Question

Let X be a RV with cumulative density function \(f(x) = x^2\) for \(0 \le x \le 1\)

Answer 1

Show that this is a valid cumulative density function. (Two conditions must be checked).
would that mean show that \(\int\limits_{- \infty}^{\infty}f(x)=1\) and that \(f(x)\ge 0\)?

Answer 2

while I'm here, part b is
What is the probability that X is between 1 and 3?

Answer 3

A) For all a and b, if a<b then f(a) <=f(b) (f is non-decreasing)
B) x->-inf implies f(x)->0
C) x->inf implies f(x)->1
D) f is right continuous
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you know i think that 1 thing comes from if the cdf is a pdf

Answer 4

i'm about your integral being 1

Answer 5

oh yeah I was reading this the other night https://www.quora.com/What-is-the-difference-between-a-probability-density-function-and-a-cumulative-distribution-function

Answer 6

I think \[P(1\le X \le 3)=f(3)-f(1)\]

Answer 7

but hey f(x)=x^2 is only defined for 0<=x<=1 ?

Answer 8

what is f when x is 3?

Answer 9

I think it's p(X=x)

Answer 10

uhh 9?

Answer 11

so f(x)=x^2 when 0<=x<=?

Answer 12

I'm just confused because it looks like you only have f(x)=x^2 for when 0<=x<1

Answer 13

f(1)-f(0) = 1-0=1

Answer 14

<=*

Answer 15

I'm confused as well, I assume my professor is going to clear it up later tonight (hopefully)

Answer 16

I can show you an example that makes more sense to me
And I'm saying it makes more since because we have a function defined for where we do want to find the probability for certain values for x.

let f(x)=0 if x<0
=x/2 if 0<=x<1
=1 if 1<=x
---------------------
\[P(-1<X \le \frac{1}{2})=f(\frac{1}{2})-f(-1)=\frac{\frac{1}{2}}{2}-0=\frac{1}{4}-0=\frac{1}{4}\]

Answer 17

maybe your f(x) was suppose to be
f(x)=0 if x<0
=x^2 if 0<=x<1
=1 of 1<=x

Answer 18

If that is so then \[P(1 \le X \le 3)=f(3)-f(1)=1-1=0\]

Answer 19

which sorta makes sense there is a 0 probability of it happening since your function isn't even defined there

Answer 20

That explains a lot, thank you.
Off to memorize some distributions now, thanks for the help :)

Answer 21

oh we could have wrote f as
f(x)=0 if x<0
=x^2 if 0<=x<=1
=1 if x>1

you know since you had that in your initial post (that middle inequality)
but it doesn't matter since x^2=1 when x=1

Answer 22

np
good luck

Answer 23

@bibby dude.