353 - Topic 04 - Bayes

704 days ago by Professor353

John Travis - Mississippi College

Bayes Theorem

For a given sample space S, there may be a natural way to partition the space into disjoint sets--so that all elements in S belong to exactly one of the sets.  For example, separating a given set of people into males and females or dividing up a pocket full of change into pennies, nickles, dimes, quarters, etc. 

For disjoint sets, a Venn diagram is traditionally written using a collection of disconnected blobs--one for each set in the partition--in a box.  However, when sets form a partition it is often simpler to write a Venn diagram as a pie chart with each set in the partition corresponding to one sector of that pie chart.

Notationally, we will describe our partition of subsets as

$S = B_{1}\cup B_{2}\cup B_{3} \cdots \cup B_{N}$.

From this partitioned sample space, one may desire a conditional probabability for some outcome A.  For example, if A is the event that a random student selected from a particular class fails the course and $S = Males \cup Females$, then $P(fails | Male)$ and $P(fails | Female)$ might be well known from historical values.  However, given that a student who failed has set up an appointment to meet with the teacher, determining the likelihood that the person will be Male (eg. $P(Male | fails)$) is often a harder thing to quantify directly.  Bayes Theorem is the answer to solving this type of problem.

Derivation of Bayes Theorem:  From the definition of conditional probability and using the notation developed above:

$P(A \cap B_{k}) = P(A) P(B_{k}| A) $ or

$P(A \cap B_{k})$ = $P(B_{k})$ $ P(A | B_{k}) $ and by transitivity

$P(A) P(B_{k}| A) $ = $P(B_{k})$ $ P(A | B_{k}) $

Since $\bigcup B_{k}$ comprises all of S, then one may compute P(A) by adding up the probabilities of its parts--indeed, $A = (A\cap B_{1}) \cup (A\cap B_{2}) \cup (A\cap B_{3}) \cup \cdots \cup (A \cap B_{N})$.  In the diagram below, this is illustrated using the probabilities inside the inner circle.

Using the second formula with this partition of A (remember the $B_k$ are all disjoint) yields:

 ${\bf P(A)} = P(A\cap B_{1}) + P(A\cap B_{2}) + \cdots + P(A \cap B_{N})$ =  

$ = P(B_{1})P(A|B_{1}) + P(B_{2})$$P(A|B_{2})$$+ \cdots + $$P(B_{N})$$P(A|B_{N})$

On the other hand, using the third formula and solving yields:

$P(B_{k}| A) $ = $P(B_{k})$ $ P(A | B_{k})/ {\bf P(A)}  $ 

Replacing the P(A) on the bottom with the bold formulation gives Bayes Theorem.

Therefore, Bayes Theorem is very useful when it is possible to determine the conditional probabilities $P(A|B_{k}), k=1 \cdots N$ but perhaps not so easy to compute $P(B_{k}| A) $.

%hide %auto ## ## Illustration of Bayes Theorem for introductory Probability and Statistics ## ## John Travis ## Mississippi College ## 06/17/11 ## ## The user inputs: ## - A list of probabilities corresponding to each sector of a partition ## - A list of prior conditional probabilities P( A | B_1), etc. ## Output: ## - Two Venn diagrams. The first describes the input data. The second describes the posterior probabilities. ## - Numerical values for everything, if desired. ## ## An updated version of this worksheet may be available at http://sagenb.mc.edu ## # This function is used to convert an input string into separate entries def g(s): return str(s).replace(',',' ').replace('(',' ').replace(')',' ').split() @interact def _(Partition_Probabilities=input_box('0.35,0.25,0.40',label="$P(B_1),P(B_2),...$"), Conditional_Probabilities=input_box('0.02,0.01,0.03',label='$P(A|B_1),P(A|B_2),...$'), print_numbers=checkbox(True,label='Numerical Results on Graphs?'), auto_update=False): Partition_Probabilities = g(Partition_Probabilities) Conditional_Probabilities = g(Conditional_Probabilities) n = len(Partition_Probabilities) n0 = len(Conditional_Probabilities) if n<>n0: print html("<font size=+2 color=red>You must have the same number of partition probabilities and conditional probabilities.</font>") else: # input data streams now are the same size! colors = rainbow(n) accum = float(0) # to test whether partition probs sum to one ends = [0] # where the graphed partition sectors change in pie chart mid = [] # middle of each pie chart sector used for placement of text p_Bk_given_A = [] # P( B_k | A ) pA = 0 # P(A) PP=[] # array to hold the numerical Partition Probabilities CP=[] # array to hold the numerical Conditional Probabilities for k in range(n): PP.append(float(Partition_Probabilities[k])) CP.append(float(Conditional_Probabilities[k])) p_Bk_given_A.append(PP[k]*CP[k] ) pA += p_Bk_given_A[k] accum = accum + PP[k] ends.append(accum) mid.append((ends[k]+accum)/2) # # Marching along from 0 to 1, saving angles for each partition sector boundary. # Later, we will multiple these by 2*pi to get actual sector boundary angles. # if abs(accum-float(1))>0.0000001: # Due to roundoff issues, this should be close enough. print html("<font size=+1 color=red>Sum of probabilities should equal 1.</font>") else: # probability data is sensible # # Draw the Venn diagram by drawing sectors from the angles determined above # First, create a circle of radius 1 to illustrate the the sample space S # Then draw each sector with varying colors and print out their names on the edge # G = circle((0,0), 1, rgbcolor='black',fill=False, alpha=0.4,aspect_ratio=True,axes=False,thickness=5) for k in range(n): G += disk((0,0), 1, (ends[k]*2*pi, ends[k+1]*2*pi), color=colors[mod(k,10)],alpha = 0.2) G += text('$B_'+str(k+1)+'$',(1.1*cos(mid[k]*2*pi), 1.1*sin(mid[k]*2*pi)), rgbcolor='black') G += circle((0,0), 0.6, facecolor='yellow', fill = True, alpha = 0.1, thickness=5,edgecolor='black') # Print the probabilities corresponding to each particular region as a list and on the graphs if print_numbers: html("$P(A) = %s$"%(str(pA),)) for k in range(n): html("$P(B_{%s} | A)$"%(str(k+1))+"$ = %s$"%str(p_Bk_given_A[k]/pA)) G += text(str(p_Bk_given_A[k]),(0.4*cos(mid[k]*2*pi), 0.4*sin(mid[k]*2*pi)), rgbcolor='black') G += text(str(PP[k] - p_Bk_given_A[k]),(0.8*cos(mid[k]*2*pi), 0.8*sin(mid[k]*2*pi)), rgbcolor='black') # This is essentially a repeat of some of the above code but focused only on creating the smaller inner circle dealing # with the set A so that the sectors now correspond in area to the Bayes Theorem probabilities accum = float(0) ends = [0] # where the graphed partition sectors change in pie chart mid = [] # middle of each pie chart sector used for placement of text for k in range(n): accum += float(p_Bk_given_A[k]/pA) ends.append(accum) mid.append((ends[k]+accum)/2) H = circle((0,0), 1, rgbcolor='black',fill=False, alpha=0,aspect_ratio=True,axes=False,thickness=0) H += circle((0,0), 0.6, facecolor='yellow',fill=True, alpha=0.1,aspect_ratio=True,axes=False,thickness=5,edgecolor='black') for k in range(n): H += disk((0,0), 0.6, (ends[k]*2*pi, ends[k+1]*2*pi), color=colors[mod(k,10)],alpha = 0.2) H += text('$B_'+str(k+1)+'|A$',(0.7*cos(mid[k]*2*pi), 0.7*sin(mid[k]*2*pi)), rgbcolor='black') # Now, print out the bayesian probabilities using the smaller set A only if print_numbers: for k in range(n): H += text(str( N(p_Bk_given_A[k]/pA,digits=4) ),(0.4*cos(mid[k]*2*pi), 0.4*sin(mid[k]*2*pi)), rgbcolor='black') html.table([[G,H],['Venn diagram of partition with A in middle','Venn diagram presuming A has occured']]) 
       
$P(B_1),P(B_2),...$ 
$P(A|B_1),P(A|B_2),...$ 
Numerical Results on Graphs? 

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