \frac{P(X>x+t)}{P(X>x)} = \frac{pe^{-p(x+t)}}{pe^{-px}}$ $= \frac{pe^{-px}e^{-pt}}{pe^{-px}} = e^{-pt}$

notice that $P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt}$

so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new
You've explained this very clearly. Thank you!

5. I made two errors. But it doesn't change the final result.

$\frac{P(X>x+t)}{P(X>x)} = \frac{\int^{\infty}_{x+t} pe^{-px} \ dx} {\int^{\infty}_{x} pe^{-px} \ dx}$ $= \frac{e^{-p(x+t)}}{e^{-px}} = \frac{e^{-px}e^{-pt}}{e^{-px}}= e^{-pt}= P(X>t)$

6. Originally Posted by Random Variable
I made two errors. But it doesn't change the final result.

$\frac{P(X>x+t)}{P(X>x)} = \frac{\int^{\infty}_{x+t} pe^{-px} \ dx} {\int^{\infty}_{x} pe^{-px} \ dx}$ $= \frac{e^{-p(x+t)}}{e^{-px}} = \frac{e^{-px}e^{-pt}}{e^{-px}}= e^{-pt}= P(X>t)$

7. Originally Posted by Random Variable
let X be the lifetime of the washing machine

$f(x) = pe^{-px}$

$P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)}$

EDIT: $= \frac{P(X>x+t)}{P(X>x)} = \frac{e^{-p(x+t)}}{e^{-px}}$ $= \frac{e^{-px}e^{-pt}}{e^{-px}} = e^{-pt}$

notice that $P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt}$

so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new
I just realised that your working is for F(x) = 1 - exp(-X*p), not F(x) = 1 - exp(-X^p). Does the solution still hold?

8. Originally Posted by bluebiro
I just realised that your working is for F(x) = 1 - exp(-X*p), not F(x) = 1 - exp(-X^p). Does the solution still hold?
Why don't you try to answer that question yourself?

Use, what's right above and.....

$P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} =\frac{P(X>x+t)}{P(X>x)} =\frac{1-F(x+t)}{1-F(x)}$

see what happens.

9. Originally Posted by matheagle
Why don't you try to answer that question yourself?

Use, what's right above and.....

$P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} =\frac{P(X>x+t)}{P(X>x)} =\frac{1-F(x+t)}{1-F(x)}$

see what happens.
Good point. I'm just worried I'll have to differentiate something.

10. Originally Posted by bluebiro
Good point. I'm just worried I'll have to differentiate something.
One bird to another....
You can't be afraid to make a mistake, i.e, fly on your own.

Show your work here and people will check it for you.

11. Originally Posted by matheagle
One bird to another....
You can't be afraid to make a mistake, i.e, fly on your own.

Show your work here and people will check it for you.
OK. But just to check, instead of $f(x) = pe^{-px}$ will f(x) be equal to $xp^{-p{x-1}}.e^{-p^{x}}$?

12. $F(x)=1-e^{-x^p}$, so $f(x)=px^{p-1}e^{-x^p}$

(where x>0, so the cow doesn't have a cow)

It's $x^p$ not $p^x$

And you don't need f(x), just plug F(x) into the equation.

13. Originally Posted by matheagle
$F(x)=1-e^{-x^p}$, so $f(x)=px^{p-1}e^{-x^p}$

(where x>0, so the cow doesn't have a cow)

It's $x^p$ not $p^x$

And you don't need f(x), just plug F(x) into the equation.
I get:

$1+e^{-x^{p+t}}$ divided by $1+e^{-x^{p}}$

$e^0 +e^{-x^{p+t}}$ divided by $e^0 +e^{-x^p}$
$e^{-x^{p+t-p}}$
$e^{-x^{p}}$