PDF of the PRODUCT of two independent random continuous variables

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zeph1

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Hello guys,

The problem is: I have two random variables X and Y, both are uniform distributions, X between [0,2] and Y between [-10,10].
The random variable Z= X*Y. I want to know the pz(z).

img1.png
I have this formula, that was deducted in class, and "y" is a a "fixed variable". I'm trying to do this exercise but i don't understand what we should do with the formula.
How can i have a px(x/y) if px only have the value 1/20 between [-10,10]? Same with py(y).

PS: Sorry guys i read that i should show more progress but i really can't get past this, i have no clue what is supposed to do.
 
zeph1 said:
The problem is: I have two random variables X and Y, both are uniform distributions, X between [0,2] and Y between [-10,10].
The random variable Z= X*Y. I want to know the pz(z).

I have this formula, that was deducted in class, and "y" is a a "fixed variable". I'm trying to do this exercise but i don't understand what we should do with the formula.
How can i have a px(x/y) if px only have the value 1/20 between [-10,10]? Same with py(y).

PS: Sorry guys i read that i should show more progress but i really can't get past this, i have no clue what is supposed to do.
Click to expand...

You've shown your thinking; that's all we can ask! (But we can, of course, ask further questions!)

In the title you said this is a sum of random variables, but your work appears to be for the product of two independent random variables. Which is it really?

I'm not sure I understand your question. You've slightly misstated the details: There is no Px(x/y) or Py(y) in the formula. Rather, you need Px(z/y) = 1/2 for z/y between 0 and 2, and Py(y) = 1/20 for y between -10 and 10. This will primarily affect the limits of integration you use in practice.
 
Dr.Peterson said:
You've shown your thinking; that's all we can ask! (But we can, of course, ask further questions!)

In the title you said this is a sum of random variables, but your work appears to be for the product of two independent random variables. Which is it really?

I'm not sure I understand your question. You've slightly misstated the details: There is no Px(x/y) or Py(y) in the formula. Rather, you need Px(z/y) = 1/2 for z/y between 0 and 2, and Py(y) = 1/20 for y between -10 and 10. This will primarily affect the limits of integration you use in practice.
Click to expand...

My mistake, didn't notice i had written sum. What i want is the product like you said.
Well i know those values will affect the limits but the problem is i don't know what i'm supposed to integrate. I know that the z/y inside px is equals to x.

Cubist said:
I think you jotted the equation down wrong, it should be Px(z - y) not Px(z / y). I guess you were only 45o out with the line :)

The following page might be helpful

Convolution of probability distributions - Wikipedia

en.wikipedia.org
Click to expand...

Like Dr.Peterson said, what i want is the product and not the sum, sorry for the mistake. z/y is correct then?

My main problems with this exercise are: I don't know what i should do with the formula(what values i should use, what i am supposed to integrate, etc...) and i think i'm not understanding how the product of continuous random variables work. I know that for example, a continuous convolution with a dirac delta function brings the function to the point where the dirac delta is but in this case of 2 uniform distributions i'm not seeing clearly how this works.
In class we just deducted the formula and the professor said not to worry too much about the process and just use the formula.
 
zeph1 said:
My main problems with this exercise are: I don't know what i should do with the formula(what values i should use, what i am supposed to integrate, etc...) and i think i'm not understanding how the product of continuous random variables work. I know that for example, a continuous convolution with a dirac delta function brings the function to the point where the dirac delta is but in this case of 2 uniform distributions i'm not seeing clearly how this works.
In class we just deducted the formula and the professor said not to worry too much about the process and just use the formula.
Click to expand...

Don't worry too much about the process and just use the formula! At this point, your task is to understand what the formula you derived means, and the way to do that is to try using it. This is a very simple example, perfect for that purpose.

This is not a convolution (though the idea is similar). Forget what you've learned about that.

As both you and I have pointed out, the PDF's are both constants over a certain interval (and zero elsewhere). So once you determine the appropriate limits of integration (over which both functions will be nonzero), you will be replacing them with 1/20 and 1/2 respectively. That's all there is to it!

Please make an attempt, even if you're sure you're wrong, and show it to us. That will give us something specific to talk about, and we can either surprise you by saying it's correct, or point out errors to help you think through it.
 
zeph1 said:
Like Dr.Peterson said, what i want is the product and not the sum, sorry for the mistake. z/y is correct then?
Click to expand...

Yes, if you meant product then please ignore my post#2 - since it assumed the sum of random variables.
 
Sem Título.png

I did this, the question mark on the left side is because that side is supposed to be equal to the right one but the result is negative, i know that it has to be positive but i don't know where to get the "-" in the process.
I assumed -0,01 instead of 0 because of the ln(0).

I tried to do the same but inverting the variable that is fixed and the result was different don't know why. In this case(in the picture) i choose the "y".
 
I'll have to look at this in more detail later, but here are a couple comments.

First, you can't just replace 0 with 0.01; you'd have to use limits to handle the improper integrals, if they're valid.

Second, did you check the limits of integration, to make sure the PDFs are both non-zero over that interval?

Third, what is that thing in [...]? And why is your result independent of z? I don't think you're handling Px(z/y) correctly.
 
I didn't check the limits...
Inside [...] are the Heaviside functions that limit py(y). The result is not independent of z, i didn't wrote the heaviside functions to limit the function range but it follows that value between [-10,10]. The area is not 1, but that might be because of replacing 0 with 0.01 instead of doing the limits.
 
Having worked through the whole problem now, I can confirm that you will not have an improper integral; the limits of integration will not include 0. You will probably want to consider two cases: z>0 and z<0. Just keep in mind that Px(z/y) can be replaced with 1/2, as long as you set the limits of integration as functions of z which will ensure that z/y is between 0 and 2.
 
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