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Finding the PDF for the Product of Random Variables using Jacobian Transformation
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I've been working through the following problem:
If $X$ has pdf $f_X(x)$ and Y, independent of $X$, has pdf $f_Y(y)$, establish a formula for probability density function for the random variable $Z = XY$.
I was able to get the solution using by differentiating the CDF of $Z$. However, I found an alternate solution that uses the transformation method. The solution for this method is as follows:
For $Z = XY$, set $W = X$. Then $Y = Z/W$, $X = W$, and $|J| = -1/w$. Then $f_Z, W(z, w) = f_X(w)f_Y(z/w)cdot|-1/w|$, thus $f_Z(z) = int_-infty^infty|-1/w|f_X(w)f_Y(z/w)dw.$
My question about this solution is how the Jacobian method can be used even though the transformation from $(X, Y)$ to $(U,V)$ is not one-to-one. (Any point $(0,y)$ maps to $(0, 0)$).
Any clarification on this will be greatly appreciated, thanks.
probability-theory
asked 2 days ago
Stephen Mac Lane
1
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I've been working through the following problem:
If $X$ has pdf $f_X(x)$ and Y, independent of $X$, has pdf $f_Y(y)$, establish a formula for probability density function for the random variable $Z = XY$.
I was able to get the solution using by differentiating the CDF of $Z$. However, I found an alternate solution that uses the transformation method. The solution for this method is as follows:
For $Z = XY$, set $W = X$. Then $Y = Z/W$, $X = W$, and $|J| = -1/w$. Then $f_Z, W(z, w) = f_X(w)f_Y(z/w)cdot|-1/w|$, thus $f_Z(z) = int_-infty^infty|-1/w|f_X(w)f_Y(z/w)dw.$
My question about this solution is how the Jacobian method can be used even though the transformation from $(X, Y)$ to $(U,V)$ is not one-to-one. (Any point $(0,y)$ maps to $(0, 0)$).
Any clarification on this will be greatly appreciated, thanks.
probability-theory
asked 2 days ago
Stephen Mac Lane
1
If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago
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up vote
0
down vote
favorite
I've been working through the following problem:
If $X$ has pdf $f_X(x)$ and Y, independent of $X$, has pdf $f_Y(y)$, establish a formula for probability density function for the random variable $Z = XY$.
I was able to get the solution using by differentiating the CDF of $Z$. However, I found an alternate solution that uses the transformation method. The solution for this method is as follows:
For $Z = XY$, set $W = X$. Then $Y = Z/W$, $X = W$, and $|J| = -1/w$. Then $f_Z, W(z, w) = f_X(w)f_Y(z/w)cdot|-1/w|$, thus $f_Z(z) = int_-infty^infty|-1/w|f_X(w)f_Y(z/w)dw.$
My question about this solution is how the Jacobian method can be used even though the transformation from $(X, Y)$ to $(U,V)$ is not one-to-one. (Any point $(0,y)$ maps to $(0, 0)$).
Any clarification on this will be greatly appreciated, thanks.
probability-theory
asked 2 days ago
Stephen Mac Lane
1
I've been working through the following problem:
If $X$ has pdf $f_X(x)$ and Y, independent of $X$, has pdf $f_Y(y)$, establish a formula for probability density function for the random variable $Z = XY$.
I was able to get the solution using by differentiating the CDF of $Z$. However, I found an alternate solution that uses the transformation method. The solution for this method is as follows:
For $Z = XY$, set $W = X$. Then $Y = Z/W$, $X = W$, and $|J| = -1/w$. Then $f_Z, W(z, w) = f_X(w)f_Y(z/w)cdot|-1/w|$, thus $f_Z(z) = int_-infty^infty|-1/w|f_X(w)f_Y(z/w)dw.$
My question about this solution is how the Jacobian method can be used even though the transformation from $(X, Y)$ to $(U,V)$ is not one-to-one. (Any point $(0,y)$ maps to $(0, 0)$).
Any clarification on this will be greatly appreciated, thanks.
probability-theory
asked 2 days ago
Stephen Mac Lane
1
asked 2 days ago
Stephen Mac Lane
1
asked 2 days ago
Stephen Mac Lane
1
asked 2 days ago
Stephen Mac Lane
1
1
If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago
add a comment |Â
If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago
If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago
If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago
add a comment |Â
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If we take the support of $(X,Y)$ simply to be $mathbb R^2$, then $(X,Y)to (Z,W)$ such that $Z=XY, W=X$ is a one-to-one transform, the support of $(Z,W)$ also being $mathbb R^2$. This is because we get a single inverse $(x,y)=(w,z/w)$. Apart from this, there is a general transformation formula that takes care of the problem when the map is not one-to-one.
– StubbornAtom
2 days ago