Repeated trials
01 Theory
Theory 1
Repeated trials
When a single experiment type is repeated many times, and we assume each instance is independent of the others, we say it is a sequence of repeated trials or independent trials.
The probability of any sequence of outcomes is derived using independence together with the probabilities of outcomes of each trial.
A simple type of trial, called a Bernoulli trial, has two possible outcomes, 1 and 0, or success and failure, or
and . A sequence of repeated Bernoulli trials is called a Bernoulli process.
- Write sequences like
for the outcomes of repeated trials of this type. - Independence implies
- Write
and , and because these are all outcomes (exclusive and exhaustive), we have . Then:
- This gives a formula for the probability of any sequence of these trials.
A more complex trial may have three outcomes,
, , and .
- Write sequences like
for the outcomes. - Label
and and . We must have . - Independence implies
- This gives a formula for the probability of any sequence of these trials.
Let
stand for the sum of successes in some Bernoulli process. So, for example, “ ” stands for the event that the number of successes is exactly 3. The probabilities of events follow a binomial distribution. Suppose a coin is biased with
, and is ‘success’. Flip the coin 20 times. Then: Each outcome with exactly 3 heads and 17 tails has probability
. The number of such outcomes is the number of ways to choose 3 of the flips to be heads out of the 20 total flips. The probability of at least 18 heads would then be:
With three possible outcomes,
Link to original, , and , we can write sum variables like which counts the number of outcomes, and and similarly. The probabilities of events like follow a multinomial distribution.
02 Illustration
Example - Multinomial: Soft drinks preferred
Multinomial: Soft drinks preferred
Folks coming to a party prefer Coke (55%), Pepsi (25%), or Dew (20%). If 20 people order drinks in sequence, what is the probability that exactly 12 have Coke and 5 have Pepsi and 3 have Dew?
Solution
The multinomial coefficient
gives the number of ways to assign 20 people into bins according to preferences matching the given numbers, and and . Each such assignment is one sequence of outcomes. All such sequences have probability
. The answer is therefore:
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Reliability
03 Theory
Theory 1
Consider some process schematically with components in series and components in parallel:
- Each component has a probability of success or failure.
- Event
indicates ‘success’ of that component (same name). - Then
is the probability of succeeding.
Success for a series of components requires success of each member.
- Series components rely on each other.
- Success of the whole is success of part 1 AND success of part 2 AND part 3, etc.
Failure for parallel components requires failure of each member.
- Parallel components represent redundancy.
- Success of the whole is success of part 1 OR success of part 2 OR part 3, etc.
For series components:
For parallel components:
If
for all components , then:
- Series components:
- Parallel components:
To analyze a complex diagram of series and parallel components, bundle each:
- pure series set as a single compound component with its own success probability (the product)
- pure parallel set as a single compound component with its own success probability (using the failure formula)
This is like the analysis of resistors and inductors.
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04 Illustration
Example - Series, parallel, series
Reliability: Series, parallel, series
Suppose a process has internal components arranged like this:
Write
for the event that component succeeds, and for the event that it fails. The success probabilities for each component are given in the chart:
1 2 3 4 5 92% 89% 95% 86% 91% Find the probability that the entire system succeeds.
Solution
(1) Conjoin components 2 and 3 in series.
Compute:
Therefore:
(2) Conjoin components (2-3) with 4 and 5 in parallel.
Compute for the complement (failure) first:
Flip back to success:
(3) Conjoin components 1 with (2-3-4-5) in series.
Compute:
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Discrete random variables
05 Theory
Theory 1
Random variable
A random variable (RV)
on a probability space is a function . So
assigns to each outcome a number. Note: The word ‘variable’ indicates that an RV outputs numbers.
Random variables can be formed from other random variables using mathematical operations on the output numbers.
Given random variables
and , we can form these new ones: Suppose
is some particular outcome. Then, for example, is by definition .
Random variables determine events.
- Given
, the event “ ” is equal to the set - That is: the set of outcomes mapped to
by - That is: the event “
took the value ” Such events have probabilities. We write them like this:
This generalized to events where
lies in some range or set, for example:
The axioms of probability translate into rules for these events.
For example, additivity leads to:
A discrete random variable has probability concentrated at a discrete set of real numbers.
- A ‘discrete set’ means finite or countably infinite.
- The distribution of probability is recorded using a probability mass function (PMF) that assigns probabilities to each of the discrete real numbers.
- Numbers with nonzero probability are called possible values.
PMF
The PMF function
, for a discrete RV, is defined by: A continuous random variable has probability spread out over the space of real numbers.
- The distribution of probability is recorded using a probability density function (PDF) which is integrated over intervals to determine probabilities.
The PDF function for
(a CRV) is written for , and probabilities are calculated like this:
For any RV, whether discrete or continuous, the distribution of probability is encoded by a function:
CDF
The cumulative distribution function (CDF) for a random variable
is defined for all by: Notes:
- Sometimes the relation to
is omitted and one sees just “ .” - Sometimes the CDF is called, simply, “the distribution function” because:
The CDF works for a discrete, continuous, or mixed RV
- PMF is for discrete only
- PDF is for continuous only
- CDF covers both and covers mixed RVs
The CDF of a discrete RV is always a stepwise increasing function. At each step up, the jump size matches the PMF value there.
From this graph of
: we can infer the PMF values based on the jump sizes:
For a discrete RV, the CDF and the PMF can be calculated from each other using formulas.
Link to originalPMF from CDF from PMF
Given a PMF
, the CDF is determined by: where
is the set of possible values of . Given a CDF
, the PMF is determined by:
06 Illustration
Example - PDF and CDF: Roll 2 dice
PDF and CDF: Roll 2 dice
Roll two dice colored red and green. Let
record the number of dots showing on the red die, the number on the green die, and let be a random variable giving the total number of dots showing after the roll, namely .
- Find the PMFs of
and of and of . - Find the CDF of
. - Find
. Solution
(1) Sample space.
Denote outcomes with ordered pairs of numbers
, where is the number showing on the red die and is the number on the green one. Require that
are integers satisfying . Events are sets of distinct such pairs.
(2) Create chart of outcomes.
Chart:
(3) Definitions of
, , and . We have
and . Therefore
.
(4) Find PMF of
. Use variable
for each possible value of , so . Find
: Therefore
for every .
(5) Find PMF of
. Same as for
:
(6) Find PMF of
. Find
: Count outcomes along diagonal lines in the chart.
Create table of
:
Create bar chart of
:
Evaluate:
.
(7) Find CDF of
. CDF definition:
Apply definition: add new PMF value at each increment:
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Example - Total heads count; binomial expansion of 1
PMF for total heads count; binomial expansion of 1
A fair coin is flipped
times. Let
be the random variable that counts the total number of heads in each sequence. The PMF of
is given by: Since the total probability must add to 1, we know this formula must hold:
Is this equation really true?
There is another way to view this equation: it is the binomial expansion
where and : Link to original
Example - Life insurance payouts
Life insurance payouts
A life insurance company has two clients,
and , each with a policy that pays $100,000 upon death. Consider events that the older client dies next year, and that the younger dies next year. Suppose and . Define a random variable
measuring the total money paid out next year in units of $1,000. The possible values for are 0, 100, 200. We calculate: Link to original