How to find which of two events -drawn from a normal distribution- is more likely?
$begingroup$
I know the probability of event A is given by:
$$Phi(f(x)+g(y)) - Phi(f(x)-g(y)), $$
and the probability of event B is
$$Phi(m(y)+n(x)) - Phi(m(y)-n(x)).$$
where $Phi$ is the cumulative distribution function of the standard normal.
I want to find the more likely of those two events as a function of $x$ and $y$. I tried a few different approaches, but I couldn't make it much simpler than this:
$$Pr(A|x,y) > Pr(B|x,y)
\ Leftrightarrow frac{1}{2 pi} intlimits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > frac{1}{2 pi} intlimits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt $$
An obvious sufficient condition for $Pr(A|x,y) > Pr(B|x,y)$ is $f(x) - g(y) < m(y) - n(x)$ and $f(x) + g(y) > m(y) + n(x)$. But I can't figure out a tractable necessary condition. I would appreciate any hints!
Edit: I can assume that $f$ and $m$ are linear, and $g$ and $n$ quadratic, if that is of any use.
inequality probability-distributions normal-distribution functional-inequalities
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
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add a comment |
$begingroup$
I know the probability of event A is given by:
$$Phi(f(x)+g(y)) - Phi(f(x)-g(y)), $$
and the probability of event B is
$$Phi(m(y)+n(x)) - Phi(m(y)-n(x)).$$
where $Phi$ is the cumulative distribution function of the standard normal.
I want to find the more likely of those two events as a function of $x$ and $y$. I tried a few different approaches, but I couldn't make it much simpler than this:
$$Pr(A|x,y) > Pr(B|x,y)
\ Leftrightarrow frac{1}{2 pi} intlimits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > frac{1}{2 pi} intlimits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt $$
An obvious sufficient condition for $Pr(A|x,y) > Pr(B|x,y)$ is $f(x) - g(y) < m(y) - n(x)$ and $f(x) + g(y) > m(y) + n(x)$. But I can't figure out a tractable necessary condition. I would appreciate any hints!
Edit: I can assume that $f$ and $m$ are linear, and $g$ and $n$ quadratic, if that is of any use.
inequality probability-distributions normal-distribution functional-inequalities
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
$endgroup$
add a comment |
$begingroup$
I know the probability of event A is given by:
$$Phi(f(x)+g(y)) - Phi(f(x)-g(y)), $$
and the probability of event B is
$$Phi(m(y)+n(x)) - Phi(m(y)-n(x)).$$
where $Phi$ is the cumulative distribution function of the standard normal.
I want to find the more likely of those two events as a function of $x$ and $y$. I tried a few different approaches, but I couldn't make it much simpler than this:
$$Pr(A|x,y) > Pr(B|x,y)
\ Leftrightarrow frac{1}{2 pi} intlimits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > frac{1}{2 pi} intlimits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt $$
An obvious sufficient condition for $Pr(A|x,y) > Pr(B|x,y)$ is $f(x) - g(y) < m(y) - n(x)$ and $f(x) + g(y) > m(y) + n(x)$. But I can't figure out a tractable necessary condition. I would appreciate any hints!
Edit: I can assume that $f$ and $m$ are linear, and $g$ and $n$ quadratic, if that is of any use.
inequality probability-distributions normal-distribution functional-inequalities
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
$endgroup$
I know the probability of event A is given by:
$$Phi(f(x)+g(y)) - Phi(f(x)-g(y)), $$
and the probability of event B is
$$Phi(m(y)+n(x)) - Phi(m(y)-n(x)).$$
where $Phi$ is the cumulative distribution function of the standard normal.
I want to find the more likely of those two events as a function of $x$ and $y$. I tried a few different approaches, but I couldn't make it much simpler than this:
$$Pr(A|x,y) > Pr(B|x,y)
\ Leftrightarrow frac{1}{2 pi} intlimits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > frac{1}{2 pi} intlimits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt $$
An obvious sufficient condition for $Pr(A|x,y) > Pr(B|x,y)$ is $f(x) - g(y) < m(y) - n(x)$ and $f(x) + g(y) > m(y) + n(x)$. But I can't figure out a tractable necessary condition. I would appreciate any hints!
Edit: I can assume that $f$ and $m$ are linear, and $g$ and $n$ quadratic, if that is of any use.
inequality probability-distributions normal-distribution functional-inequalities
inequality probability-distributions normal-distribution functional-inequalities
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
edited Dec 4 '18 at 22:56
deanavery
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
asked Dec 3 '18 at 23:14
deanaverydeanavery
63
63
add a comment |
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