Show that the function is symmetric.
$begingroup$
Consider the following proposal function
$q(vec{theta'}|vec{theta})$:
For a given vector $vec{theta}=(theta_1,theta_2,theta_3)$
generate $vec{theta'}=(theta_1',theta_2',theta_3')$ as
$$theta_1'=|theta_1+delta_1|$$ $$theta_2'=|theta_2+delta_2|$$
$$theta_3'=|theta_3+delta_3|$$
where $delta_isim N(0,0.1^2).$ Show that the proposal function is
symmetric, i.e, that $q(vec{theta'}|vec{theta}) =
q(vec{theta}|vec{theta'}).$
I have that
$$q(vec{theta'}|vec{theta}) = q(|vec{theta} + vec{delta}|)$$
but when I try to calculate $q(vec{theta}|vec{theta'})$ I find that I need to solve $vec{theta}$ from $vec{theta'}=|vec{theta}+vec{delta}|.$ Am I missing something here?
probability probability-theory
$endgroup$
add a comment |
$begingroup$
Consider the following proposal function
$q(vec{theta'}|vec{theta})$:
For a given vector $vec{theta}=(theta_1,theta_2,theta_3)$
generate $vec{theta'}=(theta_1',theta_2',theta_3')$ as
$$theta_1'=|theta_1+delta_1|$$ $$theta_2'=|theta_2+delta_2|$$
$$theta_3'=|theta_3+delta_3|$$
where $delta_isim N(0,0.1^2).$ Show that the proposal function is
symmetric, i.e, that $q(vec{theta'}|vec{theta}) =
q(vec{theta}|vec{theta'}).$
I have that
$$q(vec{theta'}|vec{theta}) = q(|vec{theta} + vec{delta}|)$$
but when I try to calculate $q(vec{theta}|vec{theta'})$ I find that I need to solve $vec{theta}$ from $vec{theta'}=|vec{theta}+vec{delta}|.$ Am I missing something here?
probability probability-theory
$endgroup$
add a comment |
$begingroup$
Consider the following proposal function
$q(vec{theta'}|vec{theta})$:
For a given vector $vec{theta}=(theta_1,theta_2,theta_3)$
generate $vec{theta'}=(theta_1',theta_2',theta_3')$ as
$$theta_1'=|theta_1+delta_1|$$ $$theta_2'=|theta_2+delta_2|$$
$$theta_3'=|theta_3+delta_3|$$
where $delta_isim N(0,0.1^2).$ Show that the proposal function is
symmetric, i.e, that $q(vec{theta'}|vec{theta}) =
q(vec{theta}|vec{theta'}).$
I have that
$$q(vec{theta'}|vec{theta}) = q(|vec{theta} + vec{delta}|)$$
but when I try to calculate $q(vec{theta}|vec{theta'})$ I find that I need to solve $vec{theta}$ from $vec{theta'}=|vec{theta}+vec{delta}|.$ Am I missing something here?
probability probability-theory
$endgroup$
Consider the following proposal function
$q(vec{theta'}|vec{theta})$:
For a given vector $vec{theta}=(theta_1,theta_2,theta_3)$
generate $vec{theta'}=(theta_1',theta_2',theta_3')$ as
$$theta_1'=|theta_1+delta_1|$$ $$theta_2'=|theta_2+delta_2|$$
$$theta_3'=|theta_3+delta_3|$$
where $delta_isim N(0,0.1^2).$ Show that the proposal function is
symmetric, i.e, that $q(vec{theta'}|vec{theta}) =
q(vec{theta}|vec{theta'}).$
I have that
$$q(vec{theta'}|vec{theta}) = q(|vec{theta} + vec{delta}|)$$
but when I try to calculate $q(vec{theta}|vec{theta'})$ I find that I need to solve $vec{theta}$ from $vec{theta'}=|vec{theta}+vec{delta}|.$ Am I missing something here?
probability probability-theory
probability probability-theory
asked Dec 6 '18 at 15:53
ParsevalParseval
2,8751718
2,8751718
add a comment |
add a comment |
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