let X be a RV with cumulative density function f(x)=x^2 for 0=

Question

let X be a RV with cumulative density function f(x)=x^2 for 0=<x=<1
A) Show that this is a valid cumulative density function.
B)What is the probability that X is between 1 and 3

myininaya · Answer

I know it has to be nondecreasing and right continuous.

myininaya · Answer

I also think the lim x->-inf f=0
and lim x->inf f=1

anonymous · Answer

indeed in addition is a probability and statistics problem.

anonymous · Answer

RV=random Variable

myininaya · Answer

I'm going to assume this if 
f(x)=0    if x<0
      =x^2 if 0<=x<=1
      = 1  if x>1
this will make answering that last question easier
since P(1<x<3)=f(3)-f(1)

myininaya · Answer

@Zarkon can we just assume that other part? or does it have to be given?

anonymous · Answer

so quiet lol

myininaya · Answer

well i'm pretty sure that is another way to write this cdf

myininaya · Answer

because any x value bigger than 1 will have to have a y value greater than or equal to 1 (since it is a nondecreasing function)

myininaya · Answer

well equal to 1 since the limit as x-> inf we have f goes to 1

myininaya · Answer

and since x->-inf gives us f->0
we have to have for values x less than 0 that the y is 0

myininaya · Answer

do you have any questions?

anonymous · Answer

is in the form of integration
for par A)

myininaya · Answer

we don't need integrating

myininaya · Answer

integration*

myininaya · Answer

we just need those four things I mentioned

myininaya · Answer

1) nondecreasing
2) right continuous
3) as x->inf, f(x)->1
4) as x->-inf, f(x)->0

myininaya · Answer

http://www.maths.qmul.ac.uk/~bb/MS_Lectures_3and4.pdf here it is in a pretty pdf

myininaya · Answer

i think you thought about integration because you were thinking of a pdf

anonymous · Answer

i see

myininaya · Answer

and my rewrite of your cdf is fine

anonymous · Answer

that its  for part A or there is more procedure ?

myininaya · Answer

you need to show those 4 things 
that is it

myininaya · Answer

for part a

anonymous · Answer

ok

anonymous · Answer

i guess both part are done

myininaya · Answer

1) nondecreasing that is easy to see
2) right continuous on [0,1] easy to see
and I think the limit parts are also easy to see

myininaya · Answer

so yep should be easy to show

anonymous · Answer

ok

zarkon · Answer

Did you get an answer for B?


and yes it should specify that f takes the value 0 for x<0 and 1 for x>1

zarkon · Answer

I was playing a game so I didn't see the posts

anonymous · Answer

lol