Probability density function of digital filter

  • Thread starter purplebird
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Main Question or Discussion Point

given that x has an exponential density function ie p(x) = exp (-x) and x(n) & x(m) are statistically independent.

Now y(n) = x(n-1)+x(n)

what is the pdf (probability density function) of y(n)
 

Answers and Replies

quadraphonics
given that x has an exponential density function ie p(x) = exp (-x) and x(n) & x(m) are statistically independent.

Now y(n) = x(n-1)+x(n)

what is the pdf (probability density function) of y(n)
The pdf of a sum of two independent variables can be obtained by the convolution of their respective pdf's.
 
ssd
268
6
Y(n) will be distributed as gamma(2,1) if X(n) has the pdf exp[-x(n)].
 
18
0
I made a mistake while typing up the question :

y(n) = [x(n-1) + x(n)]^2

so

y(n) = x(n)^2 + x(n-1)^2

So is the pdf of y(n) convolution of exp(-x(n)^2) and exp(-x(n-1)^2)

Thanks
 

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