Question

Please see page 30 here http://www.cs.utexas.edu/~jason777/School/introduction%20to%20probability%20(bertsekas,%202nd,%202008).pdf and also page 28 which defines the Total Probability Theorem.

I'm confused at the end of page 30 (the equation that starts with P(U3) =

Now, the Total Probability Theorem states a set A is required which is a partition of the sample space. But the set B which is included with U in this equation cannot be a partition since the union of its elements does not equal the sample space.

Where did this equation come from?

Answer 1

just checking: this one in the red box... yeah?

Answer 2

Yep that's right

Answer 3

im out, sorry my probability skills are poor
@amistre64 ?

Answer 4

a tree might be the best way to visualize this for me

Answer 5

u is up to date, b is behind

probability of uptodate in week 3 is the sum of probabilities of being up to date or behind in the prior week right?

Answer 6

Yeah I think so. But even putting the question aside, I'm just curious about the language used there. Since it says "According to the total probability theorem" then lists that equation. But on page 28 where this is defined, it states a set A is required, all of the elemenents of which must add up to the entire sample space. But neither U or B here could be such a set?

Answer 7

define your entire sample space for a moment

Answer 8

I'd imagine it would be U+B?

Answer 9

U and B are conditional right?

Answer 10

P(u3|u2) = P(u3 and u2) / P(u2)

so, the probability of u3 and u2, can be reworked as:

P(u3 and u2) = P(u2) * P(u3| u2)

Answer 11

our sample space i believe is a bunch of unions

Answer 12

does this make sense?

Answer 13

P(uptodate in week 3) = P(u3 and u2) + P(u3 and b2)

Answer 14

I can see how you derived the P(U3 and U2). But the entire form of this equation is different to its definition on page 28? Where it is of the form P(B) = P(A)P(B|A) +... In this question though the first term doesn't include the other set, only U?

Answer 15

its the same form to me:

Answer 16

the set of uptodate is determined by the union of prior sets. U3 depends on U3 and U2, as well as U3 and B2

if the placement of U3 is bothering, then recall that it is commutative.

Answer 17

\[X\cap Y=Y\cap X\]

Answer 18

Ohh, unless in the definition B is referring to an element or subset of A, it isn't some other set?

Answer 19

\[P(U_3)=P(U_2\cap U_3)+P(B_2\cap U_3)\]
\[\frac{P(U_2\cap U_3)}{P(U_2)}= P(U_3|U_2)\color{red}{\implies}{P(U_2\cap U_3)}={P(U_2)}~P(U_3|U_2)\]
\[\frac{P(B_2\cap U_3)}{P(B_2)}= P(U_3|B_2)\color{red}{\implies}{P(B_2\cap U_3)}={P(B_2)}~P(U_3|B_2)\]

Answer 20

that was harder to type up than it should have been :)

Answer 21

Well done =) So where am I going wrong here. I look at the definition of the total probability theorem and it nevertheless should be of the form P(sample set UNION B). And yet in this equation we still do not even use the sample set anywhere? Since I imagine any combinations of U will not be the sample set since they cannot cover all of B.

Answer 22

sample space*

Answer 23

props @amistre64 ... ur killing this!

Answer 24

you arent confusing generality for specifics are you? B is being defined in this case as 2 different things.

Answer 25

Yeah B is fine. But we don't have an A equivalent (sample space) in this equation?

Answer 26

the sample space is; uptodate, or behind in the prior week

Answer 27

sample space:

A1 = uptodate in prior week
A2 = behind in prior week
-------------------------

the event that we are taking the probability of:

B = uptodate in current week

Answer 28

this is why i asked you to define the sample space to start with ...

Answer 29

Ok sorry could you elaborate please how we are defining B differently here? I was treating both as an event.

Answer 30

lets change the letters in the general case, the definition

A = X, and B=Y

Y is the event that is taking place, the event that we want to know the probability of. agreed?

Answer 31

Ok

Answer 32

P(Y) is defined as the sum of its parts .... the union of the sample space of events

the sample space is X1, X2 ... these events cannot happen at the same time can they? they are disjoint events (you cant be uptodate and behind at the same time can you?)

Answer 33

Yep that's fine

Answer 34

P(Y) = P(X1 and Y) + P(X2 and Y)

or written another way

P(Y) = P(X1) P(Y,given X1) + P(X2) P(Y,given X2)

let the event be defined as: Y=U3, uptodate in week 3

let the sample space be defined as:

X1= U2, uptodate in week 2
X2 = B2, not uptodate in week 2

Answer 35

Great argument I follow it algebraically. But how is it that a higher iteration of U can change from a being part of the sample space to being an event?

Answer 36

the setup they give you is can example of recurrsion, the value of a given object depends on its previous value.

all they did was show you how to define the event/sample space for a given week.

U2 and B2 are events in their own right ... they are determined by the previous week.

Answer 37

spose you want to catch a flight, but its probability of being on time depends on weather, mechanical failure, bird poop in the windshield ... etc.

the sample space for a given event, is composed of seperate events, which in turn can be composed of even more events .... it becomes a multidimensional issue

Answer 38

Hmm interesting, so why don't we also include U1 and B1?

Answer 39

we do, but implicitly. they are implied in U2 and B2

Answer 40

Ahh so derive U2 and B2 it's already a superset of U1 and B1?

Answer 41

the calculation for U3, is simple to start with .... it is the construction of the prior events U2 and B2

U2 and B2 can be constructed from the prior events of U1 and B1

U1 and B1 can be constructed from the prior events of ... they gave no starting point so a closed form is not readily available

Answer 42

spose we knew that 25 weeks ago, they had an initial probability of Ui = .5 and Bi=.5

then we could calculate all the way up to the current week

Answer 43

So would U2 union B2 here = 1? The entire sample space?

Answer 44

well,we know P(Y) +P(notY) = 1

that does not mean that P(Y) = 1 in its own right

Answer 45

if P(Y) might be .4327 then the sum of its parts would have to equal .4327 (whch is the unit conversion for P(Y))

Answer 46

But by the total probability theorem these parts must add up to the sample space?

Answer 47

U2 and B2 being the sample space here.

Answer 48

U2 union B2

Answer 49

U2 and B2 are the results of week2