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Tue 9 Nov, 2004 11:10 pm
I can't find any reference as to how to do this!!!!
I know I need to find ' k ', but how would i go about doing this?
the fucntion is f(x) = xe^(-2x)
Thanks!
-Mike
f(x) is the PDF
maybe you want the CDF...in which case just integrate it
Hmmmm
Something is wrong. It seems obvious that you are using an exponential PDF. In this case the PDF is not f(x) as it is not in the format of a exp PDF. And you don't have any constraints to make the area equal to 1.
This the exp format for PDF:
/
| -kx
| ke x>=0
f(x)= <
|
| 0 elsewhere
\
You probably know this already but your equation confused me (without parameters).
Hope this helped.
IT REMOVED ALL MY SPACES!!!
How annoying
It was meant to look like this
. /
. | -kx
. | ke x>=0
. f(x)= <
. |
. | 0 elsewhere
. \
Didnt work again.
Well I hope you get the point anyway.
I apologize, my first comment was wrong.
The integral of your function from 0 to infinity is 0.25
Since integrating over a region of the PDF is supposed to tell you the probability that x is within those bounds, the integral from 0 to infinity should be 1 (and if your equation is not absolutely summable, you will need to impose artificial bounds).
So just multiply the function by 4 and that should be the PDF
visk, are you trying to say:
f(x) = -kx, x<0
f(x) = k*exp(x), x>=0
?
Because that would not be absolutely summable
visk wrote:It was meant to look like this
. /
. | -kx
. | ke x>=0
. f(x)= <
. |
. | 0 elsewhere
. \
Try using the "code" tag (without quotes) ["code"]this is in a code window["/code"]:
Code:
This is in a code window
stuh505, thats not what i was trying to do but you are right. That eqution is not summable. I misunderstood the question.
thanks for the tip merlin
anyway