Stats - Continious random distribution (PDF)

In summary, a function f(x) had the following form: f ( x) = kx^-(k+1) where 1 < x < infinity. For what values of k is f a pdf? The Attempt at a Solution started out by using the first equation in the attachment, but got stuck there. Tried using the 2nd equation, integration was long, and i encountered logs, K turned out to have a single value of 0.00693. Finally, tried using the 2nd equation, integration was long, and i encountered logs, K turned out to have a single value of 0.00693. Just because a function is semi-positive definite does not mean it qualifies as a PDF. The area
  • #1
mohdhm
42
0

Homework Statement


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 ?


Homework Equations


See attachments (the theorem is included)


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
--------------------------

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
Hi mohdhm! :smile:
mohdhm said:
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 ?

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

erm … are you sure about k = 0? :wink:
 
  • #3
What is your function? Is it (kx)^{-k-1} or k * x^{-k-1}
 
  • #4
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
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
mohdhm said:
Tried using the 2nd equation, integration was long, and i encountered logs, K turned out to have a single value of 0.00693
This function integrates easily with respect to x. Treat k as a constant when doing the integral.
 
Last edited:
  • #6
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
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
Hi mohdhm! :smile:
mohdhm said:
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 ?

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. :wink:

Try again! :smile:
 

1. What is a continuous random distribution?

A continuous random distribution is a probability distribution that describes the probabilities of all possible outcomes of a continuous random variable. This means that the variable can take on any value within a certain range, as opposed to a discrete random variable which can only take on a specific set of values.

2. What is the difference between a PDF and a CDF?

A PDF (probability density function) is a function that describes the relative likelihood of a continuous random variable taking on a specific value. It shows the probability of the variable falling within a specific range of values. A CDF (cumulative density function) is the cumulative version of the PDF, showing the probability of the variable being less than or equal to a certain value.

3. How do you calculate the mean and standard deviation of a continuous random distribution?

The mean of a continuous random distribution can be calculated by taking the integral of the product of each possible value of the variable and its corresponding probability. The standard deviation can be calculated by taking the square root of the variance, which is the integral of the squared differences between each value and the mean, multiplied by their corresponding probabilities.

4. What is the normal distribution and why is it important?

The normal distribution, also known as the Gaussian distribution, is a type of continuous random distribution that is symmetric and bell-shaped. It is important because it is a commonly occurring distribution in nature and can be used to model many real-world phenomena. It also has many mathematical properties that make it useful in statistical analysis and hypothesis testing.

5. How do you use the central limit theorem with continuous random distributions?

The central limit theorem states that the sum of a large number of independent random variables will be approximately normally distributed, regardless of the distribution of the individual variables. This can be applied to continuous random distributions by taking a large sample from the distribution and calculating the mean of the sample. The resulting distribution of sample means will be approximately normal, even if the original distribution is not.

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