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Calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$.
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Let $F$ and $G$ be the cumulative functions of $N(mu_1,sigma_1^2)$ and $N(mu_2,sigma_2^2)$. Let $U$ be a uniformly distributed random variable on $[0,1]$. How to calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$?
For standard normal random variables, $F^-1$ is named as probit function, but I cannot find a direct expression to calculate the expectation as an integration.
probability-theory normal-distribution
edited Jul 23 at 5:19
Did
242k23208443
asked Jul 23 at 3:01
Zayne
407
add a comment |Â
up vote
3
down vote
favorite
Let $F$ and $G$ be the cumulative functions of $N(mu_1,sigma_1^2)$ and $N(mu_2,sigma_2^2)$. Let $U$ be a uniformly distributed random variable on $[0,1]$. How to calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$?
For standard normal random variables, $F^-1$ is named as probit function, but I cannot find a direct expression to calculate the expectation as an integration.
probability-theory normal-distribution
edited Jul 23 at 5:19
Did
242k23208443
asked Jul 23 at 3:01
Zayne
407
2
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
1
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $F$ and $G$ be the cumulative functions of $N(mu_1,sigma_1^2)$ and $N(mu_2,sigma_2^2)$. Let $U$ be a uniformly distributed random variable on $[0,1]$. How to calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$?
For standard normal random variables, $F^-1$ is named as probit function, but I cannot find a direct expression to calculate the expectation as an integration.
probability-theory normal-distribution
edited Jul 23 at 5:19
Did
242k23208443
asked Jul 23 at 3:01
Zayne
407
Let $F$ and $G$ be the cumulative functions of $N(mu_1,sigma_1^2)$ and $N(mu_2,sigma_2^2)$. Let $U$ be a uniformly distributed random variable on $[0,1]$. How to calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$?
For standard normal random variables, $F^-1$ is named as probit function, but I cannot find a direct expression to calculate the expectation as an integration.
probability-theory normal-distribution
edited Jul 23 at 5:19
Did
242k23208443
asked Jul 23 at 3:01
Zayne
407
edited Jul 23 at 5:19
Did
242k23208443
edited Jul 23 at 5:19
Did
242k23208443
edited Jul 23 at 5:19
Did
242k23208443
242k23208443
asked Jul 23 at 3:01
Zayne
407
asked Jul 23 at 3:01
Zayne
407
asked Jul 23 at 3:01
Zayne
407
407
2
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
1
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11
add a comment |Â
2
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
1
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11
2
2
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
1
1
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11
add a comment |Â
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2
If $F$ is a CDF of a continuous random variable, $U$ is a uniformly distributed random variable on $[0, 1]$, then the CDF of $F^-1(U)$ is exactly the same as $F$ - this is known as the inverse probability integral transform: $PrF^-1(U) leq x = PrU leq F(x) = F(x)$. So the distribution of both random variable are known. Next problem is that since both have the "same sources of randomness", you may express them in terms of the same random variable $Z$, say a standard normal one.
– BGM
Jul 23 at 3:17
1
Sangchul Lee has already given the procedure for computing this in his comment on your previous post. Explicit computation in terms of $mu_i, sigma _i$ , $i=1,2$ may not be possible.
– Kavi Rama Murthy
Jul 23 at 6:11