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I am reading a game theory lecture notes. Some parts involve a continuous time Bayesian updating computation which I didn't really understand.

There are two states $\{Good,Bad\}$. At time t people has prior belief that the state of world being $Good$ as $p_t$. There is conclusive signal which can confirm that the state is good. Within time interval $[t,t+d_t)$, the probability of good signal's arrival is $\lambda\cdot K_t$. Then it wrote the updating rule as

$$p_t+dp_t=\frac{p_t(1-\lambda\cdot K_td_t)}{1-p_t\lambda\cdot K_t d_t},$$

and further conclude that

$$\frac{dp_t}{d_t}=-\lambda p_t(1-p_t)K_t.$$

I didn't understand both computation here.

For the first result, it seems the author only record $$p_{t+dt}(good|no\ news\ arrived).$$

Also, even if I accept the first equation, should I subtract $p_t$ on both sides and divided both sides by $d_t$? This does not give the result the author gave.

Severus
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1 Answers1

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Define $G$ and $B$ to be event of being in good and bad state, respectively. Let $I$ be event that no news arrives in the interval $[t, t+dt)$. As you state, we are interested in $\Pr(G|I)$, because if news arrives, then we are sure to be in good state. So, we are interested in evolution of our belief of being in good state as time passes without the arrival of news. Using Baye's theorem $$ \begin{split} \Pr(G|I) &= \frac{\Pr(I|G)\Pr(G)}{\Pr(I|G)\Pr(G)+\Pr(I|B)\Pr(B)} = \frac{(1-\lambda K_tdt)p_t}{(1-\lambda K_tdt)p_t + 1(1-p_t)} \\ &= \frac{(1-\lambda K_tdt)p_t}{1-\lambda p_tK_t dt}. \end{split} $$ The posterior probability after observing the information over time $[t, t+dt)$, be denoted by $p_t + dp_t$. Therefore, $$ p_t + dp_t = \frac{(1-\lambda K_tdt)p_t}{1-\lambda p_tK_t dt} $$ Simplifying it after ignoring $dp_tdt$, we get $$ \frac{dp_t}{dt} = -\lambda p_t(1-p_t)K_t. $$

vdesai
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