No, you cannot marginalize the prior and then multiply the marginal with g(x). Here is why:

The correct way to use Bayes theorem is to do the following (also suggested by John):

$g_p(x,y,z) \propto g(x,y,z) k(x,y,z)$

Thus,

$g_p(x) \propto \int_{y,z} \bigl(g(x,y,z) k(x,y,z) \bigr)$

Your want to do the following:

$g_p^'(x) \propto \int_{y,z} \bigl(g(x,y,z) \bigr) \int_{y,z} \bigl(k(x,y,z) \bigr)$

In general, $g_p(x)$ and $g_p^'(x)$ will not be identical.

No, you cannot marginalize the prior and then multiply the marginal with g(x). Here is why:

The correct way to use Bayes theorem is to do the following (also suggested by John):

$g_{p1}(x,y,z) \propto g(x,y,z) k(x,y,z)$

Thus,

$g_{p1}(x) \propto \int_{y,z} \bigl(g(x,y,z) k(x,y,z) \bigr)$

Your want to do the following:

$g_{p2}(x) \propto \int_{y,z} \bigl(g(x,y,z) \bigr) \int_{y,z} \bigl(k(x,y,z) \bigr)$

In general, $g_{p1}(x)$ and $g_{p2}(x)$ will not be identical.

Another idea is to integrate out y and z from K(.) to get $K_x(x)$. You could then use $K_x(x)$ as the prior for g(x). If integrating out the prior is not possible then you have to follow John's suggestion.

No, you cannot marginalize the prior and then multiply the marginal with g(x). Here is why:

The correct way to use Bayes theorem is to do the following (also suggested by John):

$g_p(x,y,z) \propto g(x,y,z) k(x,y,z)$

Thus,

$g_p(x) \propto \int_{y,z} \bigl(g(x,y,z) k(x,y,z) \bigr)$

Your want to do the following:

$g_p^'(x) \propto \int_{y,z} \bigl(g(x,y,z) \bigr) \int_{y,z} \bigl(k(x,y,z) \bigr)$

In general, $g_p(x)$ and $g_p^'(x)$ will not be identical.