# Summation of random sequences and convolution in pdf domain?

Hi all,
I have an all time doubt here. We know that if r.v z = x + y where x and y are 2 random sequences having corresponding pdfs p(x) and p(y), the pdf of z, p(z) = convolution ( p(x),p(y) ). I have seen the derivation for the continuous case although not thorough how to prove it. I wanted a proof for the discrete case (ie, x , y and z are discrete). I attempted through a straight forward method and got stuck. My method was defining a function which calculates the pmf vector from the input sequence x. I thought I would get an input output relation between input vector x and pmf vector.Even for a histogram I could not do it. Why is it so? Is it because of the nonlinearity?

My aim was to try this:

x1 ----> p(x1)
x2 ----> p(x2)

z= x1+x2 ------> p(z) = p(x1 + x2) = conv(p(x1),p(x2))

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Stephen Tashi
How are you defining a "convolution"? Your notation doesn't make it clear.

There are ideas of convolution that are more general than the convolution of probability density functions. You might not get the right answer if you are using the wrong definition of convolution.

To compute the probability that Z = z, you add up the probabilities of all the combinations of values (x,y) such that x + y = z. For discrete random variables, this is usually represented by a summation. Is that what your are doing?

@chiro - Yes Sir... You are right.

Stephen Tashi
To apply the discrete definition to histograms, you would have to get the range of the indexes in the sum correct. They wouldn't go to infinity.

What you mean by "conv(p(x1),p(x2))" is unclear. Is this a function in computer software?

If you can't explain what you are doing, try giving a simple example, like two histograms, each having 3 bins. Show the calculations that don't work.

ok here goes

by conv(p(x1),p(x2)) I meant convolution of p(x1) and p(x2).

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Stephen Tashi