Multivariate random variables - PDF Free Download

Multivariate random variables DS GA 1002 Statistical and Mathematical Models http://www.cims.nyu.edu/~cfgranda/pages/dsga1002_fall16 Carlos Fernandez-Granda

Joint distributions Tool to characterize several uncertain numerical quantities of interest within the same probabilistic model We can group the variables into random vectors X 1 X = X 2 X n

Discrete random variables Continuous random variables Joint distributions of discrete and continuous random variables

Joint probability mass function The joint pmf of X and Y is defined as p X,Y (x, y) := P (X = x, Y = y) It is the probability of X, Y being equal to x, y respectively By the definition of a probability measure p X,Y (x, y) 0 for any x R X, y R Y x R X y R Y p X,Y (x, y) = 1

Joint probability mass function The joint pmf of a discrete random vector X is p X ( x) := P (X 1 = x 1, X 2 = x 2,..., X n = x n ) It is the probability of X being equal to x By the definition of a probability measure p X ( x) 0 x 1 R 1 p X ( x) = 1 x 2 R 2 x n R n

Joint probability mass function By the Law of Total Probability, for any set S R X R y, P ((X, Y ) S) = P ( (x,y) S {X = x, Y = y} ) (union of disjoint events) = P (X = x, Y = y) (x,y) S = (x,y) S p X,Y (x, y) Similarly, for any discrete set S R n ( ) P X S = p X ( x) x S

Marginalization To compute the marginal pmf of X from the joint pmf p X,Y p X (x) = P (X = x) = P ( y RY {X = x, Y = y}) (union of disjoint events) = y R Y P (X = x, Y = y) = y R Y p X,Y (x, y) This is called marginalizing over Y

Marginalization Marginal pmf of a subvector X I, I {1, 2,..., n}, p XI ( x I ) = p X ( x) j 2 R j2 j n m R jn m j 1 R j1 {j 1, j 2,..., j n m } := {1, 2,..., n} /I

Conditional probability mass function The conditional pmf of Y given X is p Y X (y x) = P (Y = y X = x) = p X,Y (x, y), as long as p X (x) > 0 p X (x) Valid pmf parametrized by x Chain rule for discrete random variables p X,Y (x, y) = p X (x) p Y X (y x)

Conditional probability mass function The conditional pmf of a random subvector X I, I {1, 2,..., n}, given another subvector X J is p XI X J ( x I x J ) := p X ( x) p XJ ( x J ) {j 1, j 2,..., j n m } := {1, 2,..., n} /I Chain rule for discrete random vectors p X ( x) = p X1 (x 1 ) p X2 X 1 (x 2 x 1 )... p Xn X1,...,X n 1 (x n x 1,..., x n 1 ) n ( ) = p Xi X xi x {1,...,i 1} {1,...,i 1} i=1 Any order works!

Example: Flights and rain (continued) Probabilistic model for late arrivals at an airport P (late, no rain) = 2 14, P (on time, no rain) = 20 20, P (late, rain) = 3 20, P (on time, rain) = 1 20 { 1 if plane is late L = 0 otherwise { 1 it rains R = 0 otherwise

Example: Flights and rain (continued) R p L,R 0 1 p L p L R ( 0) p L R ( 1) L 0 14 20 1 2 20 1 20 3 20 15 20 5 20 7 8 1 8 1 4 3 4 p R 16 20 4 20 p R L ( 0) 14 15 1 15 p R L ( 1) 2 5 3 5

Independence of discrete random variables X and Y are independent if and only if p X,Y (x, y) = p X (x) p Y (y), for all x R X, y R Y, Equivalently p X Y (x y) = p X (x) p Y X (y x) = p Y (y) for all x R X, y R Y

Mutually independent random variables The n entries X 1, X 2,..., X n in a random vector X are mutually independent if and only if n p X ( x) = p Xi (x i ) i=1

Conditionally mutually independent random variables The components of a subvector X I, I {1, 2,..., n} are conditionally mutually independent given another subvector X J, J {1, 2,..., n}, if and only if p XI X J ( x I x J ) = i I p Xi X J (x i x J )

Pairwise independence X 1 and X 2 are outcomes of independent unbiased coin flips X 3 = { 1 if X 1 = X 2, 0 if X 1 X 2. Are X 1, X 2 and X 3 independent?

Pairwise independence X 1 and X 2 are independent by assumption

Pairwise independence X 1 and X 2 are independent by assumption The pmf of X 3 is p X3 (1) = p X3 (0) =

Pairwise independence X 1 and X 2 are independent by assumption The pmf of X 3 is p X3 (1) = p X1,X 2 (1, 1) + p X1,X 2 (0, 0) = 1 2, p X3 (0) = p X1,X 2 (0, 1) + p X1,X 2 (1, 0) = 1 2

Pairwise independence Are X 1 and X 3 independent?,

Pairwise independence Are X 1 and X 3 independent? p X1,X 3 (0, 0) =,

Pairwise independence Are X 1 and X 3 independent? p X1,X 3 (0, 0) = p X1,X 2 (0, 1) = 1 4,

Pairwise independence Are X 1 and X 3 independent? p X1,X 3 (0, 0) = p X1,X 2 (0, 1) = 1 4 = p X 1 (0) p X3 (0),

Pairwise independence Are X 1 and X 3 independent? p X1,X 3 (0, 0) = p X1,X 2 (0, 1) = 1 4 = p X 1 (0) p X3 (0), p X1,X 3 (1, 0) = p X1,X 2 (1, 0) = 1 4 = p X 1 (1) p X3 (0),

Pairwise independence Are X 1 and X 3 independent? p X1,X 3 (0, 0) = p X1,X 2 (0, 1) = 1 4 = p X 1 (0) p X3 (0), p X1,X 3 (1, 0) = p X1,X 2 (1, 0) = 1 4 = p X 1 (1) p X3 (0), p X1,X 3 (0, 1) = p X1,X 2 (0, 0) = 1 4 = p X 1 (0) p X3 (1),

Pairwise independence X 1, X 2 and X 3 are pairwise independent Are X 1, X 2 and X 3 mutually independent?

Pairwise independence X 1, X 2 and X 3 are pairwise independent Are X 1, X 2 and X 3 mutually independent? p X1,X 2,X 3 (1, 1, 1) = p X1 (1) p X2 (1) p X3 (1) =

Pairwise independence X 1, X 2 and X 3 are pairwise independent Are X 1, X 2 and X 3 mutually independent? p X1,X 2,X 3 (1, 1, 1) = P (X 1 = 1, X 2 = 1) = 1 4 p X1 (1) p X2 (1) p X3 (1) = 1 8

Pairwise independence X 1, X 2 and X 3 are pairwise independent Are X 1, X 2 and X 3 mutually independent? p X1,X 2,X 3 (1, 1, 1) = P (X 1 = 1, X 2 = 1) = 1 4 p X1 (1) p X2 (1) p X3 (1) = 1 8 No!

Discrete random variables Continuous random variables Joint distributions of discrete and continuous random variables

Continuous random variables We consider events that are composed of unions of Cartesian products of intervals (Borel sets) The joint cumulative distribution function (cdf) of X and Y is F X,Y (x, y) := P (X x, Y y) In words, probability of X and Y being smaller than x and y respectively The cdf of X is F X ( x) := P (X 1 x 1, X 2 x 2,..., X n x n )

Joint cumulative distribution function Every joint cdf satisfies lim F X,Y (x, y) = 0, x lim F X,Y (x, y) = 0, y lim F X,Y (x, y) = 1 x,y F X,Y (x 1, y 1 ) F X,Y (x 2, y 2 ) if x 2 x 1, y 2 y 1 (nondecreasing)

Joint cumulative distribution function For any two-dimensional interval P (x 1 X x 2, y 1 Y y 2 ) = P ({X x 2, Y y 2 } {X > x 2 } {Y > y 2 }) = P (X x 2, Y y 2 ) P (X x 1, Y y 2 ) P (X x 2, Y y 1 ) + P (X x 1, Y y 1 ) = F X,Y (x 2, y 2 ) F X,Y (x 1, y 2 ) F X,Y (x 2, y 1 ) + F X,Y (x 1, y 1 ) Completely characterizes the distribution of the random variables / random vector

Joint probability density function If the joint cdf is differentiable f X,Y (x, y) := 2 F X,Y (x, y) x y n F X ( x) f X ( x) := x 1 x 2 x n

Joint probability density function Probability of (X, Y ) (x, x + x ) (y, y + y ) for x, y 0 is f X,Y (x, y) x y It is a density, not a probability measure! From the monotonicity of the joint cdf f X,Y (x, y) 0 f X ( x) 0

Joint probability density function For any Borel set S R 2 P ((X, Y ) S) = In particular, x= y= S f X,Y (x, y) dx dy f X,Y (x, y) dx dy = 1

Joint probability density function For any Borel set S R n In particular, ( ) P X S = f X ( x) d x S R n f X ( x) d x = 1

Example: Triangle lake 1.5 1 E F 0.5 B C D 0 A 0.5 0.5 0 0.5 1 1.5

Example: Triangle lake 0 if x 1 < 0 or x 2 < 0, 2x 1 x 2, if x 1 0, x 2 0, x 1 + x 2 1, 2x 1 + 2x 2 x2 2 F X ( x) = x 1 2 1, if x 1 1, x 2 1, x 1 + x 2 1, 2x 2 x2 2, if x 1 1, 0 x 2 1, 2x 1 x1 2, if 0 x 1 1, x 2 1, 1, if x 1 1, x 2 1

Marginalization We can compute the marginal cdf from the joint cdf or from the joint pdf F X (x) = P (X x) = lim y F X,Y (x, y) F X (x) = P (X x) = Differentiating we obtain f X (x) = x u= y= y= f X,Y (x, y) dy f X,Y (u, y) du dy

Marginalization Marginal pdf of a subvector X I, I := {i 1, i 2,..., i m }, f XI ( x I ) = x j1 f X ( x) dx j1 dx j2 dx jn m x j2 x jn m where {j 1, j 2,..., j n m } := {1, 2,..., n} /I

Example: Triangle lake (continued) Marginal cdf of x 1 F X1 (x 1 ) = Marginal pdf of x 1 lim F x 2 X ( x) = f X1 (x 1 ) = df X 1 (x 1 ) dx 1 = 0 if x 1 < 0, 2x 1 x1 2 if 0 x 1 1, 1 if x 1 1 { 2 (1 x 1 ) if 0 x 1 1 0 otherwise

Joint conditional cdf and pdf given an event If we know that (X, Y ) S for any Borel set in R 2 F X,Y (X,Y ) S (x, y) := P (X x, Y y (X, Y ) S) = = P (X x, Y y, (X, Y ) S) P ((X, Y ) S) u x,v y,(u,v) S f X,Y (u, v) du dv (u,v) S f X,Y (u, v) du dv f X,Y (X,Y ) S (x, y) := 2 F X,Y (X,Y ) S (x, y) x y

Conditional cdf and pdf Distribution of Y given X = x? The event has zero probability!

Conditional cdf and pdf Distribution of Y given X = x? The event has zero probability! Define f Y X (y x) := f X,Y (x, y), if f X (x) > 0 f X (x) F Y X (y x) := y u= f Y X (u x) du Chain rule for continuous random variables f X,Y (x, y) = f X (x) f Y X (y x)

Conditional cdf and pdf P (x X x + x ) f X (x) = lim x 0 x 1 P (x X x + x, Y y) f X,Y (x, y) = lim x 0 x y

Conditional cdf and pdf F Y X (y x) = y lim u= x 0, y 0 1 = lim x 0 P (x X x + x ) 1 P (x X x + x ) y u= P (x X x + x, Y y) = lim x 0 P (x X x + x ) = lim x 0 P (Y y x X x + x) P (x X x + x, Y u) y P (x X x + x, Y u) y du du

Conditional pdf of a random subvector Conditional pdf of a random subvector X I, I {1, 2,..., n}, given another subvector X {1,...,n}/I is f XI X {1,...,n}/I ( xi x {1,...,n}/I ) := f X ( x) f X{1,...,n}/I ( x{1,...,n}/i ) Chain rule for continuous random vectors f X ( x) = f X1 (x 1 ) f X2 X 1 (x 2 x 1 )... f Xn X1,...,X n 1 (x n x 1,..., x n 1 ) n ( ) = f Xi X xi x {1,...,i 1} {1,...,i 1} i=1 Any order works!

Example: Triangle lake (continued) Conditioned on {x 1 = 0.75} what is the pdf and cdf of x 2?

Example: Triangle lake (continued) f X2 X 1 (x 2 x 1 )

Example: Triangle lake (continued) f X2 X 1 (x 2 x 1 ) = f X ( x) f X1 (x 1 )

Example: Triangle lake (continued) f X2 X 1 (x 2 x 1 ) = f X ( x) f X1 (x 1 ) = 1 1 x 1, 0 x 2 1 x 1

Example: Triangle lake (continued) f X2 X 1 (x 2 x 1 ) = f X ( x) f X1 (x 1 ) = 1 1 x 1, 0 x 2 1 x 1 F X2 X 1 (x 2 x 1 ) = x2 = x 2 1 x 1 f X2 X 1 (u x 1 ) du

Example: Desert Car traveling through the desert Time until the car breaks down: T State of the motor: M State of the road: R Model: M uniform between 0 (no problem) and 1 (very bad) R uniform between 0 (no problem) and 1 (very bad) M and R independent T exponential with parameter M + R

Example: Desert Joint pdf?

Example: Desert Joint pdf? f M,R,T (m, r, t)

Example: Desert Joint pdf? f M,R,T (m, r, t) = f M (m) f R M (r m) f T M,R (t m, r)

Example: Desert Joint pdf? f M,R,T (m, r, t) = f M (m) f R M (r m) f T M,R (t m, r) = f M (m) f R (r) f T M,R (t m, r) by independence

Example: Desert Joint pdf? f M,R,T (m, r, t) = f M (m) f R M (r m) f T M,R (t m, r) = f M (m) f R (r) f T M,R (t m, r) by independence { (m + r) e (m+r)t for t 0, 0 m 1, 0 r 1, = 0 otherwise

Example: Desert Car breaks down after 15 min (0.25 h), T = 0.25 Road seems OK, R = 0.2 What was the state of the motor M?

Example: Desert Car breaks down after 15 min (0.25 h), T = 0.25 Road seems OK, R = 0.2 What was the state of the motor M? f M R,T (m r, t) = f M,R,T (m, r, t) f R,T (r, t)

Example: Desert f R,T (r, t) =

Example: Desert f R,T (r, t) = 1 m=0 f M,R,T (m, r, t) dm

Example: Desert f R,T (r, t) = 1 m=0 = e tr ( 1 f M,R,T (m, r, t) dm m=0 me tm dm + r 1 m=0 ) e tm dm

Example: Desert 1 f R,T (r, t) = f M,R,T (m, r, t) dm m=0 ( 1 1 ) = e tr me tm dm + r e tm dm m=0 m=0 ( 1 (1 + t) e = e tr t t 2 + r (1 ) e t ) t

Example: Desert 1 f R,T (r, t) = f M,R,T (m, r, t) dm m=0 ( 1 1 ) = e tr me tm dm + r e tm dm m=0 m=0 ( 1 (1 + t) e = e tr t t 2 + r (1 ) e t ) t = e tr t 2 ( 1 + tr e t (1 + t + tr) ) for t 0, 0 r 1

Example: Desert f M R,T (m r, t) = f M,R,T (m, r, t) f R,T (r, t) = = e tr t 2 (m + r) e (m+r)t (1 + tr e t (1 + t + tr)) (m + r) t 2 e tm 1 + tr e t (1 + t + tr)

Example: Desert f M R,T (m r, t) = f M,R,T (m, r, t) f R,T (r, t) = = e tr t 2 (m + r) e (m+r)t (1 + tr e t (1 + t + tr)) (m + r) t 2 e tm 1 + tr e t (1 + t + tr) f M R,T (m 0.2, 0.25) = (m + 0.2) 0.25 2 e 0.25m 1 + 0.25 0.2 e 0.25 (1 + 0.25 + 0.25 0.2) = 1.66 (m + 0.2) e 0.25m for 0 m 1

State of the car 1.5 f M R,T (m 0.2, 0.25) 1 0.5 0 0 0.2 0.4 0.6 0.8 1 m

Independent continuous random variables Two random variables X and Y are independent if and only if F X,Y (x, y) = F X (x) F Y (y), for all (x, y) R 2 Equivalently, F X Y (x y) = F X (x) F Y X (y x) = F Y (y) for all (x, y) R 2

Independent continuous random variables Two random variables X and Y with joint pdf f X,Y are independent if and only if f X,Y (x, y) = f X (x) f Y (y), for all (x, y) R 2 Equivalently, f X Y (x y) = f X (x) f Y X (y x) = f Y (y) for all (x, y) R 2

Mutually independent continuous random variables The components of a random vector X are mutually independent if and only if Equivalently, F X ( x) = f X ( x) = n F Xi (x i ) i=1 n f Xi (x i ) i=1

Mutually conditionally independent random variables The components of a subvector X I, I {1, 2,..., n} are mutually conditionally independent given another subvector X J, J {1, 2,..., n}, if and only if F XI X J ( x I x J ) = i I F Xi X J (x i x J ) Equivalently, f XI X J ( x I x J ) = i I f Xi X J (x i x J )

Functions of random variables U = g (X, Y ) and V = h (X, Y ) F U,V (u, v) = P (U u, V v) = P (g (X, Y ) u, h (X, Y ) v) = f X,Y (x, y) dx dy {(x,y) g(x,y) u,h(x,y) v}

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y?

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y? F Z (z)

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y? F Z (z) = P (X + Y z)

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y? F Z (z) = P (X + Y z) = = y= y= z y x= f X (x) f Y (y) dx dy F X (z y) f Y (y) dy

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y? F Z (z) = P (X + Y z) = = y= y= z y x= f X (x) f Y (y) dx dy F X (z y) f Y (y) dy f Z (z) = d u dz lim F X (z y) f Y (y) dy u y= u

Sum of independent random variables X and Y are independent random variables, what is the pdf of Z = X + Y? F Z (z) = P (X + Y z) = = y= y= z y x= f X (x) f Y (y) dx dy F X (z y) f Y (y) dy f Z (z) = d dz lim = u y= Convolution of individual pdfs u y= u F X (z y) f Y (y) dy f X (z y) f Y (y) dy

Example: Coffee beans Company buys coffee beans from two local producers Beans from Colombia: C tons/year Beans from Vietnam: V tons/year Model: C uniform between 0 and 1 V uniform between 0 and 2 C and V independent What is the distribution of the total amount of beans B?

Example: Coffee beans f B (b) =

Example: Coffee beans f B (b) = u= f C (b u) f V (u) du

Example: Coffee beans f B (b) = = 1 2 u= 2 u=0 f C (b u) f V (u) du f C (b u) du

Example: Coffee beans f B (b) = = 1 2 = u= 2 u=0 1 2 1 2 1 2 2 u=b 1 f C (b u) f V (u) du f C (b u) du b u=0 du = b 2 if b 1 b u=b 1 du = 1 2 if 1 b 2 du = 3 b 2 if 2 b 3

Example: Coffee beans 1 f C f V 1 f B 0.5 0.5 0 0 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3

Gaussian random vector A Gaussian random vector X has a joint pdf of the form f X ( x) = 1 ( (2π) n Σ exp 1 ) 2 ( x µ)t Σ 1 ( x µ) where the mean µ R n and the covariance matrix Σ is a symmetric positive definite matrix

Linear transformation of Gaussian random vectors X is a Gaussian r.v. of dimension n with mean µ and covariance matrix Σ For any matrix A R m n and b R m Y = AX + b is Gaussian with mean A µ + b and covariance matrix AΣA T

Marginal distributions are Gaussian Gaussian random vector, [ ] [ ] X µ Z :=, with mean µ := X Y µ Y and covariance matrix Σ Z = [ ] Σ X Σ X Y Σ T X Y Σ Y X is a Gaussian random vector with mean µ X and covariance matrix Σ X

Marginal distributions are Gaussian f Y (y) f X (x) fx,y (X, Y ) 0.2 0.1 0 3 2 1 2 x 0 1 2 3 2 0 y

Discrete random variables Continuous random variables Joint distributions of discrete and continuous random variables

Discrete and continuous random variables How do we model the relation between a continuous random variable C and a discrete random variable D? Conditional cdf and pdf of C given D By the Law of Total Probability F C D (c d) := P (C c D = d) f C D (c d) := df C D (c d) dc F C (c) = f C (c) = d R D p D (d) F C D (c d) d R D p D (d) f C D (c d)

Mixture models Data are drawn from continuous distribution whose parameters are chosen from a discrete set Important example: Gaussian mixture models

Grizzlies in Yellowstone Model for the weight of grizzly bears in Yellowstone: Males: Gaussian with µ := 240 kg and σ := 40kg Females: Gaussian with µ := 140 kg and σ := 20kg There are about the same number of females and males

Grizzlies in Yellowstone The distribution of the weight of all bears W can be modeled as a Gaussian mixture with two random variables: S (sex) and W (weight)

Grizzlies in Yellowstone The distribution of the weight of all bears W can be modeled as a Gaussian mixture with two random variables: S (sex) and W (weight) f W (w)

Grizzlies in Yellowstone The distribution of the weight of all bears W can be modeled as a Gaussian mixture with two random variables: S (sex) and W (weight) 1 f W (w) = p S (s) f W S (w s) s=0

Grizzlies in Yellowstone The distribution of the weight of all bears W can be modeled as a Gaussian mixture with two random variables: S (sex) and W (weight) f W (w) = 1 p S (s) f W S (w s) s=0 = 1 2 2π (w 240) 2 e 3200 40 + (w 140) 2 e 800 20

Grizzlies in Yellowstone 2 1 10 2 f W S ( 0) f W S ( 1) f W ( ) 0 0 100 200 300 400

Continuous and discrete random variables Conditional pmf of D given C? Event {C = c} has zero probability

Continuous and discrete random variables Conditional pmf of D given C? Event {C = c} has zero probability p D C (d c) := lim 0 P (D = d, c C c + ) P (c C c + )

Continuous and discrete random variables Conditional pmf of D given C? Event {C = c} has zero probability p D C (d c) := lim 0 P (D = d, c C c + ) P (c C c + ) By the Law of Total Probability and a limit argument p D (d) = c= f C (c) p D C (d c) dc

Bayesian coin flip Bayesian methods often endow parameters of discrete distributions with a continuous marginal distribution

Bayesian coin flip Bayesian methods often endow parameters of discrete distributions with a continuous marginal distribution You suspect a coin is biased You are uncertain about the bias so you model it as a random variable with pdf What is the probability of heads? f B (b) = 2b for b [0, 1]

Bayesian coin flip 2 f B ( ) 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1

Bayesian coin flip p X (1)

Bayesian coin flip p X (1) = b= f B (b) p X B (1 b) db

Bayesian coin flip p X (1) = = b= 1 b=0 f B (b) p X B (1 b) db 2b 2 db

Bayesian coin flip p X (1) = = = 2 3 b= 1 b=0 f B (b) p X B (1 b) db 2b 2 db

Chain rule for continuous and discrete random variables P (c C c + D = d) p D (d) f C D (c d) = lim P (D = d) 0 = lim 0 P (D = d, c C c + ) P (c C c + ) P (D = d, c C c + ) = lim 0 P (c C c + ) = f C (c) p D C (d c)

Grizzlies in Yellowstone You spot a grizzly that is about 180 kg What is the probability that it is male?

Grizzlies in Yellowstone You spot a grizzly that is about 180 kg What is the probability that it is male? p S W (0 180)

Grizzlies in Yellowstone You spot a grizzly that is about 180 kg What is the probability that it is male? p S W (0 180) = p S (0) f W S (180 0) f W (180)

Grizzlies in Yellowstone You spot a grizzly that is about 180 kg What is the probability that it is male? p S W (0 180) = p S (0) f W S (180 0) f W (180) ( 1 40 exp = ( 1 40 exp 602 3200 602 3200 ) ) + 1 20 exp ( 402 800 ) = 0.545

Bayesian coin flip Coin flip is tails What is the distribution of the bias now?

Bayesian coin flip Coin flip is tails What is the distribution of the bias now? f B X (b 0)

Bayesian coin flip Coin flip is tails What is the distribution of the bias now? f B X (b 0) = f B (b) p X B (0 b) p X (0)

Bayesian coin flip Coin flip is tails What is the distribution of the bias now? f B X (b 0) = f B (b) p X B (0 b) p X (0) 2b (1 b) = 1/3 = 6b (1 b)

Bayesian coin flip 1.5 2 f B ( ) f B X ( 0) 1 0.5 0 0 0.2 0.4 0.6 0.8 1