Convolution of continuous case

  • Thread starter Gooolati
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  • #1
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Hello all,

I am currently working on studying for my P actuary exam and had some questions regarding using convolution for the continuous case of the sum of two independent random variables. I have no problem with the actual integration, but what is troubling me is finding the bounds.


Homework Statement


Consider two independent random variables X and Y. Let fx(x) = 1 - (x/2) for 0<=x<=2 and 0 otherwise. Let fy(y) = 2-2y for 0<=y<=1 and 0 otherwise. Find the probability density function of X+Y.


Homework Equations


for a = x+y

(fx*fy)(a) (the convolution) is the integral from negative infinity to infinity of fx(a-y)fy(y)dy or fy(a-x)fxdx


The Attempt at a Solution



For 0<=a<=1 I integrate using the above formula and get 2a-(3/2)a^2 + (1/6)a^3

I would have thought the bounds to be from 0 to 1 but they are actually from 0 to a. And for the other cases of a (where 1<=a<=2 and 2<=a<=3) I am having trouble finding out where the bounds are. Some insight would greatly be appreciated! Thanks!
 

Answers and Replies

  • #2
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You don't need bounds (apart from +- infinity) at all. It is useful to restrict the integral to the parts where fx and fy are non-zero as this makes it easier to evaluate the integrals, but it is not necessary.

Did you draw a sketch of the x- and y-values for different values of a? This will help.
 
  • #3
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Hi mfb thanks for the reply. Wouldn't I need the bounds for when I write my final answer as a probability density?

And for the second question, would I sketch x+y=a alongside my fx(x) and fy(y)? So that way as a=x+y has different values I could see how x and y vary? Like I found this website:

http://probabilityexam.wordpress.com/2011/05/26/examples-of-convolution-continuous-case/

and that illustrates it really well but that's the simplest case and I'm not sure if it will work for more complicated probability density's.
 
  • #4
35,655
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Wouldn't I need the bounds for when I write my final answer as a probability density?
The numbers of the bounds will appear in some way, but you don't need them as explicit bounds. The only special thing about those bounds: the functions are zero (but still regular functions) outside.

and that illustrates it really well but that's the simplest case and I'm not sure if it will work for more complicated probability density's.
It will work here. If it does not work any more, you'll have to rely on calculations to get it right.
 
  • #5
Ray Vickson
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Homework Helper
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Hello all,

I am currently working on studying for my P actuary exam and had some questions regarding using convolution for the continuous case of the sum of two independent random variables. I have no problem with the actual integration, but what is troubling me is finding the bounds.


Homework Statement


Consider two independent random variables X and Y. Let fx(x) = 1 - (x/2) for 0<=x<=2 and 0 otherwise. Let fy(y) = 2-2y for 0<=y<=1 and 0 otherwise. Find the probability density function of X+Y.


Homework Equations


for a = x+y

(fx*fy)(a) (the convolution) is the integral from negative infinity to infinity of fx(a-y)fy(y)dy or fy(a-x)fxdx


The Attempt at a Solution



For 0<=a<=1 I integrate using the above formula and get 2a-(3/2)a^2 + (1/6)a^3

I would have thought the bounds to be from 0 to 1 but they are actually from 0 to a. And for the other cases of a (where 1<=a<=2 and 2<=a<=3) I am having trouble finding out where the bounds are. Some insight would greatly be appreciated! Thanks!

If X ranges between 0 and 2, and Y ranges from 0 to 1, then X+Y ranges from 0 to 1+2 = 3. Nothing greater than 3 is needed, and nothing less than 3 will work. After all there are nonzero probabilities that X > 1.999 and Y > 0.999, so there is a nonzero probability that the sum is almost 3.
 

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