Modified prior distribution

Question

Suppose we have a prior $p(z)$ and conditional likelihood $p(x|z)$ whose integral gives the evidence $p(x)=\int p(x|z)p(z)dz$. Now let $q(x)$ be a distribution different from $p(x)$ and let $p_q(z)=\int p(z|x)q(x)dx$ be the resulting prior we get when we integrate the posterior $p(z|x)$ with respect to $q(x)$ instead of $p(x)$. Is it true that $\int p(x|z)p_q(z)dz=q(x)$ ? It seems very intuitive, yet I cannot prove it.

Answer 1

Yes if that is what your notation represents, though you use $p()$ to mean various different things

Some alternative notation might make it clearer:

• Suppose the likelihood of $z$ given the observation $x$ is proportional to the conditional probability density $p(x \mid z)$

• If the prior density for $z$ is $\pi_1(z)$, then the posterior density for $z$ given the observation $x$ is $\pi_1(z \mid x) =\frac{\pi_1(z) \,p(x \mid z) }{f_1(x)}$, where $f_1(x)=\int \pi_1(z) \,p(x \mid z) \, dz$ represents the marginal density for $x$ based on that prior distribution for $z$

• If the prior density for $z$ is $\pi_2(z)$, then the posterior density for $z$ given the observation $x$ is $\pi_2(z \mid x) =\frac{\pi_2(z) \,p(x \mid z) }{f_2(x)}$, where $f_2(x)=\int \pi_2(z) \,p(x \mid z) \, dz$ represents the marginal density for $x$ based on that prior distribution for $z$

If your underlying question is whether the marginal density for $x$ depends on the assumed prior distribution for $z$, then the answer is yes