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Copy file name to clipboardExpand all lines: lectures/likelihood_bayes.md
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@@ -76,7 +76,7 @@ $f$ or $g$.
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Before the beginning of time, nature once and for all decides whether she will draw a sequence of IID draws from $f$ or from $g$.
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We will sometimes let $q$ be the density that nature chose once and for all, so
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We let $q$ be the density that nature chose once and for all, so
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that $q$ is either $f$ or $g$, permanently.
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Nature knows which density it permanently draws from, but we the observers do not.
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def compute_div_m(f, g):
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"""
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Compute Jensen-Shannon divergence
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Compute KL(m, f) and KL(m, g)
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"""
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def m(w):
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return 0.5 * (f(w) + g(w))
@@ -602,7 +602,7 @@ Since $KL(m, f) < KL(m, g)$, $f$ is "closer" to the mixture distribution $m$.
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Hence by our discussion on KL divergence and likelihood ratio process in
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{doc}`likelihood_ratio_process`, $\log(L_t) \to \infty$ as $t \to \infty$.
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Now looking back to the key equation {eq}`eq_Bayeslaw1033`.
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Now look back at the key equation {eq}`eq_Bayeslaw1033`.
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Consider the function
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@@ -734,12 +734,12 @@ Thus, let $\{\pi_t(\omega)\}_{t=0}^\infty$ denote a particular sample path index
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\in \Omega$.
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We can think of nature as drawing an $\omega \in \Omega$ from a probability distribution
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${\textrm{Prob}}\Omega$ and then generating a single realization (or _simulation_) $\{\pi_t(\omega)\}_{t=0}^\infty$ of the process.
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${\textrm{Prob}}(\Omega)$ and then generating a single realization (or _simulation_) $\{\pi_t(\omega)\}_{t=0}^\infty$ of the process.
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The limit points of $\{\pi_t(\omega)\}_{t=0}^\infty$ as $t \rightarrow +\infty$ are realizations of a random variable that is swept out as we sample $\omega$ from $\Omega$ and construct repeated draws of $\{\pi_t(\omega)\}_{t=0}^\infty$.
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By staring at law of motion {eq}`eq_recur1` or {eq}`eq:like44`, we can figure out some things about the probability distribution of the limit points
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By staring at the law of motion {eq}`eq_recur1` or {eq}`eq:like44`, we can figure out some things about the probability distribution of the limit points
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@@ -765,8 +765,8 @@ $$
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$$
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Thus, for some realizations, $\lim_{\rightarrow + \infty} \pi_t(\omega) =1$
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while for other realizations, $\lim_{\rightarrow + \infty} \pi_t(\omega) =0$.
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Thus, for some realizations, $\lim_{t \rightarrow + \infty} \pi_t(\omega) =1$
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while for other realizations, $\lim_{t \rightarrow + \infty} \pi_t(\omega) =0$.
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Now let's remember that $\{\pi_t\}_{t=0}^\infty$ is a martingale and apply the law of iterated expectations.
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@@ -888,7 +888,7 @@ plt.show()
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The above graph indicates that
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* each of paths converges
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* each of the paths converges
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* some of the paths converge to $1$
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@@ -913,7 +913,7 @@ Evidently, by $t = 199$, $\pi_t$ has converged to either $0$ or $1$.
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The fraction of paths that have converged to $1$ is $.5$
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The fractions of paths that have converged to $0$ is also $.5$.
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The fraction of paths that have converged to $0$ is also $.5$.
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Does the fraction $.5$ ring a bell?
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@@ -954,9 +954,9 @@ $F$ more frequently along a sample path, and this pushes $\pi_t$ toward $1$.
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