Probability density function of digital filter

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SUMMARY

The discussion focuses on determining the probability density function (PDF) of the random variable y(n) defined as y(n) = [x(n-1) + x(n)]^2, where x(n) follows an exponential distribution p(x) = exp(-x) and x(n) and x(m) are statistically independent. The PDF of the sum of two independent variables is derived through convolution of their respective PDFs. The corrected expression indicates that y(n) is the sum of squares of independent exponential variables, leading to the conclusion that the PDF of y(n) is not simply the convolution of exp(-x(n)^2) and exp(-x(n-1)^2), but rather requires a different approach to derive the correct distribution.

PREREQUISITES
  • Understanding of exponential probability density functions
  • Knowledge of convolution in probability theory
  • Familiarity with gamma distribution properties
  • Basic concepts of statistical independence
NEXT STEPS
  • Study the properties of the gamma distribution and its relation to sums of independent exponential variables
  • Learn about convolution of probability density functions in detail
  • Explore the derivation of PDFs for sums of independent random variables
  • Investigate the implications of squaring random variables on their distributions
USEFUL FOR

Statisticians, data scientists, and anyone involved in probability theory or digital signal processing who seeks to understand the behavior of sums of independent random variables and their distributions.

purplebird
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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)
 
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purplebird said:
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.
 
Y(n) will be distributed as gamma(2,1) if X(n) has the pdf exp[-x(n)].
 
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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