Find likelihood for a given posterior
$begingroup$
Let $p(z)=mathcal{N}(z|0,1)$ be the standard normal density. Is there a conditional density $p(x|z)$ that could be expressed analytically and that would make the posterior $p(z|x)=mathcal{N}(z|phi(x), bullet)$ a normal distribution whose mean is a non-linear function of $x$ (say, $phi(x)=x^2$ for example) ? The variance could be arbitrary.
probability statistics normal-distribution machine-learning bayesian
asked Jan 13 at 15:22
NocturneNocturne
1,101510
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add a comment |
$begingroup$
Let $p(z)=mathcal{N}(z|0,1)$ be the standard normal density. Is there a conditional density $p(x|z)$ that could be expressed analytically and that would make the posterior $p(z|x)=mathcal{N}(z|phi(x), bullet)$ a normal distribution whose mean is a non-linear function of $x$ (say, $phi(x)=x^2$ for example) ? The variance could be arbitrary.
probability statistics normal-distribution machine-learning bayesian
asked Jan 13 at 15:22
NocturneNocturne
1,101510
$endgroup$
add a comment |
$begingroup$
Let $p(z)=mathcal{N}(z|0,1)$ be the standard normal density. Is there a conditional density $p(x|z)$ that could be expressed analytically and that would make the posterior $p(z|x)=mathcal{N}(z|phi(x), bullet)$ a normal distribution whose mean is a non-linear function of $x$ (say, $phi(x)=x^2$ for example) ? The variance could be arbitrary.
probability statistics normal-distribution machine-learning bayesian
asked Jan 13 at 15:22
NocturneNocturne
1,101510
$endgroup$
Let $p(z)=mathcal{N}(z|0,1)$ be the standard normal density. Is there a conditional density $p(x|z)$ that could be expressed analytically and that would make the posterior $p(z|x)=mathcal{N}(z|phi(x), bullet)$ a normal distribution whose mean is a non-linear function of $x$ (say, $phi(x)=x^2$ for example) ? The variance could be arbitrary.
probability statistics normal-distribution machine-learning bayesian
probability statistics normal-distribution machine-learning bayesian
asked Jan 13 at 15:22
NocturneNocturne
1,101510
asked Jan 13 at 15:22
NocturneNocturne
1,101510
asked Jan 13 at 15:22
NocturneNocturne
1,101510
asked Jan 13 at 15:22
NocturneNocturne
1,101510
asked Jan 13 at 15:22
NocturneNocturne
1,101510
1,101510
add a comment |
add a comment |
1 Answer
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$begingroup$
I take what you have written as the prior distribution for $z$ being $mathcal{N}(0,1)$ a standard normal, i.e. with mean $0$ and variance $1$.
If the conditional density for $x$ given $z$ is for example $frac 1 {xsqrt{2pi}} e^{-{left(log_e(x)-zright)^2}/{2}}$ (log-normal with parameters $z$ and $1$) for $x gt 0$, then the posterior distribution for $z$ having observed $x$ would be $mathcal{N}left(frac12log_e(x),frac12right)$
answered Jan 14 at 0:24
HenryHenry
99.7k479165
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1 Answer
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1 Answer
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$begingroup$
I take what you have written as the prior distribution for $z$ being $mathcal{N}(0,1)$ a standard normal, i.e. with mean $0$ and variance $1$.
If the conditional density for $x$ given $z$ is for example $frac 1 {xsqrt{2pi}} e^{-{left(log_e(x)-zright)^2}/{2}}$ (log-normal with parameters $z$ and $1$) for $x gt 0$, then the posterior distribution for $z$ having observed $x$ would be $mathcal{N}left(frac12log_e(x),frac12right)$
answered Jan 14 at 0:24
HenryHenry
99.7k479165
$endgroup$
add a comment |
$begingroup$
I take what you have written as the prior distribution for $z$ being $mathcal{N}(0,1)$ a standard normal, i.e. with mean $0$ and variance $1$.
If the conditional density for $x$ given $z$ is for example $frac 1 {xsqrt{2pi}} e^{-{left(log_e(x)-zright)^2}/{2}}$ (log-normal with parameters $z$ and $1$) for $x gt 0$, then the posterior distribution for $z$ having observed $x$ would be $mathcal{N}left(frac12log_e(x),frac12right)$
answered Jan 14 at 0:24
HenryHenry
99.7k479165
$endgroup$
add a comment |
$begingroup$
I take what you have written as the prior distribution for $z$ being $mathcal{N}(0,1)$ a standard normal, i.e. with mean $0$ and variance $1$.
If the conditional density for $x$ given $z$ is for example $frac 1 {xsqrt{2pi}} e^{-{left(log_e(x)-zright)^2}/{2}}$ (log-normal with parameters $z$ and $1$) for $x gt 0$, then the posterior distribution for $z$ having observed $x$ would be $mathcal{N}left(frac12log_e(x),frac12right)$
answered Jan 14 at 0:24
HenryHenry
99.7k479165
$endgroup$
I take what you have written as the prior distribution for $z$ being $mathcal{N}(0,1)$ a standard normal, i.e. with mean $0$ and variance $1$.
If the conditional density for $x$ given $z$ is for example $frac 1 {xsqrt{2pi}} e^{-{left(log_e(x)-zright)^2}/{2}}$ (log-normal with parameters $z$ and $1$) for $x gt 0$, then the posterior distribution for $z$ having observed $x$ would be $mathcal{N}left(frac12log_e(x),frac12right)$
answered Jan 14 at 0:24
HenryHenry
99.7k479165
answered Jan 14 at 0:24
HenryHenry
99.7k479165
answered Jan 14 at 0:24
HenryHenry
99.7k479165
answered Jan 14 at 0:24
HenryHenry
99.7k479165
99.7k479165
add a comment |
add a comment |
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