Sums of Independent Random Variables

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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])
 

Answers and Replies

  • #2
Ray Vickson
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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])
So U = X+Y, but what is V?

RGV
 
  • #3
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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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