[PDF] Probability Cheatsheet v2.0 4 Sept 2015 Probability Cheatsheet

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Probability Cheatsheet v2.0

Compiled by William Chen (http://wzchen.com) and Joe Blitzstein, with contributions from Sebastian Chiu, Yuan Jiang, Yuqi Hou, and Jessy Hwang. Material based on Joe Blitzstein's (@stat110) lectures (http://stat110.net) and Blitzstein/Hwang's Introduction to Probability textbook (http://bit.ly/introprobability). Licensed underCC BY-NC-SA 4.0. Please share comments, suggestions, and errors

Last Updated September 4, 2015

CountingMultiplication Rulecakewa!eSVCSVCSVCcakewa!ecakewa!ecakewa!eLet's say we have a compound experiment (an experiment with

multiple components). If the 1st component hasn1possible outcomes, the 2nd component hasn2possible outcomes, ..., and therth component hasnrpossible outcomes, then overall there are n

1n2:::nrpossibilities for the whole experiment.

Sampling Table765842931The sampling table gives the number of possible samples of sizekout of a population of sizen, under various assumptions about how the sample is collected.Order Matters Not Matter

With Replacementn

kn+k1 k

Without Replacementn!(nk)!

n k

Naive Denition of Probability

If all outcomes are equally likely, the probability of an eventA happening is: P naive(A) =number of outcomes favorable toAnumber of outcomesThinking ConditionallyIndependence Independent EventsAandBare independent if knowing whether Aoccurred gives no information about whetherBoccurred. More formally,AandB(which have nonzero probability) are independent if and only if one of the following equivalent statements holds:

P(A\B) =P(A)P(B)

P(AjB) =P(A)

P(BjA) =P(B)

Conditional IndependenceAandBare conditionally independent givenCifP(A\BjC) =P(AjC)P(BjC). Conditional independence does not imply independence, and independence does not imply conditional independence.

Unions, Intersections, and Complements

De Morgan's LawsA useful identity that can make calculating probabilities of unions easier by relating them to intersections, and vice versa. Analogous results hold with more than two sets. (A[B)c=Ac\Bc (A\B)c=Ac[Bc

Joint, Marginal, and Conditional

Joint ProbabilityP(A\B) orP(A;B) { Probability ofAandB. Marginal (Unconditional) ProbabilityP(A) { Probability ofA. Conditional ProbabilityP(AjB) =P(A;B)=P(B) { Probability of

A, given thatBoccurred.

Conditional ProbabilityisProbabilityP(AjB) is a probability function for any xedB. Any theorem that holds for probability also holds for conditional probability.

Probability of an Intersection or Union

Intersections via Conditioning

P(A;B) =P(A)P(BjA)

P(A;B;C) =P(A)P(BjA)P(CjA;B)

Unions via Inclusion-Exclusion

P(A[B) =P(A) +P(B)P(A\B)

P(A[B[C) =P(A) +P(B) +P(C)

P(A\B)P(A\C)P(B\C)

+P(A\B\C): Simpson's ParadoxDr. HibbertDr. Nickheartband-aidIt is possible to have

P(AjB;C)< P(AjBc;C) andP(AjB;Cc)< P(AjBc;Cc)

yet alsoP(AjB)> P(AjBc):Law of Total Probability (LOTP) LetB1;B2;B3;:::Bnbe apartitionof the sample space (i.e., they are disjoint and their union is the entire sample space).

P(A) =P(AjB1)P(B1) +P(AjB2)P(B2) ++P(AjBn)P(Bn)

P(A) =P(A\B1) +P(A\B2) ++P(A\Bn)

ForLOTP with extra conditioning, just add in another eventC!

P(AjC) =P(AjB1;C)P(B1jC) ++P(AjBn;C)P(BnjC)

P(AjC) =P(A\B1jC) +P(A\B2jC) ++P(A\BnjC)

Special case of LOTP withBandBcas partition:

P(A) =P(AjB)P(B) +P(AjBc)P(Bc)

P(A) =P(A\B) +P(A\Bc)

Bayes' Rule

Bayes' Rule, and with extra conditioning (just add inC!)

P(AjB) =P(BjA)P(A)P(B)

P(AjB;C) =P(BjA;C)P(AjC)P(BjC)

We can also write

P(AjB;C) =P(A;B;C)P(B;C)=P(B;CjA)P(A)P(B;C)

Odds Form of Bayes' Rule

P(AjB)P(AcjB)=P(BjA)P(BjAc)P(A)P(Ac)

Theposterior oddsofAare thelikelihood ratiotimes theprior odds. Random Variables and their DistributionsPMF, CDF, and Independence Probability Mass Function (PMF)Gives the probability that a discreterandom variable takes on the valuex. p

X(x) =P(X=x)

01234
0.0 0.2 0.4 0.6 0.8 1.0 x pmf l l l l lThe PMF satises p

X(x)0 andX

xp

X(x) = 1

Cumulative Distribution Function (CDF)Gives the probability that a random variable is less than or equal tox. F

X(x) =P(Xx)01234

0.0 0.2 0.4 0.6 0.8 1.0 x cdf l ll ll ll ll lThe CDF is an increasing, right-continuous function with F

X(x)!0 asx! 1andFX(x)!1 asx! 1

IndependenceIntuitively, two random variables are independent if knowing the value of one gives no information about the other. Discrete r.v.sXandYare independent if forallvalues ofxandy

P(X=x;Y=y) =P(X=x)P(Y=y)

Expected Value and IndicatorsExpected Value and Linearity Expected Value(a.k.a.mean,expectation, oraverage) is a weighted average of the possible outcomes of our random variable. Mathematically, ifx1;x2;x3;:::are all of the distinct possible values thatXcan take, the expected value ofXis

E(X) =P

ix iP(X=xi) X 3 2 6 10 1 1 5 4 Y 4 2 8 23
Ð3 0 9 1 X + Y 7 4 14 33
Ð2 1 14 5 ! x i ! y i +! (x i + y i )=E(X)E(Y)+E(X + Y)= i=1ni=1ni=1n n1n1n1LinearityFor any r.v.sXandY, and constantsa;b;c;

E(aX+bY+c) =aE(X) +bE(Y) +c

Same distribution implies same meanIfXandYhave the same distribution, thenE(X) =E(Y) and, more generally,

E(g(X)) =E(g(Y))

Conditional Expected Valueis dened like expectation, only conditioned on any eventA.

E(XjA) =P

xxP(X=xjA)Indicator Random Variables Indicator Random Variableis a random variable that takes on the value 1 or 0. It is always an indicator of some event: if the event occurs, the indicator is 1; otherwise it is 0. They are useful for many problems about counting how many events of some kind occur. Write I A=(

1 ifAoccurs,

0 ifAdoes not occur.

Note thatI2

A=IA;IAIB=IA\B;andIA[B=IA+IBIAIB.

DistributionIABern(p) wherep=P(A).

Fundamental BridgeThe expectation of the indicator for eventAis the probability of eventA:E(IA) =P(A).

Variance and Standard Deviation

Var(X) =E(XE(X))2=E(X2)(E(X))2

SD(X) =qVar(X)

Continuous RVs, LOTUS, UoUContinuous Random Variables (CRVs) What's the probability that a CRV is in an interval?Take the dierence in CDF values (or use the PDF as described later).

P(aXb) =P(Xb)P(Xa) =FX(b)FX(a)

ForX N(;2), this becomes

P(aXb) = b

a What is the Probability Density Function (PDF)?The PDFf is the derivative of the CDFF. F

0(x) =f(x)

A PDF is nonnegative and integrates to 1. By the fundamental theorem of calculus, to get from PDF back to CDF we can integrate:

F(x) =Z

x 1 f(t)dt -4-2024 0.00 0.10 0.20 0.30 x PDF -4-2024 0.0 0.2 0.4 0.6 0.8 1.0 x CDFTo nd the probability that a CRV takes on a value in an interval, integrate the PDF over that interval.

F(b)F(a) =Z

b a f(x)dx How do I nd the expected value of a CRV?Analogous to the discrete case, where you sumxtimes the PMF, for CRVs you integrate xtimes the PDF.

E(X) =Z

1 1 xf(x)dxLOTUS Expected value of a function of an r.v.The expected value ofX is dened this way:

E(X) =X

xxP(X=x) (for discreteX)

E(X) =Z

1 1 xf(x)dx(for continuousX) TheLaw of the Unconscious Statistician (LOTUS)states that you can nd the expected value of afunction of a random variable, g(X), in a similar way, by replacing thexin front of the PMF/PDF by g(x) but still working with the PMF/PDF ofX:

E(g(X)) =X

xg(x)P(X=x) (for discreteX)

E(g(X)) =Z

1 1 g(x)f(x)dx(for continuousX) What's a function of a random variable?A function of a random variable is also a random variable. For example, ifXis the number of bikes you see in an hour, theng(X) = 2Xis the number of bike wheels you see in that hour andh(X) =X

2=X(X1)2

is the number of pairsof bikes such that you see both of those bikes in that hour. What's the point?You don't need to know the PMF/PDF ofg(X) to nd its expected value. All you need is the PMF/PDF ofX.

Universality of Uniform (UoU)

When you plug any CRV into its own CDF, you get a Uniform(0,1) random variable. When you plug a Uniform(0,1) r.v. into an inverse CDF, you get an r.v. with that CDF. For example, let's say that a random variableXhas CDF

F(x) = 1ex;forx >0

By UoU, if we plugXinto this function then we get a uniformly distributed random variable.

F(X) = 1eXUnif(0;1)

Similarly, ifUUnif(0;1) thenF1(U) has CDFF. The key point is that for any continuous random variableX, we can transform it into a Uniform random variable and back by using its CDF.

Moments and MGFsMoments

Moments describe the shape of a distribution. LetXhave meanand standard deviation, andZ= (X)=be thestandardizedversion ofX. Thekth moment ofXisk=E(Xk) and thekth standardized moment ofXismk=E(Zk). The mean, variance, skewness, and kurtosis are important summaries of the shape of a distribution.

MeanE(X) =1

VarianceVar(X) =22

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