3.7, 3.8, 3.9, 3.11 Functions of multiple random variables - PowerPoint PPT Presentation

3.7, 3.8, 3.9, 3.11 Functions of multiple random variables (continuous) Prof. Tesler Math 186 Winter 2019 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 1 / 30

Mass density (review from Calculus and Physics) y B ρ ( x , y ) x Two-dimensional version: Consider a shape B ⊆ R 2 . Make very thin horizontal and vertical cuts. Let ρ ( x , y ) be the density at ( x , y ) . This is the mass per unit area. It can be measured in g/cm 2 . In 3D, it would be g/cm 3 . ρ ( x , y ) � 0 everywhere. The area of a differential patch is dA = dx dy = dy dx . The mass of a differential patch is ρ ( x , y ) dA (density times area). �� The total mass of B is ρ ( x , y ) dA B Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 2 / 30

Continuous joint probability density function Joint probability density function of two variables We require: f X , Y ( x , y ) � 0 for all points ( x , y ) . � ∞ � ∞ f X , Y ( x , y ) dx dy = 1 − ∞ − ∞ Probability of an event The probability of event B ⊆ R 2 is y B �� P ( B ) = f X , Y ( x , y ) dA x B Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 3 / 30

Uniform probability on a region C Uniform probability on a region C means that all points inside C have equal probability density, and all points outside C have probability density 0 : � 1 if ( x , y ) ∈ C area ( C ) f X , Y ( x , y ) = otherwise 0 y Let C be the disk of radius 2 3 centered at the origin: C � 2 if x 2 + y 2 � 4 1 x 4 π f X , Y ( x , y ) = −3 3 otherwise 0 −3 �� 4 π dA = 1 1 4 π · area ( C ) = 1 Total probability = 4 π · 4 π = 1 C Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 4 / 30

Probability of an event y 3 D 2 x −3 3 −3 �� 4 π · 4 π 4 π dA = 1 1 4 π area ( D ) = 1 2 = 1 P ( X > 0 ) = 2 D Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 5 / 30

Marginal densities 3 y √ 4 − x 2 y = x −3 3 √ 4 − x 2 y = − −3 Form an x -strip: hold x constant and vary y . √ The perimeter is x 2 + y 2 = 4 , so y = ± 4 − x 2 on the perimeter. The strip is vertical, so the − solution is at the bottom and the + solution is at the top. √ √ 4 − x 2 � y � 4 − x 2 . The part of the strip within the shape is − Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 6 / 30

Marginal densities The marginal density at x : Form an x -strip . Hold x constant and integrate over all y . 3 y � ∞ f X ( x ) = − ∞ f X , Y ( x , y ) dy √ 4 − x 2 y = � √ 4 − x 2 1 = 4 π dy √ 4 − x 2 − x √ −3 3 4 − x 2 = 2 √ 4 π y = − 4 − x 2 −3 � √ 4 − x 2 if − 2 � x � 2 2 π f X ( x ) = otherwise 0 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 7 / 30

Marginal densities The marginal density at y is similar. Form a y -strip . Hold y constant and integrate over all x . The strip is horizontal and goes left to right instead of bottom to top. y 3 � ∞ f Y ( y ) = − ∞ f X , Y ( x , y ) dx � � 4 − y 2 4 − y 2 x = − x = � √ 4 − y 2 1 − √ = 4 π dx x 4 − y 2 = 2 √ −3 3 4 − y 2 4 π −3 � √ 4 − y 2 if − 2 � y � 2 f Y ( y ) = 2 π otherwise 0 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 8 / 30

Independence Random variables X , Y , Z , . . . are independent if their joint pdf factorizes as follows, for all x , y , z , . . .. f X , Y , Z ,... ( x , y , z , . . . ) = f X ( x ) f Y ( y ) f Z ( z ) · · · Technicality: Exceptions are allowed, as long as the probability of an exception is 0. For example, in a continuous distribution: The probability of a point is 0. In 2D, the probability of a discrete set of points or curves is 0. Exceptions only happen with continuous distributions, not discrete. Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 9 / 30

Independence Summary of previous formulas � if x 2 + y 2 � 4 1 3 y 4 π f X , Y ( x , y ) = otherwise 0 � √ 4 − x 2 / ( 2 π ) if − 2 � x � 2 x f X ( x ) = −3 3 otherwise 0 � � 4 − y 2 / ( 2 π ) if − 2 � y � 2 f Y ( y ) = −3 otherwise 0 Check independence: � � ( 4 − x 2 )( 4 − y 2 ) / ( 4 π 2 ) if − 2 � x � 2 and − 2 � y � 2 f X ( x ) f Y ( y ) = otherwise 0 This is different than f X , Y ( x , y ) . The formula is different, and it’s nonzero inside a square instead of inside a circle. So X , Y are dependent. Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 10 / 30

Expected values Definition For a function g ( X , Y ) of continuous random variables, the expected value is � ∞ � ∞ E ( g ( X , Y )) = g ( x , y ) f X , Y ( x , y ) dA − ∞ − ∞ This is similar to the definition in the discrete case, but using integrals instead of sums. Compute E ( X ) for the circle example �� �� x x E ( X ) = 4 π dA + 4 π dA = 0 right semicircle left semicircle The two integrals are negatives of each other, so they sum to 0. Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 11 / 30

Compute E ( R ) in the circle example √ X 2 + Y 2 . Compute E ( R ) : In polar coordinates, recall R = �� x 2 + y 2 · 1 � � � X 2 + Y 2 � E ( R ) = E = 4 π dA C This is easier in polar coordinates than in Cartesian coordinates. Switch to polar coordinates, and note that the integral separates: � 2 π � 2 � � 2 π � � � 2 �� � r 4 π · r dr d θ = 1 r r 2 dr E ( R ) = 4 π dA = d θ 4 π 0 0 0 0 C Evaluate the integrals: � 2 π � 2 r = 2 = 2 3 − 0 3 r 2 dr = r 3 � θ = 2 π = 8 � � d θ = θ = 2 π − 0 = 2 π � � � 3 3 3 θ = 0 � 0 0 r = 0 Plug in their values: � 8 � = 16 π E ( R ) = 1 12 π = 4 4 π ( 2 π ) 3 3 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 12 / 30

Variance The variance formula is the same for continuous as for discrete: Var ( X ) = E (( X − µ ) 2 ) = E ( X 2 ) − ( E ( X )) 2 However, expected value is computed using an integral instead of a sum. Compute Var ( R ) and SD ( R ) : � 2 π � 2 � � 2 π � � � 2 �� r 2 r 2 � 4 π · r dr d θ = 1 r 3 dr E ( R 2 ) = 4 π dA = d θ 4 π 0 0 0 0 C � 2 π � 2 = 2 4 − 0 4 2 r 3 dr = r 4 � � d θ = 2 π = 4 � 4 4 � 0 0 r = 0 E ( R 2 ) = 1 4 π ( 2 π )( 4 ) = 2 Var ( R ) = E ( R 2 ) − ( E ( R )) 2 = 2 − ( 4 / 3 ) 2 = 2 / 9 � SD ( R ) = 2 / 9 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 13 / 30

Mass density in physics vs. continuous pdf Physics Probability Mass density Probability density function ρ ( x , y ) � 0 f X , Y ( x , y ) � 0 Mass of shape D ⊆ R 2 : Probability of event D ⊆ R 2 : �� �� and P ( R 2 ) = 1 M = ρ ( x , y ) dA � 0 P ( D ) = f X , Y ( x , y ) dA D D Center of mass ( ¯ x , ¯ y ) Expected value �� �� x · ρ ( x , y ) dA E ( X )= x · f X , Y ( x , y ) dA = numerator of ¯ x . D x = ¯ �� R 2 ρ ( x , y ) dA The denominator of ¯ x is 1, since M = 1 . D y formula is similar ¯ E ( Y ) formula is similar When the total mass is 1, we have ( ¯ x , ¯ y ) = ( E ( X ) , E ( Y )) . We used R 2 . For R n , use ( x 1 , . . . , x n ) instead of ( x , y ) . Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 14 / 30

Determining the constant y 1 x 2 Question: Determine the formula of the probability density if it is proportional to x + 4 y inside the rectangle and is 0 outside. We have f X , Y ( x , y ) = c ( x + 4 y ) inside the rectangle and 0 outside, for some constant c . Find c so that the total probability is 1 : � 2 � 1 P = c ( x + 4 y ) dy dx = 1 0 0 The inside integral is � y = 1 y = 0 = c ( x ( 1 − 0 ) + 2 ( 1 2 − 0 2 )) = c ( x + 2 ) c ( xy + 2 y 2 ) � � 2 Plug that back in: P = 0 c ( x + 2 ) dx Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 15 / 30

Determining the constant y 1 x 2 Continue evaluating: 2 � 2 �� � x 2 = 0 c ( x + 2 ) dx = c 2 + 2 x P � � x = 0 � � 2 2 − 0 2 = c + 2 ( 2 − 0 ) = c · ( 2 + 4 ) = 6 c 2 To get P = 1 , solve 6 c = 1 , so c = 1 / 6 . Plug this value of c into the formula f X , Y ( x , y ) = c ( x + 4 y ) . Thus, the pdf is � x + 4 y inside rectangle: 0 � x � 2 and 0 � y � 1 6 f X , Y ( x , y ) = outside rectangle 0 Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 16 / 30

Marginal densities for rectangle example y For 0 � x � 2 : 1 � 1 1 dy = xy + 2 y 2 � x + 4 y � f X ( x ) = x 2 � 6 6 � 0 y = 0 = x ( 1 − 0 ) + 2 ( 1 2 − 0 2 ) = x + 2 6 6 Otherwise, f X ( x ) = 0 . Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 17 / 30

Marginal densities for rectangle example y For 0 � y � 1 : 1 � x 2 � 2 2 �� x + 4 y 12 + 4 xy � f Y ( y ) = dx = x 2 � 6 6 � 0 x = 0 � 2 2 − 0 2 � + 4 ( 2 − 0 ) y = 12 6 6 = 4 y + 1 = 4 12 + 8 y 3 Otherwise, f Y ( y ) = 0 . Prof. Tesler Ch. 3. Joint random variables (continuous) Math 186 / Winter 2019 18 / 30