fixing images

bayes-nets/d-separation.md

@@ -63,11 +63,11 @@ An analogous proof can be used to show the same thing for the case where $$X$$ h
 
 ## 6.4.2 Common Cause
 
-![Common Cause with no observations](../assets/images/cause_free.PNG)
+<img src="{{ site.baseurl }}/assets/images/cause_free.PNG" alt="Common Cause with no observations" />
 
 *Figure 3: Common Cause with no observations.*
 
-![Common Cause with Y observed](../assets/images/cause_observed.PNG)
+<img src="{{ site.baseurl }}/assets/images/cause_observed.PNG" alt="Common Cause with Y observed" />
 
 *Figure 4: Common Cause with Y observed.*
 
@@ -100,11 +100,11 @@ $$P(X | Z, y) = \frac{P(X, Z, y)}{P(Z, y)} = \frac{P(X|y) P(Z|y) P(y)}{P(Z|y) P(
 
 ## 6.4.3 Common Effect
 
-![Common Effect with no observations](../assets/images/effect_free.PNG)
+<img src="{{ site.baseurl }}/assets/images/effect_free.PNG" alt="Common Effect with no observations" />
 
 *Figure 5: Common Effect with no observations.*
 
-![Common Effect with Y observed](../assets/images/effect_observed.PNG)
+<img src="{{ site.baseurl }}/assets/images/effect_observed.PNG" alt="Common Effect with Y observed" />
 
 *Figure 6: Common Effect with Y observed.*
 
@@ -134,7 +134,7 @@ Common Effect can be viewed as ``opposite'' to Causal Chains and Common Cause 
 
 This same logic applies when conditioning on descendants of $$Y$$ in the graph. If one of $$Y$$'s descendant nodes is observed, as in Figure 7, $$X$$ and $$Z$$ are not guaranteed to be independent.
 
-![Common Effect with child observations](../assets/images/effect_children.PNG)
+<img src="{{ site.baseurl }}/assets/images/effect_children.PNG" alt="Common Effect with child observations" />
 
 *Figure 7: Common Effect with child observations.*
 
@@ -171,8 +171,8 @@ Any path in a graph from $$X$$ to $$Y$$ can be decomposed into a set of 3 consec
 
 **Active triples**: We can enumerate all possibilities of active and inactive triples using the three canonical graphs we presented below in the figures.
 
-![Active triples](../assets/images/active.PNG){:width="49%"}
-![Inactive triples](../assets/images/inactive.PNG){:width="49%"}
+<img src="{{ site.baseurl }}/assets/images/active.PNG" alt="Active triples" />
+<img src="{{ site.baseurl }}/assets/images/inactive.PNG" alt="Inactive triples" />
 
 ## 6.4.5 Examples
 
@@ -193,7 +193,7 @@ $$R \perp\!\!\!\perp B | T'$$ -- Not guaranteed
 </p>
 $$R \perp\!\!\!\perp T' | T$$ -- Guaranteed
 
-![Example 2](../assets/images/lrbdtt.png){:width="50%"} 
+<img src="{{ site.baseurl }}/assets/images/lrbdtt.png" alt="Example 2" />
 
 This graph contains combinations of all three canonical graphs (can you list them all?). 
 
@@ -211,7 +211,7 @@ $$L \perp\!\!\!\perp B | T'$$ -- Not guaranteed
 </p>
 $$L \perp\!\!\!\perp B | T, R$$ -- Guaranteed
 
-![Example 3](../assets/images/rtds.png)
+<img src="{{ site.baseurl }}/assets/images/rtds.png" alt="Example 3" />
 
 This graph contains combinations of all three canonical graphs. 
 <p>

bayes-nets/elimination.md

@@ -22,11 +22,11 @@ An alternate approach is to eliminate hidden variables one by one. To **eliminat
 
 A **factor** is defined simply as an _unnormalized probability_. At all points during variable elimination, each factor will be proportional to the probability it corresponds to, but the underlying distribution for each factor won't necessarily sum to 1 as a probability distribution should. The pseudocode for variable elimination is here:
 
-![Variable Elimination](../assets/images/VarElim.png)
+<img src="{{ site.baseurl }}/assets/images/VarElim.png" alt="Variable Elimination" />
 
 Let's make these ideas more concrete with an example. Suppose we have a model as shown below, where $$T$$, $$C$$, $$S$$, and $$E$$ can take on binary values. Here, $$T$$ represents the chance that an adventurer takes a treasure, $$C$$ represents the chance that a cage falls on the adventurer given that they take the treasure, $$S$$ represents the chance that snakes are released if an adventurer takes the treasure, and $$E$$ represents the chance that the adventurer escapes given information about the status of the cage and snakes.
 
-![Variable Elimination](../assets/images/another_bayes_nets.png)
+<img src="{{ site.baseurl }}/assets/images/another_bayes_nets.png" alt="Variable Elimination" />
 
 <p>
 </p>

bayes-nets/structure.md

@@ -13,11 +13,11 @@ In this class, we will refer to two rules for Bayes Net independences that can b
 
 - **Each node is conditionally independent of all its ancestor nodes (non-descendants) in the graph, given all of its parents.**
 
-  ![Parents](../assets/images/parents.png)
+<img src="{{ site.baseurl }}/assets/images/parents.png" alt="Parents" />
 
 - **Each node is conditionally independent of all other variables given its Markov blanket.** A variable’s Markov blanket consists of parents, children, and children’s other parents.
 
-  ![Markov Blanket](../assets/images/blanket.png)
+<img src="{{ site.baseurl }}/assets/images/blanket.png" alt="Markov Blanket" />
 
 Using these tools, we can return to the assertion in the previous section: that we can get the joint distribution of all variables by joining the CPTs of the Bayes Net.
 
@@ -28,7 +28,7 @@ This relation between the joint distribution and the CPTs of the Bayes net works
 <p></p>
 Let's revisit the previous example. We have the CPTs $$P(B)$$ , $$P(E)$$ , $$P(A |B,E)$$ , $$P(J | A)$$ and $$P(M | A)$$ , and the following graph:
 
-![Basic Bayes Net Examples](../assets/images/basic_bayes_nets.png)
+<img src="{{ site.baseurl }}/assets/images/basic_bayes_nets.png" alt="Basic Bayes Net Examples" />
 
 For this Bayes net, we are trying to prove the following relation: