What is the PDF of z where z = x-y

[nikozm]

Hello,

I 'm trying to express the PDF of z (z ≥ 0) where z = x-y (and let x,y ≥ 0)

Thank you in advance

 

[mathman]

http://www.statlect.com/sumdst1.htm

Try the above. You should Google "sum of independent random variables".

For your question you need two things, x and y independent, and let w = -y so you can use standard sum formula for x+w.

 

[nikozm]

Thank you for your answer.

I 'm trying to express the distribution of x-y in integral form. I m not sure about the integration bounds though.

I presume that the standard convolution PDF can be used: let z= x-y, then:
fz(z) =∫ fx(x)*fy(z+x) dx, but with what integration lower and upper bounds ??

Assume that x,y ≥ 0.

Any help would be useful

 

[mathman]

Thank you for your answer.

I 'm trying to express the distribution of x-y in integral form. I m not sure about the integration bounds though.

I presume that the standard convolution PDF can be used: let z= x-y, then:
fz(z) =∫ fx(x)*fy(z+x) dx, but with what integration lower and upper bounds ??

Assume that x,y ≥ 0.

Any help would be useful

The integral is over the whole real line. Since the random variables are assumed non-negative, the integration need only be for non-negative x.

 

[nikozm]

I found a slide (page 20/36) in:
http://www.wiwi.uni-muenster.de/05/download/studium/advancedstatistics/ws1314/Chapter_4.pdf

According to the above, the integral goes:

f_z(z)=∫f_x(y+z)*f_y(y) dy with lower integration bound zero and upper bound z

or the upper bound should be infinity ?

(note that z is also nonnegative)

Thank you in advance

 

[mathman]

I am confused about your notation. You seem to have switched x and y between the posts, so I am not sure how you are defining z.

As far as the upper limit is concerned, infinity is always correct. However because the random variables are non-negative, one of the f's may be 0 past some point, so the integral doesn't need to go any further.