Simple justification that function sends a nonzero probability density to a nonzero probability density
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There is a certain nice (like analytic) function $g$ such that for random variables $X$ and $Y$, we have $Y=g(X)$. We know that the pdf $p_X(0)neq 0$. I'd like to justify that $p_Y(g(0))neq 0$ with a simple argument. I know that $$p_Y(y)=sum_{f(x_k)=y}frac{p_X(x)}{|dy/dx|}Big|_{x=x_k}$$ holds, but I'd like to avoid using this formula to make our argument as prerequisite-less as possible. If necessary, you can assume $X$ is the uniform distribution on a certain segment on $mathbb{R}$.
probability
asked Jan 10 at 20:52
J. DoeJ. Doe
213
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add a comment |
$begingroup$
There is a certain nice (like analytic) function $g$ such that for random variables $X$ and $Y$, we have $Y=g(X)$. We know that the pdf $p_X(0)neq 0$. I'd like to justify that $p_Y(g(0))neq 0$ with a simple argument. I know that $$p_Y(y)=sum_{f(x_k)=y}frac{p_X(x)}{|dy/dx|}Big|_{x=x_k}$$ holds, but I'd like to avoid using this formula to make our argument as prerequisite-less as possible. If necessary, you can assume $X$ is the uniform distribution on a certain segment on $mathbb{R}$.
probability
asked Jan 10 at 20:52
J. DoeJ. Doe
213
$endgroup$
$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
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– Henry
Jan 10 at 21:12
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Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
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– Henry
Jan 10 at 21:12
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I'm sorry, but I don't think I can come up with a simple argument with your inputs.
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– J. Doe
Jan 10 at 23:23
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Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
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– Kavi Rama Murthy
Jan 10 at 23:56
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@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23
add a comment |
$begingroup$
There is a certain nice (like analytic) function $g$ such that for random variables $X$ and $Y$, we have $Y=g(X)$. We know that the pdf $p_X(0)neq 0$. I'd like to justify that $p_Y(g(0))neq 0$ with a simple argument. I know that $$p_Y(y)=sum_{f(x_k)=y}frac{p_X(x)}{|dy/dx|}Big|_{x=x_k}$$ holds, but I'd like to avoid using this formula to make our argument as prerequisite-less as possible. If necessary, you can assume $X$ is the uniform distribution on a certain segment on $mathbb{R}$.
probability
asked Jan 10 at 20:52
J. DoeJ. Doe
213
$endgroup$
There is a certain nice (like analytic) function $g$ such that for random variables $X$ and $Y$, we have $Y=g(X)$. We know that the pdf $p_X(0)neq 0$. I'd like to justify that $p_Y(g(0))neq 0$ with a simple argument. I know that $$p_Y(y)=sum_{f(x_k)=y}frac{p_X(x)}{|dy/dx|}Big|_{x=x_k}$$ holds, but I'd like to avoid using this formula to make our argument as prerequisite-less as possible. If necessary, you can assume $X$ is the uniform distribution on a certain segment on $mathbb{R}$.
probability
probability
asked Jan 10 at 20:52
J. DoeJ. Doe
213
asked Jan 10 at 20:52
J. DoeJ. Doe
213
asked Jan 10 at 20:52
J. DoeJ. Doe
213
asked Jan 10 at 20:52
J. DoeJ. Doe
213
asked Jan 10 at 20:52
J. DoeJ. Doe
213
213
$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
I'm sorry, but I don't think I can come up with a simple argument with your inputs.
$endgroup$
– J. Doe
Jan 10 at 23:23
$begingroup$
Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:56
$begingroup$
@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23
add a comment |
$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
I'm sorry, but I don't think I can come up with a simple argument with your inputs.
$endgroup$
– J. Doe
Jan 10 at 23:23
$begingroup$
Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:56
$begingroup$
@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23
$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
I'm sorry, but I don't think I can come up with a simple argument with your inputs.
$endgroup$
– J. Doe
Jan 10 at 23:23
$begingroup$
I'm sorry, but I don't think I can come up with a simple argument with your inputs.
$endgroup$
– J. Doe
Jan 10 at 23:23
$begingroup$
Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:56
$begingroup$
Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:56
$begingroup$
@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23
$begingroup$
@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23
add a comment |
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$begingroup$
If you knew $X$ was uniform on $[0,1]$ and $g(x)=x$, how would you know what the density of $Y$ would be at $0$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
Or if $g(x)=k$ for some constant $k$, does $Y$ have a density at $k$?
$endgroup$
– Henry
Jan 10 at 21:12
$begingroup$
I'm sorry, but I don't think I can come up with a simple argument with your inputs.
$endgroup$
– J. Doe
Jan 10 at 23:23
$begingroup$
Density functions are not unique. They are defined only upto sets of measure zero. So the question has no meaning.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:56
$begingroup$
@KaviRamaMurthy You're right. I think my actual problem was reformulated to a well-posed one, which I resolved. Thank you, anyway.
$endgroup$
– J. Doe
Jan 11 at 1:23