# Week 4 Multiple Integrals

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Information about Week 4 Multiple Integrals

Published on March 6, 2014

Author: anhtuantran509

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AMATH 460: Mathematical Methods for Quantitative Finance 4. Multiple Integrals Kjell Konis Acting Assistant Professor, Applied Mathematics University of Washington Kjell Konis (Copyright © 2013) 4. Multiple Integrals 1 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 2 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 3 / 58

Double Integrals Review of the deﬁnite integral of a single-variable function f(x) a b n−1 b f (x ) dx = lim a n→∞ Kjell Konis (Copyright © 2013) f (a + i∆x ) ∆x i=0 4. Multiple Integrals ∆x = b−a n 4 / 58

13(1) + 7(l) + 10(1) + 4(l) = 34 Double Integrals the approximating rectangular boxes shown in Figure 7. 0 FIGURE 7 FIGURE 7 GUR 16 2 We get better approximations t We get better approximations to the volum squares. Figure 8 shows how the squares. Figure 8 shows how the columns s 2 l6—v2v ‘I (a) n 4.1 n = ii = (a) 41.5 4.1 = ii = 41.5 (hI ?fl a 8. (hI 1 ?fl 44.875 1 a 8. FIGURE 8 FIGURE 8 Thesum numl sum approximations to the volume under: oximations to the volume in Example I if weThe Riemann the m n approximations to the volume16under: increase Riemann ws how the columns start to look more dA = actual solid at f (i∆x , j∆y ) ∆A f (x , y ) like the lim m,n→∞ [0,2]×[0,2] i=1 j=1 = ∆x = 2−0 m Kjell Konis (Copyright © 2013) ∆y = 2−0 n 4. Multiple Integrals — v2 — 44.875 2v 16 = v2 become — ∆A = ∆x ∆y 5 / 58 —

Double Integrals In general, double integral over a rectangle R = [a, b] × [c, d] m f (x , y ) dA = R n lim m,n→∞ f (a + i∆x , c + j∆y ) ∆A i=1 j=1 If f (x , y ) ≥ 0 ∀(x , y ) ∈ R, then V = f (x , y ) dA R is the volume of the region above R and below surface z = f (x , y ) Kjell Konis (Copyright © 2013) 4. Multiple Integrals 6 / 58

Properties of Double Integrals Linearity Properties: [f (x , y ) + g(x , y )] dA = R f (x , y ) dA + R cf (x , y ) dA = c g(x , y ) dA R f (x , y ) dA R R Comparison: If f (x , y ) ≥ g(x , y ) ∀(x , y ) ∈ R then f (x , y ) dA ≥ R Kjell Konis (Copyright © 2013) g(x , y ) dA R 4. Multiple Integrals 7 / 58

Iterated Integrals Suppose f (x , y ) is continuous on the rectangle R = [a, b] × [c, d] Partial integration: ﬁx x , integrate f (x , y ) as a function of y alone d A(x ) = f (x , y ) dy c An iterated integral is the integral of A(x ) wrt x b b d A(x ) dx = a f (x , y ) dy dx a c Usually the brackets are omitted b d b d f (x , y ) dy dx = a c f (x , y ) dy dx a c Iterating the other way d b d b f (x , y ) dx dy f (x , y ) dx dy = c Kjell Konis (Copyright © 2013) a c 4. Multiple Integrals a 8 / 58

Double Integrals vs. Iterated Integrals Big Question: What is the relationship between a double integral and an iterated integral? b f (x , y ) dA R Kjell Konis (Copyright © 2013) d f (x , y ) dy dx ? a 4. Multiple Integrals c 9 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 10 / 58

Fubini’s Theorem If f (x , y ) is continuous on the rectangle R = [a, b] × [c, d] then b d f (x , y ) dA = R d b f (x , y ) dy dx = a c f (x , y ) dx dy c a The order of iteration does not matter Kjell Konis (Copyright © 2013) 4. Multiple Integrals 11 / 58

Example Let R = [1, 3] × [2, 5] and f (x , y ) = 2y − 3x . Compute 5 f (x , y ) dA = R 2 1 5 = 2 5 = 2 5 = = 2 3 R f (x , y ) dA (2y − 3x ) dx dy 3 2xy − x 2 2 6y − 27 2 3 dy 1 − 2y − 3 2 dy 4y − 12 dy 2y 2 − 12y 5 2 = Kjell Konis (Copyright © 2013) 50 − 60 − 8 − 24 = 6 4. Multiple Integrals 12 / 58

Example (continued) 3 f (x , y ) dA = R 1 2 3 = 1 1 3 = = = Kjell Konis (Copyright © 2013) 1 (2y − 3x ) dy dx (y 2 − 3xy ) 5 dx 2 3 = 5 [(25 − 15x ) − (4 − 6x )] dx 21 − 9x dx 9 21x − x 2 2 63 − 3 1 81 9 − 21 − = 42 − 36 = 6 2 2 4. Multiple Integrals 13 / 58

Double Integrals Non-Rectangular Regions If f (x , y ) is continuous on a region D that can be described D = {(x , y ) : a ≤ x ≤ b, g1 (x ) ≤ y ≤ g2 (x ) } then b f (x , y ) dA = D a g2 (x ) f (x , y ) dy dx g1 (x ) If f (x , y ) is continuous on a region D that can be described D = {(x , y ) : c ≤ y ≤ d, h1 (y ) ≤ x ≤ h2 (y ) } then d f (x , y ) dA = D Kjell Konis (Copyright © 2013) c h2 (x ) f (x , y ) dx dy h1 (x ) 4. Multiple Integrals 14 / 58

Example Let D = {(x , y ) : |x | + |y | ≤ 1} diamond w/ corners at (0, ±1) and (±1, 0) Compute the integral of f (x , y ) = 1 over D 1 dA D 0 1+x = 1 1 dy dx + −1 −1−x 1+x 0 = y −1 Kjell Konis (Copyright © 2013) 0 1 dy dx x −1 1−x 1 dx + −1−x 1−x 0 4. Multiple Integrals y dx x −1 15 / 58

Example (continued) 0 1 1+x − −1−x = dx + −1 0 = 1 2 + 2x dx + −1 = 2x + x 2 0 0 1−x − x −1 dx 2 − 2x dx + 2x − x 2 1 0 −1 = 0 0+0 − −2+1 + 2−1 − 0−0 =1+1 =2 Kjell Konis (Copyright © 2013) 4. Multiple Integrals 16 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 17 / 58

Change of Variables: Single Variable Case Let f (x ) be a continuous function Let g(s) be a continuously diﬀerentiable and invertible function • Implies g(s) either strictly increasing or strictly decreasing g(s) maps the interval [c, d] into the interval [a, b], i.e., s ∈ [c, d] → x = g(s) ∈ [a, b] Integration by substitution says: s=g −1 (b) x =b f (x ) dx = x =a Kjell Konis (Copyright © 2013) s=g −1 (a) 4. Multiple Integrals f (g(s)) g (s) ds 18 / 58

Change of Variables: Functions of 2 Variables Let f (x , y ) be a continuous function Want to compute: f (x , y ) dA D Let Ω be a domain such that the mapping x = x (s, t) y = y (s, t) of a point (s, t) ∈ Ω to a point (x , y ) ∈ D is one-to-one and onto • x (s, t) and y (s, t) continuously diﬀerentiable That is, (s, t) ∈ Ω ←→ (x , y ) = (x (s, t), y (s, t)) ∈ D Want to ﬁnd a function h(s, t) such that f (x , y ) dx dy = D Kjell Konis (Copyright © 2013) h(s, t) ds dt Ω 4. Multiple Integrals 19 / 58

Change of Variables Get started f (x , y ) = f (x (s, t), y (s, t)) In the single variable case, if x = g(s) then dx = g (s) ds In the 2-variable case, (x , y ) = (x (s, t), y (s, t)) is a vector-valued function of 2 variables The gradient of (x (s, t), y (s, t)) is the 2 × 2 array  ∂x  ∂s  D(x (s, t), y (s, t)) =   ∂y ∂s Kjell Konis (Copyright © 2013) 4. Multiple Integrals ∂x  ∂t    ∂y  ∂t 20 / 58

Jacobian The 2-variable equivalent of dx = g (s) ds is dx dy = ∂x ∂y ∂x ∂y − ∂s ∂t ∂t ∂s ds dt and the quantity in the square brackets is called the Jacobian 2 dimensional change of variables formula: f (x , y ) dx dy = D Kjell Konis (Copyright © 2013) f (x (s, t), y (s, t)) Ω 4. Multiple Integrals ∂x ∂y ∂x ∂y ds dt − ∂s ∂t ∂t ∂s 21 / 58

Example Example from previous section Let D = {(x , y ) : |x | + |y | ≤ 1} diamond w/ corners at (0, ±1) and (±1, 0) Compute the integral of f (x , y ) = 1 over D 0 1+x 1 dA = D Kjell Konis (Copyright © 2013) 1 1 dy dx + −1 −1−x 4. Multiple Integrals 0 1−x 1 dy dx x −1 22 / 58

Example (continued) Consider the change of variables: s = x + y, t =x −y Solve for x and y in terms of s and t: x= s +t , 2 y= s −t 2 Compute the partial derivatives of the change of variables: ∂x 1 = , ∂s 2 ∂x 1 = , ∂t 2 ∂y 1 = , ∂s 2 ∂y 1 =− ∂t 2 Compute the Jacobian: ∂x ∂y 1 ∂x ∂y 1 − = − ∂s ∂t ∂t ∂s 2 2 Kjell Konis (Copyright © 2013) − 4. Multiple Integrals 1 1 1 11 =− − =− 22 4 4 2 23 / 58

Example (continued) Multivariate change of variables formula: 1 dx dy = D 1 Ω ∂x ∂y ∂x ∂y − ds dt = ∂s ∂t ∂t ∂s Ω 1 ds dt 2 Finally, domain of integration (Ω): t=x−y s=x+y Kjell Konis (Copyright © 2013) 4. Multiple Integrals 24 / 58

Example (continued) Domain of integration: Ω = [−1, 1] × [−1, 1] 0 1+x 1 1 dA = 1 dy dx + −1 −1−x Ω D 0 1−x 1 dy dx 1 ds dt 2 = t=1 s=1 t=−1 s=−1 = t=1 = t=−1 s 2 x −1 1 ds dt 2 s=1 dt s=−1 t=1 = 1 dt t=−1 = 2 Kjell Konis (Copyright © 2013) 4. Multiple Integrals 25 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 26 / 58

Change of Variables Example Evaluate x dx dy where y D D = (x , y ) ∈ R2 : x ≥ 0, y= 1 ≤ xy ≤ 2, y = 2x y 1≤ ≤2 x Can assume x > 0 1 x y=x y= 2 x ≤y ≤ x ≤ y ≤ 2x D= x dx dy = Kjell Konis (Copyright © 2013) x 1 x x 1 D 2 √ 2 2 √ 2x 1 x x dy dx + 4. Multiple Integrals 1 2 x 2 x dy dx x 27 / 58

1 = √ 2 2 √ 2 2 = = x dy dx + y =2x xy 1 y= x √ 2 2 2 2x − 1 dx + 2 3 x −x 3 1 √ 2 2 2 −1 − 3 1 2 2 y= x  xy 1 2 x dy dx x √   dx + 2 x 2 1 √ 1 = 1 x  1 = √ 2x   dx y =x 2 − x 2 dx √ 1 + 2x − x 3 3 2 1 √ √ 2 ( 2)3 2 − 3 3 2 2 + √ 1 1 √ 2 2 − ( 2)3 − 2 − 3 3 √ √ √ √ √ √ 1 2 3 2 12 2 4 2 5 −12 + 10 2 −6 + 5 2 = − − + + − − = = 3 6 6 6 6 3 6 3 Kjell Konis (Copyright © 2013) 4. Multiple Integrals 28 / 58

Again, Using a Change of Variables Consider the change of variables y y t= x s = xy , D = (x , y ) ∈ R2 : x ≥ 0, y = 2x 1 ≤ xy ≤ 2, 1≤ y= y ≤2 x y=x Ω = [1, 2] × [1, 2] x 1 x x t=2 s=2 x dx dy = D Kjell Konis (Copyright © 2013) y= 2 x dx dy t=1 s=1 4. Multiple Integrals 29 / 58

Again, Using a Change of Variables First, solve for functions x = x (s, t) and y = y (s, t): √ s x= , y = st t Partial derivatives for the change of variables are: ∂x ∂s ∂ 1 −1 s2t 2 ∂s = = ∂ 1 1 s2t2 ∂s ∂x ∂t 1 −1 −1 1 s 2t 2 = √ 2 2 st = ∂y ∂s = √ 1 −1 1 t s 2t2 = √ 2 2 s Kjell Konis (Copyright © 2013) = ∂ 1 −1 s2t 2 ∂t √ s 1 1 −3 = − s2t 2 = − √ 2 2t t ∂y ∂t ∂ 1 1 s2t2 ∂t = 4. Multiple Integrals = √ 1 1 −1 s s2t 2 = √ 2 2 t 30 / 58

Again, Using a Change of Variables Jacobian for the change of variables: = √ √ s s √ − − √ 2 st 2 t 2t t = √ t √ 2 s 1 1 + 4t 4t = ∂x ∂y ∂x ∂y − ∂s ∂t ∂t ∂s 1 2t 1 √ Change of variables formula: t=2 s=2 x dx dy = D Kjell Konis (Copyright © 2013) t=1 s=1 4. Multiple Integrals s 1 ds dt t 2t 31 / 58

t=2 s=2 s=1 1 2 = t=1 s 1 ds dt = t 2t 1 2 = = 1 3 1 3 t=2 t=1 t=2 t=1 t=2 s=2 1 s=1 2 3 −3 s2t 2 3 = = Kjell Konis (Copyright © 2013) s=2 dt s=1 √ 3 3 2 2t − 2 − t − 2 dt t=1 t=2 √ 3 2 2 − 1 t − 2 dt t=1 √ = 3 s 2 t − 2 ds dt 1 2 2−1 −2t − 2 3 t=2 t=1 √ √ 2 2−1 (2 − 2) 3 √ 5 2−6 3 4. Multiple Integrals 32 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 33 / 58

Separable Functions Take another look at the example in the last section t=2 t=1 s=2 s=1 s 1 ds dt = t 2t 1 2 t=2 s=2 t=1 1 3 s 2 t − 2 ds dt s=1 Since t (and any function of t) is constant while integrating wrt s t=2 t=1 s=2 s=1 s 1 ds dt = t 2t 1 2 t=2 s=2 3 t− 2 1 s 2 ds dt s=1 t=1 The double integral is the product of 2 single-variable deﬁnite integrals t=2 t=1 s=2 s=1 Kjell Konis (Copyright © 2013) s 1 ds dt = t 2t 1 2 s=2 1 t=2 s 2 ds s=1 4. Multiple Integrals 3 t − 2 dt t=1 34 / 58

Separable Functions 1 2 s=2 1 t=2 s 2 ds s=1 3 t − 2 dt = t=1 = 1 2 2 3 s2 3 −2 3 s2 3 s=2 1 t=2 −2t − 2 s=1 s=2 t=1 1 t=2 t− 2 s=1 t=1 = − 2 √ 2 2−1 3 = − √ 2 1 2−2 2− √ +1 3 2 1 √ −1 2 √ √ 1 4−4 2− 2+2 3 √ 5 2−6 3 = − = Kjell Konis (Copyright © 2013) 4. Multiple Integrals 35 / 58

Separable Functions Let R = [a, b] × [c, d] be a rectangle Let f (x , y ) be a continuous, separable function f (x , y ) = g(x ) h(y ) g(x ) and h(x ) continuous Double integral of f over R is the product of single-variable integrals d f (x , y ) dx dy b = R g(x ) h(y ) dx dy c a d = b h(y ) c g(x ) dx dy a b d g(x ) dx = a Kjell Konis (Copyright © 2013) 4. Multiple Integrals h(y ) dy c 36 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 37 / 58

Polar Coordinates Describe points in R2 using r ∈ [0, ∞) θ ∈ [0, 2π) y (x0,y0) q r0 r = distance from origin θ1 θ = angle counter clockwise from positive x -axis θ0 x (x , y ) ←→ (r , θ) x (r , θ) = r cos(θ) q r1 (x1,y1) y (r , θ) = r sin(θ) Can simplify integration problems Kjell Konis (Copyright © 2013) 4. Multiple Integrals 38 / 58

Change of Variables to Polar Coordinates The change of variables is: x (r , θ) = r cos(θ) y (r , θ) = r sin(θ) The partial derivatives of the change of variables are: ∂x = cos(θ) ∂r ∂x = −r sin(θ) ∂θ ∂y = sin(θ) ∂r ∂y = r cos(θ) ∂θ The Jacobian is: ∂x ∂y ∂x ∂y − ∂r ∂θ ∂θ ∂r = cos(θ) · r cos(θ) − (−r sin(θ)) · sin(θ) = r cos2 (θ) + sin2 (θ) = r Kjell Konis (Copyright © 2013) 4. Multiple Integrals 39 / 58

Change of Variables to Polar Coordinates The change of variable formula to polar coordinates: f (x , y ) dx dy = f (r cos(θ), r sin(θ)) r dr dθ D D Integrate f (x , y ) over a disk of radius R centered at the origin: 2π f (x , y ) dx dy = D(0,R) R 0 0 f (r cos(θ), r sin(θ)) r dr dθ Integrate f (x , y ) over the plane R2 : 2π R2 f (x , y ) dx dy = 0 ∞ 0 f (r cos(θ), r sin(θ)) r dr dθ The latter can be useful for integrating probability density functions Kjell Konis (Copyright © 2013) 4. Multiple Integrals 40 / 58

Example Let D = {(x , y ) : x 2 + y 2 ≤ 1} (disk of radius 1 centered at origin) Compute the integral (1 − x 2 − y 2 ) dx dy D Try once using xy-coordinates Then try once using polar coordinates Kjell Konis (Copyright © 2013) 4. Multiple Integrals 41 / 58

√ 1 f (x , y ) dy dx = −1 D √ y =− 1−x 2 √ √ 6( 1 − x 2 )3 − 2( 1 − x 2 )3 3 1 = −1 4 3 √ y = 1−x 2   dx √ 2( 1 − x 2 )3 2(1 − x ) 1 − x 2 − 3 −1 1 dx dx ( 1 − x 2 )3 dx −1 . . . Kjell Konis (Copyright © 2013) (1 − x 2 − y 2 ) dy dx 2 = = 1−x 2 y3  (1 − x 2 )y − 3 −1 1 = √  1 = − 1−x 2 . . . 1 x 6 1 − x 2 (5 − 2x 2 ) + 3 arcsin(x ) 4. Multiple Integrals 1 −1 42 / 58

Example (in polar coordinates) 2π f (x , y ) dy dx = D 0 1 2π = 1 − r 2 cos2 (θ) − r 2 sin2 (θ) r dr dθ 0 0 1 1 − r 2 cos2 (θ) + sin2 (θ) r dr dθ 0 2π = 0 0 2π = 0 2π = Kjell Konis (Copyright © 2013) 0 1 r − r 3 dr dθ r2 r4 − 2 4 1 dθ 0 1 π dθ = 4 2 4. Multiple Integrals 43 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 44 / 58

The Standard Normal Density The standard normal density x2 1 φ(x ) = √ e − 2 2π The function Φ used in the Black-Scholes formula x Φ(x ) = φ(t) dt −∞ 1 Raises 2 2 questions: √1 2π come from? 1 Where does the 2 Why not use a closed-form expression for Φ? Where does the √1 come from? 2π 21 2 Kjell Konis (Copyright © 2013) 4. Multiple Integrals 45 / 58

The Standard Normal Density Since φ is a probability density function ∞ ∞ φ(x ) dx = φ(x ) dx = 1 −∞ Implies that −∞ ∞ x2 e − 2 dx = √ 2π −∞ Change of variables ∞ 2 e −x dx = √ π −∞ 2 The problem is e −x does not have an antiderivative Kjell Konis (Copyright © 2013) 4. Multiple Integrals 46 / 58

Change to Polar Coordinates Let ∞ M= 2 e −x dx −∞ Want to show that M = √ π Can also express M as ∞ M= 2 e −y dy −∞ Now want to show that M 2 = π M2 = ∞ 2 e −x dx −∞ ∞ 2 e −y dy −∞ This is the double integral of a separable function Kjell Konis (Copyright © 2013) 4. Multiple Integrals 47 / 58

Change to Polar Coordinates Unseparate the double integral ∞ M2 = ∞ 2 e −x dx −∞ ∞ = −∞ e −x 2 −∞ ∞ = 2 e −y dy ∞ 2 e −y dy dx −∞ e −x 2 −∞ ∞ 2 e −y dy −∞ ∞ ∞ e −(x = 2 +y 2 ) dy dx −∞ −∞ = R2 e −(x 2 +y 2 ) dy dx Change to polar coordinates 2π = Kjell Konis (Copyright © 2013) 0 ∞ 0 2 +[r e −[r cos(θ)] sin(θ)]2 4. Multiple Integrals r dr dθ 48 / 58

Change to Polar Coordinates M2 = 2π ∞ 0 0 2π = ∞ 0 0 2π = ∞ 0 0 2π = dθ 0 2 +[r e −[r cos(θ)] e −r r dr dθ 2 − θ 0 = [2π − 0] − Kjell Konis (Copyright © 2013) 2 [cos2 (θ)+sin2 (θ)] r dr dθ let u = −r 2 e −r r dr dθ 2π = sin(θ)]2 1 2 u=−∞ 1 − eu 2 1 2 e u (−2r dr ) du = −2r dr u=0 −∞ 0 lim e t − 1 t→−∞ 4. Multiple Integrals = 2π · 1 =π 2 49 / 58

Change to Polar Coordinates In summary: Started out with ∞ M= 2 e −x dx −∞ Showed that M 2 = π and thus that M = Can conclude that ∞ 2 e −x dx = √ √ π π −∞ Kjell Konis (Copyright © 2013) 4. Multiple Integrals 50 / 58

Outline 1 Double Integrals 2 Fubini’s Theorem 3 Change of Variables for Double Integrals 4 Change of Variables Example 5 Double Integrals of Separable Functions 6 Polar Coordinates 7 A Culturally Important Integral 8 Marginal Density of a Bivariate Normal Distribution Kjell Konis (Copyright © 2013) 4. Multiple Integrals 51 / 58

Marginal Density of a Bivariate Normal Distribution Bivariate normal density function: fX ,Y (x , y ; µx , µy , σx , σy , ρ) (x − µx )2 2ρ(x − µx )(y − µy ) (y − µy )2 − +   2 2   σx σx σy σy 1  exp −   2(1 − ρ2 ) 2πσx σy 1 − ρ2     The marginal density is ∞ fY (y ) = −∞ fX ,Y (x , y ) dx Let X and Y be returns on an asset in consecutive periods Assume that µx = µy = µ and σx = σy = σ Kjell Konis (Copyright © 2013) 4. Multiple Integrals 52 / 58

Marginal Density ∞ fX (x ) = ∞ = −∞ −∞ fX ,Y (x , y ) dy (x − µ)2 − 2ρ(x − µ)(y − µ) + (y − µ)2 1 exp − dy 2σ 2 (1 − ρ2 ) 2πσ 2 1 − ρ2 Guess that fX (x ) is normal with mean µ and variance σ 2 fX (x ) = √ 1 (x − µ)2 exp − 2σ 2 2πσ ∞ × C exp −∞ where C = √ (x − µ)2 (x − µ)2 − 2ρ(x − µ)(y − µ) + (y − µ)2 − dy 2σ 2 2σ 2 (1 − ρ2 ) 1 2πσ 1 − ρ2 Kjell Konis (Copyright © 2013) 4. Multiple Integrals 53 / 58

Marginal Density Lets look at the quantity in the square brackets (x − µ)2 (x − µ)2 − 2ρ(x − µ)(y − µ) + (y − µ)2 − 2σ 2 2σ 2 (1 − ρ2 ) = (1 − ρ2 )(x − µ)2 − (x − µ)2 + 2ρ(x − µ)(y − µ) − (y − µ)2 2σ 2 (1 − ρ2 ) = −ρ2 (x − µ)2 + 2ρ(x − µ)(y − µ) − (y − µ)2 2σ 2 (1 − ρ2 ) = − = − ρ2 (x − µ)2 − 2ρ(x − µ)(y − µ) + (y − µ)2 2σ 2 (1 − ρ2 ) ρ(x − µ) − (y − µ) 2σ 2 (1 − ρ2 ) Kjell Konis (Copyright © 2013) 2 = − y − (µ + ρ(x − µ)) 4. Multiple Integrals 2 2(σ 1 − ρ2 )2 54 / 58

Marginal Density fX (x ) = √ (x − µ)2 1 exp − 2σ 2 2πσ ∞ × C exp −∞ (x − µ)2 (x − µ)2 − 2ρ(x − µ)(y − µ) + (y − µ)2 dy − 2σ 2 2σ 2 (1 − ρ2 ) Lets look just at the integrand √ 1 2π(σ 1 − ρ2 ) exp − y − (µ + ρ(x − µ)) 2 2(σ 1 − ρ2 )2 Let m = (µ + ρ(x − µ)) and s = (σ 1 − ρ2 ), the integrand becomes √ 1 (y − m)2 exp − 2s 2 2πs Kjell Konis (Copyright © 2013) 4. Multiple Integrals 55 / 58

Marginal Density The integral becomes ∞ √ −∞ (y − m)2 1 exp − 2s 2 2πs dy Integral of a normal density over the real line, thus equal to 1 The guess for the marginal density was correct fX (x ) = √ = √ 1 (x − µ)2 exp − 2σ 2 2πσ ∞ √ −∞ 1 (y − m)2 exp − 2s 2 2πs dy 1 (x − µ)2 exp − 2σ 2 2πσ Kjell Konis (Copyright © 2013) 4. Multiple Integrals 56 / 58

Bonus: Conditional Density Function The function that we integrated out is the conditional density Denoted by fY |X (y |x ) fY |X (y |x ) = √ Kjell Konis (Copyright © 2013) 1 2π(σ 1 − ρ2 ) exp − 4. Multiple Integrals y − (µ + ρ(x − µ)) 2 2(σ 1 − ρ2 )2 57 / 58

http://computational-finance.uw.edu Kjell Konis (Copyright © 2013) 4. Multiple Integrals 58 / 58

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