W03 Notes

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.

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A simple type of trial, called a Bernoulli trial, has two possible outcomes, 1 and 0, or success and failure, or $T$ and $F$. A sequence of repeated Bernoulli trials is called a Bernoulli process.

• Write sequences like $TFFTTF$ for the outcomes of repeated trials of this type.
• Independence implies

$$P[TFFTTF]=P[T]\cdot P[F]\cdot P[F]\cdot P[T]\cdot P[T]\cdot P[F]$$

• Write $p=P[T]$ and $q=P[F]$, and because these are all outcomes (exclusive and exhaustive), we have $q=1-p$. Then:

$$P[TFFTTF]\gg \gg pqqppq\gg \gg p^3{q}^3$$

• This gives a formula for the probability of any sequence of these trials.

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A more complex trial may have three outcomes, $A$, $B$, and $C$.

• Write sequences like $ABBACABCA$ for the outcomes.
• Label $p=P[A]$ and $q=P[B]$ and $r=P[C]$. We must have $p+q+r=1$.
• Independence implies

$$P[ABBACABCA]\gg \gg pqqprpqrp\gg \gg p^4{q}^3{r}^2$$

• This gives a formula for the probability of any sequence of these trials.

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Let $S$ stand for the sum of successes in some Bernoulli process. So, for example, “$S=3$” stands for the event that the number of successes is exactly 3. The probabilities of $S$ events follow a binomial distribution.

Suppose a coin is biased with $P[H]=20\%$, and $H$ is ‘success’. Flip the coin 20 times. Then:

$$P[S=3]\gg \gg \left(\genfrac{}{}{0px}{}{20}{3}\right)\cdot {(0.2)}^3\cdot {(0.8)}^{17}$$

Each outcome with exactly 3 heads and 17 tails has probability ${(0.2)}^3\cdot {(0.8)}^{17}$. 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:

$$\begin{array}{c}P[S\ge 18]\gg \gg P[S=18]+P[S=19]+P[S=20]\\ \\ \gg \gg \left(\genfrac{}{}{0px}{}{20}{18}\right)\cdot {(0.2)}^{18}\cdot {(0.8)}^2+\left(\genfrac{}{}{0px}{}{20}{19}\right)\cdot {(0.2)}^{19}\cdot {(0.8)}^1+\left(\genfrac{}{}{0px}{}{20}{20}\right)\cdot {(0.2)}^{20}\cdot {(0.8)}^0\end{array}$$

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With three possible outcomes, $A$, $B$, and $C$, we can write sum variables like $S_A$ which counts the number of $A$ outcomes, and $S_B$ and $S_C$ similarly. The probabilities of events like $\text{“}(S_A,S_B,S_C)=(2,3,5)\text{”}$ follow a multinomial distribution.

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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 ${\displaystyle \left(\genfrac{}{}{0px}{}{20}{\mathrm{12,5,3}}\right)}$ gives the number of ways to assign 20 people into bins according to preferences matching the given numbers, $C=12$ and $P=5$ and $D=3$.

Each such assignment is one sequence of outcomes. All such sequences have probability ${(0.55)}^{12}\cdot {(0.25)}^5\cdot {(0.2)}^3$.

The answer is therefore:

$$\left(\genfrac{}{}{0px}{}{20}{\mathrm{12,5,3}}\right)\cdot {(0.55)}^{12}\cdot {(0.25)}^5\cdot {(0.2)}^3\gg \gg \frac{20!}{12!5!3!}\cdot {(0.55)}^{12}\cdot {(0.25)}^5\cdot {(0.2)}^3$$Link to original

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 $W_i$ indicates ‘success’ of that component (same name).
• Then $P[W_i]$ is the probability of $W_i$ succeeding.

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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, stack successes:

$$P[W]=P[W_1{W}_2{W}_3]=P[W_1]\cdot P[W_2]\cdot P[W_3]$$

For parallel components, stack failures:

$$\begin{array}{c}P[W^c]=P[W_1^c{W}_2^c{W}_3^c]\\ \\ \gg \gg (1-P[W_1])(1-P[W_2])(1-P[W_3])\end{array}$$

E.g. if $P[W_i]=p$ for all components $i$, then:

• Series components: $P[W]=p^3$
• Parallel components: $P[W]=1-{(1-p)}^3$

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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 $W_i$ for the event that component $i$ succeeds, and $W_i^c$ 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

For intersections, use $P[AB]=P[A]P[B]$ (independence) and for unions, use $P[A\cup B]=1-P[A^c{B}^c]$.

So $P[{(W_2{W}_3)}^c]=1-(.89)(.95)$ and:

$$\begin{array}{c}\text{success}=W_1(W_2{W}_3\cup W_4\cup W_5)\\ \\ \gg \gg P[\text{success}]=(.92)(1-(1-(.89)(.95))(.14)(.09))\gg \gg \approx 0.918\end{array}$$Link to original

Random variables

05 Theory

Theory 1

Random variable

A random variable (RV) $X$ on a probability space $(S,ℱ,P)$ is a function $X:S\to ℝ$.

So $X$ 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 $X$ and $Y$, we can form these new ones:

$$\frac{1}{2}(X+Y),X\cdot Y,\cos X,X^2,\text{etc.}$$

Suppose $s\in S$ is some particular outcome. Then, for example, $(X+Y)(s)$ is by definition $X(s)+Y(s)$.

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Random variables determine events.

• Given $a\in ℝ$, the event “$X=a$” is equal to the set $X^{-1}(a)$
• That is: the set of outcomes mapped to $a$ by $X$
• That is: the event “$X$ took the value $a$”

Such events have probabilities. We write them like this:

$$P[X=a]\gg \gg P[X^{-1}(a)]$$

This generalized to events where $X$ lies in some range or set, for example:

$$P[a\le X<b],P[X\in \{\mathrm{2,4,5,6,9}\}]$$

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The axioms of probability translate into rules for these events.

For example, additivity leads to:

$$P[X<0]+P[X=0]+P[0<X\le 3]+P[3<X]=1$$

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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 $P_X(k)$, for $X$ a discrete RV, is defined by:

$$P_X(k)=P[X=k]$$

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.

PDF

The PDF function for $Y$ (a CRV) is written $f_Y(x)$ for $x\in ℝ$, and probabilities are calculated like this:

$$P[a\le Y\le b]={\int }_a^b{f}_Y(x)dx$$

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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 $X$ is defined for all $x\in ℝ$ by: $F_X(x)=P[X\le x]$

Notes:

• Sometimes the relation to $X$ is omitted and one sees just “$F(x)$.”
• 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

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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 $F_X(x)$:

we can infer the PMF values based on the jump sizes:

$P_X(-1)$	$P_X(0)$	$P_X(1)$	$P_X(2)$	$P_X(3)$	$P_X(4)$
$0$	$1/8$	$3/8$	$3/8$	$1/8$	$0$

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For a discrete RV, the CDF and the PMF can be calculated from each other using formulas.

PMF from CDF

Given a PMF $P_X(x)$, the CDF is determined by:

$$F_X(x)=\sum _{k_i\le x}{P}_X(k_i)$$

where $\{k_1,k_2,\dots \}$ is the set of possible values of $X$.

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06 Illustration

Example - PDF and CDF: Roll 2 dice

PDF and CDF: Roll 2 dice

Roll two dice colored red and green. Let $X_R$ record the number of dots showing on the red die, $X_G$ the number on the green die, and let $S$ be a random variable giving the total number of dots showing after the roll, namely $S=X_R+X_G$.

• Find the PMFs of $X_R$ and of $X_G$ and of $S$.
• Find the CDF of $S$.
• Find $P[S=8]$.

Solution

(1) Construct sample space:

Denote outcomes with ordered pairs of numbers $(i,j)$, where $i$ is the number showing on the red die and $j$ is the number on the green one.

Therefore $i,j\in \mathbb{N}$ are integers satisfying $1\le i,j\le 6$.

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(2) Create chart of outcomes:

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(3) Define random variables:

We have $X_R(i,j)=i$ and $X_G(i,j)=j$.

Therefore $S(i,j)=i+j$.

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(4) Find PMF of $X_R$:

Use variable $k$ for each possible value of $X_R$, so $k=1,2,\dots ,6$. Find $P_{X_R}(k)$:

$$\begin{array}{c}{P}_{X_R}(k)=P[X_R=k]\\ \\ \gg \gg \frac{|\text{outcomes with }k\text{ on red}|}{|\text{all outcomes}|}\gg \gg \frac{6}{36}=\frac{1}{6}\end{array}$$

Therefore $P_{X_R}(k)=1/6$ for every $k$.

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(5) Find PMF of $X_G$ similarly:

$$P_{X_G}(k)=\frac{1}{6}\text{for all}k$$

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(6) Find PMF of $S$:

$$P_S(k)=P[S=k]\gg \gg \frac{|\text{outcomes with sum }k|}{|\text{all outcomes}|}$$

Count outcomes along diagonal lines in the chart.

$k$	2	3	4	5	6	7	8	9	10	11	12
$P_S(k)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

Evaluate: $P[S=8]=5/36$.

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(7) Find CDF of $S$:

CDF definition: $F_S(x)=P[S\le x]$

Apply definition: add new PMF value at each increment:

$$F_S(x)=\{\begin{array}{cc}0& \phantom{1\le }x<2\\ 1/36& 2\le x<3\\ 3/36& 3\le x<4\\ 6/36& 4\le x<5\\ 10/36& 5\le x<6\\ 15/36& 6\le x<7\\ 21/36& 7\le x<8\\ 26/36& 8\le x<9\\ 30/36& 9\le x<10\\ 33/36& 10\le x<11\\ 35/36& 11\le x<12\\ 36/36& 12\le x\end{array}$$ Link to original

Example - Total heads count; binomial expansion of 1

PMF for total heads count; binomial expansion of 1

A fair coin is flipped $n$ times.

Let $X$ be the random variable that counts the total number of heads in each sequence.

The PMF of $X$ is given by:

$$P_X(k)=\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$

Since the total probability must add to 1, we know this formula must hold:

$$\begin{array}{cc}1& =\sum _{\text{possible }k}{P}_X(k)\\ & \\ \gg \gg 1& =\sum _{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n\end{array}$$

Is this equation really true?

There is another way to view this equation: it is the binomial expansion ${(x+y)}^n$ where $x=\frac{1}{2}$ and $y=\frac{1}{2}$:

$${(\frac{1}{2}+\frac{1}{2})}^n=\sum _{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$ Link to original

Example - Life insurance payouts

Life insurance payouts

A life insurance company has two clients, $A$ and $B$, each with a policy that pays $100,000 upon death. Consider events $D_1$ that the older client dies next year, and $D_2$ that the younger dies next year. Suppose $P[D_1]=0.10$ and $P[D_2]=0.05$.

Define a random variable $X$ measuring the total money paid out next year in units of $1,000. The possible values for $X$ are 0, 100, 200. Now calculate:

$$\begin{array}{cccc}& P[X=0]\gg \gg & P[D_1^c]P[D_2^c]\gg \gg 0.95\cdot 0.90\gg \gg 0.86& \text{<span class="tml-eqn"></span>}\\ & & & \text{<span class="tml-eqn"></span>}\\ & P[X=100]\gg \gg & 0.05\cdot 0.90+0.95\cdot 0.10\gg \gg 0.14& \text{<span class="tml-eqn"></span>}\\ & & & \text{<span class="tml-eqn"></span>}\\ & P[X=200]\gg \gg & 0.05\cdot 0.10\gg \gg 0.005& \text{<span class="tml-eqn"></span>}\end{array}$$Link to original

Example - Probabilities via CDF

Probabilities via CDF

Suppose the CDF of $X$ is given by $F_X(x)={\displaystyle \frac{1}{1+e^{-x}}}$. Compute:

(a) $P[X\le 1]$ $$ (b) $P[X<1]$ $$ (c) $P[-0.5\le X\le 0.2]$ $$ (d) $P[-2\le X]$

Solution

(a)

$$F_X(1)=\frac{1}{1+e^{-1}}\gg \gg \approx 0.731$$

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(b) Same as (a) because $P[X=1]=0$ (single point in a continuous distribution).

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(c)

$$F_X(0.2)-F_X(-0.5)\gg \gg \frac{1}{1+e^{-0.2}}-\frac{1}{1+e^{0.5}}\gg \gg \approx 0.172$$

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(d)

$$1-F_X(-2)\gg \gg 1-\frac{1}{1+e^2}\gg \gg \approx 0.881$$

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