# Cummulative distribution function

• Mar 18th 2010, 01:39 AM
gralla55
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!
• Mar 18th 2010, 01:43 AM
mr fantastic
Quote:

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?
• Mar 18th 2010, 01:58 AM
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?
• Mar 18th 2010, 01:59 AM
mr fantastic
Quote:

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 $\displaystyle \int_0^x f(t;b) \, dt$ for x > 0.
• Mar 18th 2010, 02:13 AM
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)...
• Mar 18th 2010, 02:16 AM
mr fantastic
Quote:

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.
• Mar 18th 2010, 02:20 AM
gralla55
So, the integral should be -1/t * e^(-t/b)?
• Mar 18th 2010, 07:38 AM
matheagle
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
• Mar 22nd 2010, 03:24 PM
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) ) ?
• Mar 22nd 2010, 07:59 PM
mr fantastic
Quote:

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.

$\displaystyle F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt$ for x > 0 and you should surely know how to evaluate this integral!
• Mar 22nd 2010, 08:28 PM
matheagle
Quote:

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

$\displaystyle F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt$ 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.
• Mar 23rd 2010, 02:59 AM
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?
• Mar 23rd 2010, 03:07 AM
mr fantastic
Quote:

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?

• Mar 23rd 2010, 04:51 AM
gralla55
Finally! :D

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?
• Mar 23rd 2010, 07:21 PM
mr fantastic
Quote:

Originally Posted by gralla55
Finally! :D

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?

$\displaystyle \Pr(X > x_0 + x | X > x_0) = \frac{\Pr(X > x_0 + x)}{\Pr(X > x_0)}$.

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