Cheat Sheet

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Ω = sample space P = probability measure P(A) >= 0 for every event A - P(Ω) = 1 - P(A B) = P(A) + P(B) if A,B disjoint - P(A B) = P(A) + P(B) - P(A B) if A,B not disjoint - P(A) = 1 - P(A ) - P(0) = 0 - Properties of unions/intersections A B = B A , same for - A (B C) = (A B) C = A B C , same for - A (B C) = (A B) (A C) - A (B C) = (A B) (A C) - Properties of complements (A B) = (A ) (B ) - (A B) = (A ) (B ) - A = (A B ) (A B) - Events are disjoint when A B = Ø - No outcomes are shared - At most only one of the events can occur - Conditional Probability P(A | B) = ! ( # ∩ & ) ! ( & ) - P(A | A) = 1 - P(Ω | B) = 1 - P(A B | C) = P(A | C) + P(B | C) if A,B disjoint - P(A | B) = 1 - P(A | B) - P(A | B) = ࠵?(࠵? | ࠵?) ! ( # ) ! ( & ) - ࠵?(࠵?) = ࠵?( ࠵? | ࠵? ( ) ∗ ࠵?(࠵? ( ) + ࠵?(࠵? | ࠵?) ∗ ࠵?(࠵?) Given 2 events A, B with P(B) > 0 - A = (A B1) (A B2) ... (A Bn) if B1,B2,...Bn form a partition - Independence "Knowing that A has occurred doesn't make B any more or less likely to occur" - P(A B) = P(A)P(B) implies independence - P(A | B) = P(A) if A,B independent - Ø is independent of itself - If A,B are disjoint, then A,B are not independent if 0 < P(A), P(B) < 1 - Pairwise independent if events are independent in groups of 2 If A,B,C are P.I. then... P(A B) = P(A)P(B) P(B C) = P(B)P(C) P(A C) = P(A)P(C) - Jointly independent if any number of events are independent Jointly implies pairwise not vice versa If A,B,C are J.I. then... P(A B C) = P(A)P(B)P(C) as well as all P.I. formulas - Counting Equally likely outcomes (P is same for all, finite sample space Ω) Then, P(E) = # *+,(*-./ 01 2 # *+,(*-./ ,*,34 01 Ω - Types of Counting Ordered Order matters: § Ex: SSN, phone # § Unordered Order doesn't matter § Ex: Poker hands § With replacement Ex: Rolling a 2 in the first dice roll doesn't prevent another 2 from being rolled § Without replacement Ex: Drawing the ace of spades prevents it from being drawn again § - Counting problems Ordered with replacement "How many ways can we assign k states to N objects?" --> ࠵? 5 Ex: How many distinct SSNs are there? —> 10 6 Ex: How many sequences of N coin flips are possible? —> 2 5 - Ordered without replacement "How many ways can we order k distinct objects from a pool of N?" Ex: How many ways can k students out of N line up to exit the room? —> 5! ( 589 ) ! ࠵?࠵? ࠵?! (࠵?࠵?࠵?) - Unordered without replacement "How many ways can k unordered objects be drawn from a pool of N?" Ex: How many 5 card poker hands are possible (52 card deck)? —> ( 52࠵?5 ) = :;! :!<=! Note: (Ordered w/o replacement) = (Unordered w/ replacement) * (# of ways of ordering k objects) - Unordered with replacement - __________TEST 2 ___________ PMF P(X = k) - Valid if all values add up to 1, P(X = k) > 0 for all k - Joint PMF P(X = x, Y = y) comma means - Marginal PMF Same idea for Y - There can be (X,Y) and (X^, Y^) such that joint PMFs are different but marginal PMFs are the same - Conditional PMF Given event A: - P(X = x | A) = ! ({ ?@A } ∩ # ) ! ( # ) - - Example: - - Given random variables X, Y: - ࠵? ?|D ( ࠵? | ࠵?) = ࠵? ( ࠵? = ࠵? | ࠵? = ࠵? ) = !(?@A, D@F) !(D@F) Joint PMF / Marginal PMF - ࠵? ( ࠵? = ࠵?, ࠵? = ࠵? ) = ࠵? ( ࠵? = ࠵? | ࠵? = ࠵?) ∗ ࠵?(࠵? = ࠵?) - ࠵? ( ࠵? = ࠵?, ࠵? = ࠵? ) = ࠵? ( ࠵? = ࠵? | ࠵? = ࠵?) ∗ ࠵? ( ࠵? = ࠵? ) - ࠵?(࠵? = ࠵? | ࠵? = ࠵?) ∗ ࠵?(࠵? = ࠵?) = ࠵?(࠵? = ࠵? | ࠵? = ࠵?) ∗ ࠵?(࠵? = ࠵?) - ࠵?(࠵? = ࠵? | ࠵? = ࠵?) = ࠵?(࠵? = ࠵? | ࠵? = ࠵?) ∗ ࠵?(࠵? = ࠵?) ࠵?(࠵? = ࠵?) - Expected value - "k times probability of X = k for all possible values of k" - - E[f(X)] != f(E[X]) - Given constant a, b E[a] = a E[aX] = aE[X] E[aX + b] = aE[X] + b - Joint Expectation - E[X + Y] = E[X] + E[Y] Left side requires joint PMF, right side only requires marginal PMFs - "Indicator" random variables ࠵? 0 = 1 if *criteria 1* § ࠵? 0 = 0 if *criteria 2* § - Conditional expectation Given event A: - - Given random variables X, Y: - - Law of total expectation: - - Variance (࠵?࠵?࠵?࠵?࠵?) ; Var(X) = ࠵? [( ࠵? − ࠵? [ ࠵? ]) ; ] - Var(X) = ࠵? [ ࠵? ; ] ( ࠵? [ ࠵? ]) ; E[X] is a constant "Expected/mean squared error" - Var(X) >= 0 Var(X) = 0 when X is degenerate (all probability is on one point —> ex. P(X = 0) = 1) - sd(X) = sqrt(Var(X)) - Var(-X) = Var(X) - Given constant a, b Var(a) = 0 Var(aX + b) = ࠵? ; ࠵?࠵?࠵? ( ࠵? ) - Given Y = aX + b, for constants a, b Var(Y) = ࠵? ; ࠵?࠵?࠵? ( ࠵? ) - Covariance ( ࠵?࠵?࠵?࠵?࠵? ) ; ࠵?࠵?࠵? ( ࠵?, ࠵? ) = ࠵? [( ࠵? − ࠵? [ ࠵? ])( ࠵? − ࠵? [ ࠵? ])] - ࠵?࠵?࠵?(࠵?, ࠵?) = ࠵?[࠵?࠵?] − ࠵?[࠵?]࠵?[࠵?] Note: E[X], E[Y] are constants Cov(X, Y) can be + or - - Cov(X, X) = Var(X) - Cov(X, Y) = Cov(Y, X) - Given constants a, b, c, d: - Cov(aX + b, cY + d) = (ac)Cov(X, Y) - Cov(X + Y, Z) = Cov(X, Z) + Cov(Y, Z) - Given Y = aX + b, for constants a, b ࠵?࠵?࠵? ( ࠵?, ࠵? ) = ࠵? ∗ ࠵?࠵?࠵?(࠵?) - Correlation -1 <= ρ(X, Y) <= 1 ρ(X, Y) > 0 —> Positive correlation ρ(X, Y) < 0 —> Negative correlation - Given Y = aX + b, for constants a, b ρ(X, Y) = 1 if a > 0 ρ(X, Y) = -1 if a < 0 - Independence Random variables X and Y are independent if: - P(X = x, Y = y) = P(X = x) * P(Y = y) for all possible x, y Given X, Y are independent: P(X = x | Y = y) = P(X = x) P(X A, Y B) = P(X A) * P(Y B) Ex: P(X < 3, Y > 2) = P(X < 3) * P(Y > 2) § Doesn't work when X,Y mixed —> P(X+Y > 4) or P(X > Y) § - E[f(X) * g(Y)] = E[f(X)] * E[g(Y)] E[XY] = E[X] * E[Y] Var(X + Y) = Var(X) + Var(Y) Cov(X, Y) = 0 Thus, X, Y are "uncorrelated" § For many RVs: X1, X2, ... Xn - Independent if ࠵? ( ࠵? G = ࠵? G , ... , ࠵? 1 = ࠵? 1 ) = ࠵? ( ࠵? G = ࠵? G ) ∗ ⋯ ∗ ࠵?(࠵? 1 = ࠵? 1 ) - for all possible ࠵? G , ࠵? ; , ... , ࠵? 1 Then: - ࠵? [ ࠵? G , ࠵? ; ... , ࠵? 1 ] = ࠵? [ ࠵? G ] ࠵? [ ࠵? ; ] ... ࠵? [ ࠵? 1 ] ࠵?࠵?࠵? ( ࠵? G + ࠵? ; + ⋯ + ࠵? 1 ) = ࠵?࠵?࠵? ( ࠵? G ) + ࠵?࠵?࠵? ( ࠵? ; ) + ⋯ + ࠵?࠵?࠵?(࠵? 1 ) Bernoulli : X ~ Ber(p) X takes value 0 or 1 X = 1 —> Success X = 0 —> Failure - p = probability of success 0 <= p <= 1 - P(X = 1) = p - P(X = 0) = 1 - p - E[X] = p - Var(X) = p(1 - p) - Geometric : X ~ Geom(p) "The number of independent trials until 1 success (inclusive)" - p = probability of success 0 < p < 1 - P(X = k) = ࠵? ( 1 − ࠵? ) 98G k = 1,2,3,... - E[X] = G H - Var[X] = G8H H ! - Binomial : X ~ Bin(n, p) "Number of successes out of n independent trials each with success probability p" - p = probability of success 0 <= p <= 1 - n = number of trials n = 1,2,3,... - P(X = k) = ࠵?࠵?࠵? ∗ ࠵? 9 ( 1 − ࠵? ) 189 k = 0,1,2,...,n - E[X] = np - Given: X ~ Bin(n, p) Y ~ Bin(m, p) - Then for Z = X + Y Z ~ Bin(n + m, p) Binomial = sum of independent Bernoullis ࠵? G , ࠵? ; , ... ࠵? 1 are independent, " " ~ Ber(p) ࠵? 0 = 1 if ࠵? ,I trial succeeds § ࠵? 0 = 0 otherwise § Then: Counts the number of successes § Gives mean § Thanks to independence § - Poisson : X ~ Poisson(λ) "Number of arrivals at rate λ within a unit of time" - λ = average rate per unit time λ > 0 - P(X = k) = (࠵? 8 λ ) λ " 9! k = 0,1,2,... - Limiting case of binomial with many trials (n very large), each of which is very unlikely/rare Xn ~ Bin(n , λ/n) for each integer n > λ ^ approximately - E[X] = λ This justifies calling λ the "expected arrival rate" - Var[X] = λ - Given: Z ~ Poisson(λ + μ) P(Poisson(λ + μ) = n) = P(Z = n) P(Z = n) = . #( λ % μ ) 1! ( λ + μ ) 1 - Cheat Sheet Tuesday, October 4, 2022 6:50 PM
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