- #1
dexterdev
- 194
- 1
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))
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))