Sums of Independent Random Variables

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SUMMARY

The discussion centers on the calculation of the probability density function (pdf) for the sum of independent random variables, specifically using convolution and the Jacobian transformation. The established formula for the pdf is f[sub:X+Y](a) = ∫ f[sub:X](a-y)f[sub:Y](y)dy. The participant questions whether applying the density function for a function of a random variable, along with the Jacobian inverse, is a simpler method. The hint provided suggests that if U = X + Y, then V can be taken as Y, indicating a potential simplification in the approach.

PREREQUISITES
  • Understanding of probability density functions (pdf)
  • Knowledge of convolution in probability theory
  • Familiarity with Jacobian transformations
  • Basic concepts of independent random variables
NEXT STEPS
  • Study the properties of convolution for independent random variables
  • Learn about Jacobian transformations in multivariable calculus
  • Explore examples of pdf calculations for sums of random variables
  • Investigate alternative methods for finding pdfs of functions of random variables
USEFUL FOR

Students in probability theory, statisticians, and anyone studying random variables and their distributions will benefit from this discussion.

jrk012
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Homework Statement



So, I know the pdf for independent random variables is found by using the convolution; the pdf is f[sub:X+Y](a) = ∫ f[sub:X](a-y)f[sub:Y](y)dy, but can I just use the density function for a function of a random variable instead; that is: f[sub:X+Y](x[u,v], y[u,v])(Jacobian Inverse(x[u,v], y[u,v])) and then integrate it? It seems much easier that way.


Homework Equations



f[sub:X+Y](a) = ∫ f[sub:X](a-y)f[sub:Y](y)dy

f[sub:X+Y](x[u,v], y[u,v])(Jacobian Inverse(x[u,v], y[u,v]))

Jacobian Determinant: (∂u/∂x)(∂v/∂y)-(∂u/∂y)(∂v/∂x)


The Attempt at a Solution



More of a question on coursework than homework or a specific problem. However when I find the density function when U=X+Y I get f(x[u-v], y[v])
 
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jrk012 said:

Homework Statement



So, I know the pdf for independent random variables is found by using the convolution; the pdf is f[sub:X+Y](a) = ∫ f[sub:X](a-y)f[sub:Y](y)dy, but can I just use the density function for a function of a random variable instead; that is: f[sub:X+Y](x[u,v], y[u,v])(Jacobian Inverse(x[u,v], y[u,v])) and then integrate it? It seems much easier that way.


Homework Equations



f[sub:X+Y](a) = ∫ f[sub:X](a-y)f[sub:Y](y)dy

f[sub:X+Y](x[u,v], y[u,v])(Jacobian Inverse(x[u,v], y[u,v]))

Jacobian Determinant: (∂u/∂x)(∂v/∂y)-(∂u/∂y)(∂v/∂x)


The Attempt at a Solution



More of a question on coursework than homework or a specific problem. However when I find the density function when U=X+Y I get f(x[u-v], y[v])

So U = X+Y, but what is V?

RGV
 
They said "Hint: V=Y" so I wasn't sure if that was supposed to be assumed known or if there were different ways to do it. Sorry I forgot to put that in there!
 

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