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!

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- March 18th 2010, 01:39 AMgralla55Cummulative 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! - March 18th 2010, 01:43 AMmr fantastic
- March 18th 2010, 01:58 AMgralla55
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? - March 18th 2010, 01:59 AMmr fantastic
- March 18th 2010, 02:13 AMgralla55
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)...

- March 18th 2010, 02:16 AMmr fantastic
- March 18th 2010, 02:20 AMgralla55
So, the integral should be -1/t * e^(-t/b)?

- March 18th 2010, 07:38 AMmatheagle
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 - March 22nd 2010, 03:24 PMgralla55
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) ) ? - March 22nd 2010, 07:59 PMmr fantastic
- March 22nd 2010, 08:28 PMmatheagle
- March 23rd 2010, 02:59 AMgralla55
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? - March 23rd 2010, 03:07 AMmr fantastic
- March 23rd 2010, 04:51 AMgralla55
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? - March 23rd 2010, 07:21 PMmr fantastic