Marginalization of transformed probability distribution The Next CEO of Stack...
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Marginalization of transformed probability distribution
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$begingroup$
Suppose we have a probability distribution $P(x_1, x_2)$ and its transformation $frac{1}{C} P^2(x_1, x_2)$ with some appropriate normalization constant $C$. Does the following hold:
$$
int_{-infty}^infty frac{1}{C} P^2(x_1, x_2) dx_2 = frac{1}{C_2} P^2(x_1)
$$
with some other appropriate normalization constant $C_2$?
So in other words, do I end up in the same probability distribution when I first marginalize and then square, compared to when I first square and the nmarginalize.
probability-distributions marginal-distribution
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
$endgroup$
add a comment |
$begingroup$
Suppose we have a probability distribution $P(x_1, x_2)$ and its transformation $frac{1}{C} P^2(x_1, x_2)$ with some appropriate normalization constant $C$. Does the following hold:
$$
int_{-infty}^infty frac{1}{C} P^2(x_1, x_2) dx_2 = frac{1}{C_2} P^2(x_1)
$$
with some other appropriate normalization constant $C_2$?
So in other words, do I end up in the same probability distribution when I first marginalize and then square, compared to when I first square and the nmarginalize.
probability-distributions marginal-distribution
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
$endgroup$
add a comment |
$begingroup$
Suppose we have a probability distribution $P(x_1, x_2)$ and its transformation $frac{1}{C} P^2(x_1, x_2)$ with some appropriate normalization constant $C$. Does the following hold:
$$
int_{-infty}^infty frac{1}{C} P^2(x_1, x_2) dx_2 = frac{1}{C_2} P^2(x_1)
$$
with some other appropriate normalization constant $C_2$?
So in other words, do I end up in the same probability distribution when I first marginalize and then square, compared to when I first square and the nmarginalize.
probability-distributions marginal-distribution
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
$endgroup$
Suppose we have a probability distribution $P(x_1, x_2)$ and its transformation $frac{1}{C} P^2(x_1, x_2)$ with some appropriate normalization constant $C$. Does the following hold:
$$
int_{-infty}^infty frac{1}{C} P^2(x_1, x_2) dx_2 = frac{1}{C_2} P^2(x_1)
$$
with some other appropriate normalization constant $C_2$?
So in other words, do I end up in the same probability distribution when I first marginalize and then square, compared to when I first square and the nmarginalize.
probability-distributions marginal-distribution
probability-distributions marginal-distribution
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
edited Mar 19 at 16:56
Luca Thiede
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
asked Mar 18 at 2:18
Luca ThiedeLuca Thiede
1486
1486
add a comment |
add a comment |
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