Probability of a point from (generalized) Student's T will be greater than point from another (generalized)...
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Consider the Generalized Student's T distribution, $T(mu,sigma^2,nu)$.
Suppose $X_1 sim T(mu_1,sigma_1^2,nu_1)$ and $X_2 sim T(mu_2,sigma_2^2,nu_2)$, with $X_1$ and $X_2$ independent. What is $P(X_1 > X_2) = P(X_1 - X_2 > 0)$? Clearly, $E[X_1+X_2] = mu_1+mu_2$, but the rest isn't so clear.
Is there easily computable solution similar to: Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
probability
asked Nov 30 '18 at 23:43
gdouggdoug
1234
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add a comment |
$begingroup$
Consider the Generalized Student's T distribution, $T(mu,sigma^2,nu)$.
Suppose $X_1 sim T(mu_1,sigma_1^2,nu_1)$ and $X_2 sim T(mu_2,sigma_2^2,nu_2)$, with $X_1$ and $X_2$ independent. What is $P(X_1 > X_2) = P(X_1 - X_2 > 0)$? Clearly, $E[X_1+X_2] = mu_1+mu_2$, but the rest isn't so clear.
Is there easily computable solution similar to: Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
probability
asked Nov 30 '18 at 23:43
gdouggdoug
1234
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I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
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– I like Serena
Dec 1 '18 at 10:10
add a comment |
$begingroup$
Consider the Generalized Student's T distribution, $T(mu,sigma^2,nu)$.
Suppose $X_1 sim T(mu_1,sigma_1^2,nu_1)$ and $X_2 sim T(mu_2,sigma_2^2,nu_2)$, with $X_1$ and $X_2$ independent. What is $P(X_1 > X_2) = P(X_1 - X_2 > 0)$? Clearly, $E[X_1+X_2] = mu_1+mu_2$, but the rest isn't so clear.
Is there easily computable solution similar to: Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
probability
asked Nov 30 '18 at 23:43
gdouggdoug
1234
$endgroup$
Consider the Generalized Student's T distribution, $T(mu,sigma^2,nu)$.
Suppose $X_1 sim T(mu_1,sigma_1^2,nu_1)$ and $X_2 sim T(mu_2,sigma_2^2,nu_2)$, with $X_1$ and $X_2$ independent. What is $P(X_1 > X_2) = P(X_1 - X_2 > 0)$? Clearly, $E[X_1+X_2] = mu_1+mu_2$, but the rest isn't so clear.
Is there easily computable solution similar to: Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
probability
probability
asked Nov 30 '18 at 23:43
gdouggdoug
1234
asked Nov 30 '18 at 23:43
gdouggdoug
1234
asked Nov 30 '18 at 23:43
gdouggdoug
1234
asked Nov 30 '18 at 23:43
gdouggdoug
1234
asked Nov 30 '18 at 23:43
gdouggdoug
1234
1234
$begingroup$
I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
$endgroup$
– I like Serena
Dec 1 '18 at 10:10
add a comment |
$begingroup$
I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
$endgroup$
– I like Serena
Dec 1 '18 at 10:10
$begingroup$
I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
$endgroup$
– I like Serena
Dec 1 '18 at 10:10
$begingroup$
I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
$endgroup$
– I like Serena
Dec 1 '18 at 10:10
add a comment |
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I don't think so. The proof that 2 normal distributions add up to a normal distribution is actually quite involved.
$endgroup$
– I like Serena
Dec 1 '18 at 10:10