Some simple probability formulas with examples

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A known relationship that is usually given axiomatically:

P(B|A) = \frac{{P(AB)}}{{P(A)}}

Upon rearrangement gives the multiplication rule of probability:

P(AB) = P(A)P(B|A) = P(B)P(A|B)

Now observe a cool set up that is handy to keep in mind for proving the law of total probability and Bayes’ theorem.

Imagine that B happens with one and only one of n mutually exclusive events A_1, A_2,..., A_n, i.e.:

 B = \sum\limits_{i = 1}^n {B{A_i}}

By addition rule:

B = \sum\limits_{i = 1}^n {P(B{A_i})}.

Now by multiplication rule:

B = \sum\limits_{i = 1}^n {P({A_i})P(B|{A_i})}.

This is the law of total probability

From the same set up imagine that we want to find the probability of even A_i if B is known to have happened. By the multiplication rule:

P(A_i B) = P(B)P(A_i|B) = P(A_i)P(B|A_i)

By neglecting P(A_i B) and dividing the rest through P(B) we get:

P\left( {{A_i}|B} \right){\rm{ = }}\frac{{P({A_i})P(B|{A_i})}}{{P(B)}}

And applying the law of total probability to the bottom we have the Bayes’ equation

P\left( {{A_i}|B} \right){\rm{ = }}\frac{{P({A_i})P(B|{A_i})}}{{\sum\limits_{j = 1}^n {P({A_j})P(B|{A_j})} }}

Bunch of examples:

Problem: P_t (k) is a known probability of receiving k phone calls during time interval t. Also k=0,1,2,.... Assuming that a number of received calls during two adjeicent time periods are independent find the probability of receiving s calls for the time interval that equal 2t.

Solution: Let A_{b.b + t}^k be an event consisted of k call in the interval b till b+t. Then clearly

A_{0,2t}^s = A_{0,t}^0A_{t,2t}^s + ... + A_{0,t}^sA_{t,2t}^0

which means that the event A_{0,2t}^s can be seen as sum of s+1 mutually exclusive events, such that in the first interval of duration t number of calls received is i and in the second interval of the same duration number of received calls is s-i (i=0,1,2,...,s). By rule of addition

P(A_{0,2t}^s) = \sum\limits_{i = 0}^s {P(A_{0,t}^iA_{t,2t}^{s - i})}.

By the rule of multiplication

P(A_{0,t}^iA_{t,2t}^{s - i}) = P(A_{0,t}^i)P(A_{t,2t}^{s - i})

If we change the notation so that

{P_{2t}}(s) = P(A_{0,2t}^s)

then

{P_{2t}}(s) = \sum\limits_{i = 0}^s {{P_t}(s) \cdot P(s - i)}.

It is known that under quite general conditions

{P_t}(k) = \frac{{{{(at)}^k}}}{{k!}}\exp \{ - at\} {\rm{ }}(k = 0,1,2...)

(Recall that the Poisson distribution is an appropriate model if the following assumptions are true. (a) k is the number of times an event occurs in an interval and k can take values 0,1,2,.... (b) The occurrence of one event does not affect the probability that a second event will occur. That is, events occur independently. (c) The rate at which events occur is constant. The rate cannot be higher in some intervals and lower in other intervals (that kinda a lot to take on faith really). (d) Two events cannot occur at exactly the same instant; instead, at each very small sub-interval exactly one event either occurs or does not occur. (e) The probability of an event in a small sub-interval is proportional to the length of the sub-interval.  Or instead of those assumptions, the actual probability distribution is given by a binomial distribution and the number of trials is sufficiently bigger than the number of successes one is asking about (binomial distribution approaches Poisson).)

Parametrisation then gives

{P_{2t}}(s) = \sum\limits_{i = 0}^s {\frac{{{{(at)}^s}}}{{i!(s - i)!}}\exp \{ - 2at\} } {\rm{ = }}{(at)^s}\exp \{ - 2at\} \sum\limits_{i = 0}^s {\frac{1}{{i!(s - i)!}}}

Note that

\sum\limits_{i = 0}^s {\frac{1}{{i!(s - i)!}} = \frac{1}{{s!}}\sum\limits_{i = 0}^s {\frac{{s!}}{{i!(s - i)!}}} = \frac{1}{{s!}}{{(1 + 1)}^s} = \frac{{{2^s}}}{{s!}}}

Then

{P_{2t}}(s) = \frac{{{{(2at)}^s}\exp \{ - 2at\} }}{{s!}}{\rm{ }}(s = 0,1,2,...)

The key point is that if for time interval t we have that parametrized formula for 2t we have the one above. It holds true for any multiples of t as well.

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Published by Sergey Alexeev

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