But violence tone burst zero two lines of tears.
T3 magical directed graph go blind series.
The results of my DP + Gaussian life and death do not.
I found that push along.
Solution to a problem is still granted to the inverse push.
I**
Then I study a little.
First of all to know this stuff.
Conditional probability formula: $ P (B | A) = frac {P (AB)} {P (B)} $
This is a fantastic proof.
Probably means A and the overall general account divided by the B accounts B to get A's.
Then there is a magic formula of total probability.
It is also well understood, it is the enumeration A condition appears.
Finally, to a Bayesian formula
Also understand, is generally regarded as the cause of A, B, or condition.
So common are blind to go to a series of questions as a vector diagram to understand it.
I was wrong to write about out of Forwards equation $ f (x) = frac {f (y)} {deg (y)} + 1 $
Is able to enumerate all the points of x, multiplied by the transition probabilities, and then add all brought up, it seems no problem.
But there is a problem, what is the probability that I look like that from each point to the probability of x from y point,
Then add up? ? ? ? ?
Obviously does not know why add up, it does not take account of the weight of each case.
Obviously wrong.
errrr explain Bayesian theory.
It is acceptable, for each point, to reach his point can be considered its "cause"
So think from a departure and arrival point A is the probability of A is $ \ sum \ limits_ {j = 1} ^ {indegA} P (A | B_i) P (B_i) $
Is the total probability formula, the enumeration A condition occurs that can all point to point A $ B_i $
You can know when to push me along probability should be, set the current enumeration to $ B_i $, P | probability (A B_i) that occur under conditions A, B.
That I went to the A and B accounts came from all the A probability ($ frac {P (AB)} {P (B)} $).
However, this is a Bayesian formula $ P (A | B_i) = frac {P (B_i | A) P (B_i)} {\ sum \ limits_ {j = 1} ^ n P (A | B_j) P (B_j) } $
The total is below the cumulative probability of A, then the probability is on top of the probability of A at B's, this is the degree B, in fact, this is the inverse of the probability of push,
That is, I stood probability do I go next B are equal.
When $ P (A) $ and $ P (B) $ are not equal, the probability that the two are not equivalent.