# Math Help - Cummulative distribution function

1. ## Cummulative distribution function

Let X be an exponentially distributed stocastic variable with probability density:

f(x;b) = (1/b)e^(-1/b), for x larger than or equal to 0, else 0)

How can I find the cummulative distribution function F(x)?

Thanks!

2. Originally Posted by gralla55
Let X be an exponentially distributed stocastic variable with probability density:

f(x;b) = (1/b)e^(-1/b), for x larger than or equal to 0, else 0) Mr F says: Where is the x-dependency in your pdf!?

How can I find the cummulative distribution function F(x)?

Thanks!
You integrate the pdf. Where are you stuck?

3. Sorry, correct pdf is of course:

f(x;b) = (1/b)e^(-x/b)

So if I intergrate with regards to x it should give:

-1/x * e^(-x/b)

I think, and then I intergrate with regards to b?

4. Originally Posted by gralla55
Sorry, correct pdf is of course:

f(x;b) = (1/b)e^(-x/b)

So if I intergrate with regards to x it should give:

-1/x * e^(-x/b)

I think, and then I intergrate with regards to b?
F(x) = 0 when x < 0 and $\int_0^x f(t;b) \, dt$ for x > 0.

5. I don't get how "t" is suddently in the picture, but if I continue to intergrate with regards to "b", I get (b/x^2)e^(-x/b)...

6. Originally Posted by gralla55
I don't get how "t" is suddently in the picture, but if I continue to intergrate with regards to "b", I get (b/x^2)e^(-x/b)...
t is a dummy variable since you have to get a function of x after integrating. Replace x with t in f(x;b) and then integrate wrt t like I said. b is a parameter, not a variable. You do not integrate wrt b.

7. So, the integral should be -1/t * e^(-t/b)?

8. no, you are integrating wrt t
and t is just a dummy variable
you cannot integrate x wrt x
x is a limit in the integration

9. I'm still confused... =(

f(x;b) = (1/b)e^(-x/b) , repacing x with t

f(t;b) = (1/b)e^(-t/b), intergrating wrt t

F(t) = -1/t * e^(-t/b), adding limits?

F(x) = -1/x * e^(-x/b) - lim x->0 ( -1/x * e^(-0/b) ) ?

10. Originally Posted by gralla55
I'm still confused... =(

f(x;b) = (1/b)e^(-x/b) , repacing x with t

f(t;b) = (1/b)e^(-t/b), intergrating wrt t

F(t) = -1/t * e^(-t/b), adding limits?

F(x) = -1/x * e^(-x/b) - lim x->0 ( -1/x * e^(-0/b) ) ?
I don't see what the trouble can possibly be.

$F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt$ for x > 0 and you should surely know how to evaluate this integral!

11. Originally Posted by mr fantastic
I don't see what the trouble can possibly be.

$F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt$ for x > 0 and you should surely know how to evaluate this integral!
I wouldn't be that confident.
I have a calculus three student who didn't know what the square root of 36 was.
And I'm confident she couldn't integrate this either.

12. Haha, well I do know the square root of 36.

The anti-derivative of e^kx is (1/k)e^kx. So the antiderivative when k = -1/b should be -b*e^(-x/b). Multiplying with 1/b negates the b and I'm left with just:

-e^(-x/b)

All right, for some reason I got the antiderivative wrong the last time, which made evaluating the intergral alot harder than it should have been... so proceeding to evaluate:

(-e^(-x/b)) - (-e^(0/b) = (-e^(-x/b)) + 1 = 1 - e^(-x/b)

Is this correct?

13. Originally Posted by gralla55
Haha, well I do know the square root of 36.

The anti-derivative of e^kx is (1/k)e^kx. So the antiderivative when k = -1/b should be -b*e^(-x/b). Multiplying with 1/b negates the b and I'm left with just:

-e^(-x/b)

All right, for some reason I got the antiderivative wrong the last time, which made evaluating the intergral alot harder than it should have been... so proceeding to evaluate:

(-e^(-x/b)) - (-e^(0/b) = (-e^(-x/b)) + 1 = 1 - e^(-x/b)

Is this correct?

14. Finally!

Now to prove that the exponential distribution is memoryless, that is:

P(X > x0 + x|X > x0) = P(X > x)

Which should be simple enough, I'll just have to do something to the left side so that it becomes the right?

15. Originally Posted by gralla55
Finally!

Now to prove that the exponential distribution is memoryless, that is:

P(X > x0 + x|X > x0) = P(X > x)

Which should be simple enough, I'll just have to do something to the left side so that it becomes the right?
$\Pr(X > x_0 + x | X > x_0) = \frac{\Pr(X > x_0 + x)}{\Pr(X > x_0)}$.

Review Bayes Theorem. Get the required probabilities by integrating the pdf.