Finding k in a probability density function

In summary, the author is trying to solve a divergent integrals problem and is not sure how to do it. He has tried to do it on paper and is having trouble. He asks for help from the reader.
  • #1
mnphys
10
0

Homework Statement



Let X, Y, and Z have the joint probability density function f(x,y,z) = kxy2z for 0 < x, y < 1, and 0 < z < 2 (it is defined to be 0 elsewhere). Find k.

Homework Equations



Not sure how to type this in bbcode but: Integrate f(x,y,z) = kxy2z over the ranges of x (zero to infinity) , y (negative infinity to 1), and z (zero to two) and set k so that the result is equal to 1 (by the definition of a PDF).

The Attempt at a Solution


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The problem with this is that I keep running into divergent integrals, and I'm not sure how to avoid this. I have done all my work on paper and trying to type it all out into Word's equation editor is driving me insane but I can take photos of the work I've done if you want proof that I have DEFINITELY tried to work this out... like for pages and pages.

For example, if I start by integrating f(x,y,z) = xy2z respective to x, I end up with a non-divergent improper integral (the second term is equal to 0, but the first term takes the limit of x2y2z/2 as x approaches infinity). Let's move on and substitute "t" for x but remember that "t" is approaching infinity. If I then integrate respective to y, it gets worse - substituting "u" for y as y approaches negative infinity, that looks like t2z/6 - t2u3z/6 (remembering that t is approaching infinity and u is approaching negative infinity). Now integrating with respect to z leaves me with a really ugly equation that involves multiple terms all approaching infinity... yuck.

So... how do I resolve these divergent integrals? The book's answer is 3, and if I just set y and x's limits of integration to 1 (upper bound) and 0 (lower bound) then I get that answer, but... I can't do that, can I? Or I could change the order of integration, but no matter what order I try, I end up with at least one term involving taking the limit of a positive exponential as it approaches infinity.

Please, don't think I'm asking you to solve this for me - I really am not. Any hint as to how I can avoid these divergent integrals, or what I'm doing wrong, is all I'm asking for. If anyone would like me to upload photos of my work on this, I can do that.
 
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  • #2
mnphys said:

Homework Statement



Let X, Y, and Z have the joint probability density function f(x,y,z) = kxy2z for 0 < x, y < 1, and 0 < z < 2 (it is defined to be 0 elsewhere).

Are you sure that you've read and written the question correctly? One possible way to read the above is that Z has a range of (0, 2) and both X and Y have a range of (0,1)...

mnphys said:
So... how do I resolve these divergent integrals? The book's answer is 3, and if I just set y and x's limits of integration to 1 (upper bound) and 0 (lower bound) then I get that answer, but... I can't do that, can I?

If you have my above interpretation, then yes you can do this.
 
  • #3
The book does say "0 < x" and "y < 1" but maybe you're right and that should be assumed?

It does work out perfectly if you use 0 to 1 as the ranges.
 
  • #4
I agree. It is common to interpret ##0 < x,~y < 1## to mean both ##0 < x < 1## and ##0<y<1##.
 
  • #5
Ok. I'll roll with that... Thanks, guys (or possibly gals)!
 

1. What is "k" in a probability density function?

In a probability density function, "k" represents the constant of proportionality that ensures the total area under the curve is equal to one. It is often referred to as the normalization constant.

2. How do you find the value of "k" in a probability density function?

To find the value of "k", you can use the equation k = 1 / integral of the probability density function. Alternatively, you can use numerical methods such as the trapezoidal or Simpson's rule to approximate the value of "k".

3. Can "k" have a negative value in a probability density function?

No, "k" cannot have a negative value in a probability density function. This is because the function must always be positive and the total area under the curve must be equal to one. Therefore, "k" must be a positive constant.

4. What is the significance of "k" in a probability density function?

The value of "k" in a probability density function determines the shape and scaling of the curve. A larger value of "k" will result in a narrower and taller curve, while a smaller value of "k" will result in a wider and shorter curve. It is an important parameter to consider when analyzing and comparing different probability density functions.

5. Is the value of "k" unique for every probability density function?

Yes, the value of "k" is unique for every probability density function. This is because it is dependent on the specific shape and scaling of the curve. Two probability density functions with different shapes and scales will have different values of "k".

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