(t) = exp Z t 0 r du : The market price of risk equations are 11 = 1 r 21 + q 1 222 = 2 r (MPR) 203 204 The
solution to these equations is 1 = 1 r 1 ; 2 = 1(2 r) 2(1 r) 12p 1 2 ; provided 1 < 1. Suppose 1 < 1. Then
(MPR) has a unique solution (1; 2); we d
(T ) ; we have X(t) = Y (t); 0 t T ; and in particular, X(T ) = V : Every F(T )-measurable random variable can
be hedged; the market is complete. 19.2 Hedging when = 1 The case = 1 is analogous. Assume that = 1.
Then dS1 = S1[1 dt + 1 dB1] dS2 = S2[2 dt +
+ 1 2 2 . If ~(!) < 1, then lim
#0 e ~(!) = 1; if ~(!) = 1, then e ~(!) = 0 for every
> 0, so lim
#0 e ~(!) = 0: Therefore, lim
#0 e ~(!) = 1< 0, then IP f < 1: (Recall that x > 0). 25.5 First passage times: Second method (Based on
martingales) Let > 0 be
to match dY , we must have 2 = 0: 208 Chapter 20 Pricing Exotic Options 20.1 Reflection principle for
Brownian motion Without drift. Define M(T ) = max 0tT B(t): Then we have: IP fM(T ) > m; B(T ) < bg = IP
fB(T ) > 2m bg = 1 p 2T Z 1 2mb exp ( x2 2T ) dx
rK1fS(t)< 1; 3. ertv(S(t) is the smallest process with properties 1 and 2. Explanation of property 3. Let Y
be a supermartingale satisfying Y (t) ert(K S(t)+ ; 0 t < 1: (8.1) Then property 3 says that Y (t) ertv(S(t);
0 t < 1: (8.2) We use (8.1) to prove
EXAMPLE 16.8.5 Let F = y 2 , x, z2 , and let the curve C be the intersection of the cylinder x 2 + y 2 = 1
with the plane y + z = 2, oriented counter-clockwise when viewed from above. We compute C F dr in
two ways. First we do it directly: a vector functi
b a f (t) dt = f(b) f(a) = f(b) f(a). This theorem, like the Fundamental Theorem of Calculus, says
roughly that if we integrate a derivatve-like functon (f or f) the result depends only on the values of
the original functon (f) at the endpoints. If a vect
guidance, an individual childs play interest can develop into a classroom-wide, extended investigation or
project that includes rich mathematical learning [7882]. In classrooms in which teachers are alert to all
these possibilities, childrens Copyright 20
Surface Integrals 443 We write the hemisphere as r(, ) = cos sin ,sin sin , cos , 0 /2
and 0 2. So r = sin sin , cos sin , 0 and r = cos cos ,sin cos , sin . Then r
r = cos sin2 , sin sin2 , cos sin and |r r| = |sin | = sin , since we are interested
only
homogeneous equation. Solve this for g, then use the relationship g = y to find y. 21. Suppose that y(t) is
a solution to ay + by + cy = 0, y(t0) = 0, y(t0) = 0. Show that y(t) = 0. 17.6 Second Order Linear
Equations Now we consider second order equations
initial value problem y = t 2 + 1, y(1) = 4 has solution f(t) = t 3 /3 + t + 8/3. The general first order
equation is rather too general, that is, we cant describe methods that will work on them all, or even a
large portion of them. We can make progress w
words, work is computed using a particular line integral of the form 424 Chapter 16 Vector Calculus we
have considered. Alternately, we sometimes write C F r dt = C f, g, h x , y , z dt = C ( f dx dt
+ g dy dt + h dz dt ) dt = C f dx + g dy + h dz = C f d
at n, min at /2 + n 5.3.18. max at /2+2n, min at 3/2+2n 5.4.1. concave up everywhere 5.4.2.
concave up when x < 0, concave down when x > 0 5.4.3. concave down when x < 3, concave up when x >
3 5.4.4. concave up when x < 1/ 3 or x > 1/ 3, concave down when
y/2, Q = x/2. EXAMPLE 16.4.3 An ellipse centered at the origin, with its two principal axes aligned with
the x and y axes, is given by x 2 a 2 + y 2 b 2 = 1. We find the area of the interior of the ellipse via Greens
theorem. To do this we need a vector e
applications. Both are most easily understood by thinking of the vector field as representing a flow of a
liquid or gas; that is, each vector in the vector field should be interpreted as a velocity vector. Roughly
speaking, divergence measures the tendenc
surface, oriented outward, then D F N dS = E F dV. 16.9 The Divergence Theorem 451 Proof.
Again this theorem is too difficult to prove here, but a special case is easier. In the proof of a special case
of Greens Theorem, we needed to know that we could de
across the boundary of the region, from inside to out, and the second sums the divergence (tendency to
spread) at each point in the interior. The theorem roughly says that the sum of the microscopic
spreads is the same as the total spread across the bound
surprising then that a great many early childhood programs have a considerable distance to go to
achieve high-quality mathematics education for children age 3-6. In 2000, with the growing evidence
that the early years significantly affect mathematics lear
the initial value problem y = (t, y), y(t0) = y0, for t t0. Under reasonable conditions on , we know
the solution exists, represented by a curve in the t-y plane; call this solution f(t). The point (t0, y0) is of
course on this curve. We also know the slo
The characteristic polynomial is x 2 + 4x + 5 with roots (4 16 20)/2 = 2 i. Thus the general
solution is y = A cos(t)e 2t + B sin(t)e 2t . Suppose we know that y(0) = 1 and y(0) = 2. Then as before
we form two simultaneous equations: from y(0) = 1 we get
solution to y + p(t)y = f(t) is v(t)h(t) + AeP (t) = v(t)e P (t) + AeP (t) . EXAMPLE 17.3.1 Find the solution of
the initial value problem y + 3y/t = t 2 , y(1) = 1/2. First we find the general solution; since we are
interested in a solution with a given
coefficients. The solution to the homogeneous equation is Aet + Btet and so the simultaneous equations
are ue t + vtet = 0 ue t + vtet + vet = e t t 2 . Subtracting the equations gives ve t = e t t 2 v = 1 t 2 v =
1 t . Then substituting we get ue t = vt
explain why there is no such f. 7. Find an f so that f = yz, xz, xy, or explain why there is no such f.
8. Evaluate C (10x 4 2xy 3 ) dx3x 2 y 2 dy where C is the part of the curve x 5 5x 2 y 2 7x 2 = 0 from
(0, 0) to (3, 2). 9. Let F = yz, xz, xy. Find t
developmentally appropriate teaching, illuminating various avenues to understanding and guiding
teachers in providing activities appropriate for children as individuals and as a group. 6. Provide for
childrens deep and sustained interaction with key mathe
we can get an idea of how solution curves must look. Such a plot is called a slope field. A slope field for
= t y 2 is shown in figure 17.4.3; compare this to figure 17.4.1. With a little practice, one can sketch
reasonably accurate solution curves based