# Homework Help: Stats - Continious random distribution (PDF)

1. Oct 4, 2008

### mohdhm

1. The problem statement, all variables and given/known data
A function f( x ) had the following form : f ( x) = kx^-(k+1) where 1 < x < infinity

a) for what values of k is f a pdf ?

2. Relevant equations
See attachments (the theorem is included)

3. The attempt at a solution
started out by using the first equation in the attachment, for stuck with kx^-(k+1) = 0 and Got stuck there... well the only way k will satisfy the equation is that if it is equal to 0... but i'm not sure if that is the answer.

Tried using the 2nd equation, integration was long, and i encountered logs, K turned out to have a single value of 0.00693
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ok i figured that my first attempt was right and that k is > = 0 . But i just want to confirm this, and i don't really feel like deleting my post after writing it out

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2. Oct 5, 2008

### tiny-tim

Hi mohdhm!
erm … are you sure about k = 0?

3. Oct 5, 2008

### dirk_mec1

What is your function? Is it (kx)^{-k-1} or k * x^{-k-1}

4. Oct 5, 2008

### mohdhm

to dirk_mec1: it is the 2nd function you listed, sorry for not being clear.

tiny-tim: well k=0 works for all positive values of x and satisfies the first equation. So i guess i'm Sure. also, any positive number k will yield a positive answer which also satisfies the equation. but i have this... conflicting hunch, i want to use the integral, but i tried and it was just a big mess, am i on the right track ?

5. Oct 5, 2008

### D H

Staff Emeritus
Just because a function is semi-positive definite does not mean it qualifies as a PDF. The area under the function must be one. In your example, if k=0, f(x;k) reduces to f(x)=0. The area under this function is definitely not one.

Edit
This function integrates easily with respect to x. Treat k as a constant when doing the integral.

Last edited: Oct 5, 2008
6. Oct 5, 2008

### mohdhm

lol... yeah i did treat K as a constant. Well i can give this integral yet another go, might post my work too since i haven't done integrals for a while

7. Oct 5, 2008

### mohdhm

DH you are absolutely wrong, there is no way that integral with bounds of negative infinity to positive infinity, can work out. The whole thing just blows up when i get to the logs part. i get a negative value when converting bases, and the whole thing turns into a mess. So what's up guys, can anyone drop me a hint that can lead me on the right path ?

8. Oct 6, 2008

### tiny-tim

Hi mohdhm!
erm … the -∞ part of the integral doesn't matter …

f(x) is only defined, in the question, for 1 ≤ x ≤ +∞ …

in other words, f(x) = 0 for -∞ < x ≤ 1.

Try again!