<p>Let me denote the events with simpler symbols</p>
<ul>
<li><span class="math-container">$A = X_{t+1}$</span></li>
<li><span class="math-container">$C = e_{1:t} = e_1, \dots e_t$</span></li>
<li><span class="math-container">$D = e_{t+1}$</span></li>
</ul>
<p>So, we can rewrite</p>
<p><span class="math-container">$$
P(X_{t+1} \mid e_{1:t}, e_{t+1}) = \alpha P(e_{t+1} \mid X_{t+1},  e_{1:t}) P(X_{t+1} \mid  e_{1:t})\tag{1}\label{1}
$$</span></p>
<p>as follows</p>
<p><span class="math-container">$$
P(A \mid C, D) = \alpha P(D \mid A,  C) P(A \mid  C)
\tag{2}\label{2}
$$</span></p>
<p>Now, note that <span class="math-container">$C, D$</span> is a shorthand for <span class="math-container">$C \cap D = B$</span>, so <span class="math-container">$P(A \mid C, D) = P(A \mid B)$</span>.</p>
<p>So, by the Bayes rule, we have</p>
<p><span class="math-container">\begin{align}
P(A \mid B) 
&amp;= \frac{P(B \mid A) P(A) }{P(B)} \\
&amp;= \frac{\color{red}{P(C, D \mid A)} P(A) }{P(C, D)}
\tag{3}\label{3}
\end{align}</span></p>
<p>Let's expand the red term</p>
<p><span class="math-container">\begin{align}
\color{red}{P(C, D \mid A)}  
&amp;= P(B \mid A) \\
&amp;= \frac{P(B, A) }{P(A)}\\
&amp;= \color{blue}{\frac{P(C, D, A) }{P(A)}}
\tag{4}\label{4}
\end{align}</span></p>
<p>Now, let <span class="math-container">$E = C, A = C \cap A$</span>. So, let's continue writing \ref{4}</p>
<p><span class="math-container">\begin{align}
\color{blue}{\frac{P(C, D, A) }{P(A)}}
&amp;=\frac{P(D, E) }{P(A)} \\
&amp;= \frac{P(D \mid E) P(E) }{P(A)} \\
&amp;= \frac{P(D \mid C, A) P(C, A) }{P(A)} \\
&amp;= \color{green}{\frac{P(D \mid C, A) P(A \mid C) P(C) }{P(A)}}
\tag{5}\label{5}
\end{align}</span>
Let's put \ref{5} back into \ref{3}</p>
<p><span class="math-container">\begin{align}
P(A \mid B) 
&amp;= \frac{\color{green}{\frac{P(D \mid C, A) P(A \mid C) P(C) }{P(A)}} P(A) }{P(C, D)} \\
&amp;= \frac{ P(D \mid C, A) P(A \mid C) P(C) }{P(C, D)} \\
&amp;= \frac{ P(D \mid C, A) P(A \mid C) P(C) }{P(D \mid C) P(C)} \\
&amp;= \frac{ P(D \mid C, A) P(A \mid C) }{P(D \mid C)} \\
&amp;= \frac{1}{P(D \mid C)} P(D \mid C, A) P(A \mid C) \\
&amp;= \alpha P(D \mid C, A) P(A \mid C)
\tag{6}\label{6}
\end{align}</span></p>

CC BY-SA 4.0
2022-04-02T18:51:55.730Z
I got lost at that line, let A = C and A after (4).
mLstudent33
CC BY-SA 4.0
2022-04-02T18:58:36.023Z
@mLstudent33 It's let $E = C, A$. But $C, A$ is a shorthand for $C \cap A$ (i.e. the intersection of C and A, so I am denoting $C \cap A$ by $E$ in order to apply the regular definition of joint and conditional probabilities. Note that events are sets. I've edited the answer to add another step that clarifies what I did.
nbro
CC BY-SA 4.0
2022-04-02T19:41:57.530Z
Thanks but I still do not see how 5 can be substituted for P(B | A).  Oops I see it now
mLstudent33
CC BY-SA 4.0
2022-04-02T19:47:28.613Z
P(C,D | A) is P(C,D | A) :)
mLstudent33
CC BY-SA 4.0
2022-04-02T23:55:49.633Z
Are you returning back or the question is impressive enough?
hanugm
CC BY-SA 4.0
2022-04-03T09:30:38.237Z
@hanugm I've restarted to answer questions every once in a while. I answer the ones that I either can answer quickly or I like. This was one that I liked because it was about the Bayes rule ;)
nbro
CC BY-SA 4.0
2022-04-02T11:08:44.203Z
2022-04-05T07:55:40.613Z
2022-04-05T07:55:40.613Z
nbro
<p>On page 572 of Norvig &amp; Russell's AI book (edition 3)</p>
<p><a href="https://i.sstatic.net/yMBRU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yMBRU.png" alt="Bayes rule filtering" /></a></p>
<p>Going from the first line to the second line in one shot like that, I am lost.</p>
<p>Can someone walk me through it step by step?</p>
<p>I tried but got:</p>
<p><span class="math-container">$$
\frac{P(X_{t+1}) * P(e_{1:t} | X_{t+1}) * P(e_{t+1} | X_{t+1}) }{P(e_{1:t}, e_{t+1})}
$$</span></p>
<p>and then do not know how to turn it into that second line that says (using Bayes's rule)</p>

CC BY-SA 4.0
2022-04-01T08:50:50.257Z
You can use mathjax to format the equations.
nbro
<norvig-russell><bayes-theorem>
Why is $P(X_{t+1} \mid e_{1:t}, e_{t+1}) = \alpha P(e_{t+1} \mid X_{t+1}, e_{1:t}) P(X_{t+1} \mid e_{1:t})$ true in Norvig & Russell's book?
mLstudent33
# Why is $P(X_{t+1} \mid e_{1:t}, e_{t+1}) = \alpha P(e_{t+1} \mid X_{t+1}, e_{1:t}) P(X_{t+1} \mid e_{1:t})$ true in Norvig & Russell's book?

(Asked by: username_0 on 2022-04-01)

On page 572 of Norvig & Russell's AI book (edition 3)

[![Bayes rule filtering](https://i.sstatic.net/yMBRU.png)](https://i.sstatic.net/yMBRU.png)

Going from the first line to the second line in one shot like that, I am lost.

Can someone walk me through it step by step?

I tried but got:

$$
\frac{P(X\_{t+1}) \* P(e\_{1:t} | X\_{t+1}) \* P(e\_{t+1} | X\_{t+1}) }{P(e\_{1:t}, e\_{t+1})}
$$

and then do not know how to turn it into that second line that says (using Bayes's rule)

**Tags:** norvig-russell, bayes-theorem

> **Comments:**
> **username_1**: You can use mathjax to format the equations.  

---

## 1 answers

### Answer by username_1 on 2022-04-02. Score: 1 | Accepted answer

Let me denote the events with simpler symbols

* $A = X\_{t+1}$
* $C = e\_{1:t} = e\_1, \dots e\_t$
* $D = e\_{t+1}$

So, we can rewrite

$$
P(X\_{t+1} \mid e\_{1:t}, e\_{t+1}) = \alpha P(e\_{t+1} \mid X\_{t+1}, e\_{1:t}) P(X\_{t+1} \mid e\_{1:t})\tag{1}\label{1}
$$

as follows

$$
P(A \mid C, D) = \alpha P(D \mid A, C) P(A \mid C)
\tag{2}\label{2}
$$

Now, note that $C, D$ is a shorthand for $C \cap D = B$, so $P(A \mid C, D) = P(A \mid B)$.

So, by the Bayes rule, we have

\begin{align}
P(A \mid B)
&= \frac{P(B \mid A) P(A) }{P(B)} \\
&= \frac{\color{red}{P(C, D \mid A)} P(A) }{P(C, D)}
\tag{3}\label{3}
\end{align}

Let's expand the red term

\begin{align}
\color{red}{P(C, D \mid A)}
&= P(B \mid A) \\
&= \frac{P(B, A) }{P(A)}\\
&= \color{blue}{\frac{P(C, D, A) }{P(A)}}
\tag{4}\label{4}
\end{align}

Now, let $E = C, A = C \cap A$. So, let's continue writing \ref{4}

\begin{align}
\color{blue}{\frac{P(C, D, A) }{P(A)}}
&=\frac{P(D, E) }{P(A)} \\
&= \frac{P(D \mid E) P(E) }{P(A)} \\
&= \frac{P(D \mid C, A) P(C, A) }{P(A)} \\
&= \color{green}{\frac{P(D \mid C, A) P(A \mid C) P(C) }{P(A)}}
\tag{5}\label{5}
\end{align}
Let's put \ref{5} back into \ref{3}

\begin{align}
P(A \mid B)
&= \frac{\color{green}{\frac{P(D \mid C, A) P(A \mid C) P(C) }{P(A)}} P(A) }{P(C, D)} \\
&= \frac{ P(D \mid C, A) P(A \mid C) P(C) }{P(C, D)} \\
&= \frac{ P(D \mid C, A) P(A \mid C) P(C) }{P(D \mid C) P(C)} \\
&= \frac{ P(D \mid C, A) P(A \mid C) }{P(D \mid C)} \\
&= \frac{1}{P(D \mid C)} P(D \mid C, A) P(A \mid C) \\
&= \alpha P(D \mid C, A) P(A \mid C)
\tag{6}\label{6}
\end{align}

> **Comments:**
> **username_0**: I got lost at that line, let A = C and A after (4).  
> **username_1**: @username_0 It's let $E = C, A$. But $C, A$ is a shorthand for $C \cap A$ (i.e. the intersection of C and A, so I am denoting $C \cap A$ by $E$ in order to apply the regular definition of joint and conditional probabilities. Note that events are sets. I've edited the answer to add another step that clarifies what I did.  
> **username_0**: Thanks but I still do not see how 5 can be substituted for P(B | A).  Oops I see it now  
> **username_0**: P(C,D | A) is P(C,D | A) :)  
> **username_2**: Are you returning back or the question is impressive enough?  
> **username_1**: @username_2 I've restarted to answer questions every once in a while. I answer the ones that I either can answer quickly or I like. This was one that I liked because it was about the Bayes rule ;)  

---

