Re: can it be a distrubution function?
- From: "tamiry" <tamir.yehuda@xxxxxxxxx>
- Date: 4 Nov 2006 03:40:12 -0800
thanks for the response, let me see if i got it straight, this time.
1. the probability distribution is a function P, matching for each
event x the probability of it.
2. a random variable X is a function from sample space to the reals
(satisfying some condition).
3. the (comulative) distribution is a function matching for each real
value x, the probabilty of X being at the interval (-inf,x], i.e P(-inf
<= X <= x]), when X(S) is the sample space.
4. the (probability) density is a function such that
(integral)[a,b]f(x) = P(a <= X <= b).
yes?
karl wrote:
tamiry wrote:so it's just computing the difference of f at the boundary points...
Hi,
this week's HW had this distribution function
f =
1-(3/4)exp(-x) x >= 0
0 otherwise
then I was asked to compute P(0 <= x <= 2). if i understand correctly,
i need to compute the integral of f(x) at the given interval.
Why, if f is the distribution function? I guess, you are confusing
density and distribution function.
since it's a distribution, it should fulfill f(+inf) = 1, which it
as a
preliminary stage i tried to compute the integral of f on the entire
sample space. i expected it to be 1. what i got was:
(integral)[0,+inf](1-(3/4)exp(-x)) = (x + (3/4)exp(-x)) between 0 and
+inf = inf !
this integral does not converge, because of the "x". so, am i doing
something wrong or that f cannot be a distribution function ?
As I said, the integral has to be 1 if f is a DENSITY function. Look what conditions
a function must fulfill to be a DISTRIBUTION function.
does.
Ciao
Karl
.
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