Probability Density Function for F(x)=k(1-1/x2)

[XodoX]

Homework Statement

F(x)=k(1-1/x²), 1\leqx<2

Homework Equations

The Attempt at a Solution

How do I get the probability density function here? Simply take the derivative of this function ?

1\int2 = k(1-1/x²)

Supposed to be 1 at the bottom and two at the top.

 

[Dick]

You want to find a value of k so that the integral of your function over [1,2) is equal to 1.

 

[XodoX]

Well, it says specify the probability density function. So I'm guessing just show it, don't solve it. But I don't get what you mean, though. Solve for k and that needs to be equal to 1 then?

 

[Dick]

Well, it says specify the probability density function. So I'm guessing just show it, don't solve it. But I don't get what you mean, though. Solve for k and that needs to be equal to 1 then?

A probability density function need to have integral 1 over it's domain. Integrate your function from 1 to 2 factoring the k out. Then set the result equal to 1. Find k.

 

[XodoX]

Well, but in my post it already was 1 and 2. Ok, then factor the k out.

Do you mean like this?

k∫(1-...)dx=1

 

[Ray Vickson]

If F(x) is supposed to be the CDF, you need to find the value of k that makes that true. Then you differentiate to find the PDF. If F(x) is supposed to be a PDF, you need to integrate and equate that to 1, to find k. It is not clear which is the case.

RGV

 

[XodoX]

F(x) is the distribution function for a continuous random variable. Sorry, guess I forgot to mention it.

 

[Dick]

F(x) is the distribution function for a continuous random variable. Sorry, guess I forgot to mention it.

Ray Vickson is right. If F(x) is a cumulative distribution function, then you want to determine k by setting F(2)=1. Then differentiate to find the probability density function. I had thought it was the probability density function, in which case the procedure you outlined in post 5 is correct.

 

[XodoX]

Ray Vickson is right. If F(x) is a cumulative distribution function, then you want to determine k by setting F(2)=1. Then differentiate to find the probability density function. I had thought it was the probability density function, in which case the procedure you outlined in post 5 is correct.

Thanks. But why 2? Why not also 1 ? And why does it have to equal one again?

 

[Dick]

Thanks. But why 2? Why not also 1 ? And why does it have to equal one again?

Read http://en.wikipedia.org/wiki/Cumulative_distribution_function their pictures are for functions defined on (-infinity,infinity). You'll need to imagine the domain is [1,2) instead.

 

[XodoX]

Ok, so I just plug in 2 for x but I still use 1 and 2 when I take the integral, so the domain stays the same.

 

[Ray Vickson]

Ok, so I just plug in 2 for x but I still use 1 and 2 when I take the integral, so the domain stays the same.

I don't understand why you want to integrate anything. Somebody has already done the integration \int_1^x f(t) dt for you, and their answer is F(x). All you need to do is find the right value of k.

RGV

 

[XodoX]

Because that's what the book says I have to do to get the probability density function..

 

[Ray Vickson]

Because that's what the book says I have to do to get the probability density function..

How do you get the density function f(x) from the CDF F(x)?

RGV

 

[XodoX]

I don't know.

 

[Ray Vickson]

I don't know.

This is so basic, so if you don't know it there is something seriously wrong. My recommendation: quit the course now, you have no chance of passing.

RGV

 

[HallsofIvy]

You said, in your very first post,

How do I get the probability density function here? Simply take the derivative of this function ?

Yes! What everyone was telling you before was how to find k so that you know what the cumulative function is.

 

[XodoX]

This is so basic, so if you don't know it there is something seriously wrong. My recommendation: quit the course now, you have no chance of passing.

RGV

You have been extremely arrogant. If you don't have anything meaningful to say, then just don't say anything. You are not helpful or contributing anything. You are the most arrogant poster I have seen on here.