Math 501–1, Spring 2006 University of Utah

Solutions to Problem Assignment #5
Math 501–1, Spring 2006
University of Utah
Problems:
1. A gambling book recommends the following “winnin strategy” for the game of roulette:
It recommends that the gambler bet $1 on red. If red appears (this has probability
18/38), then the gambler should take her $1 profit and quit. Else, she should make
additional $1 bets on red on each of the next two spins of the roulette wheel, and then
quit. Let X denote the gambler’s winnings when she quits.
(a) Find P{X > 0}.
Solution. Let Ri denote the event, “red on the ith trial.” Because P(Ri) = 18/38 and
P(Rc
i ) = 20/38,
P{X = 1} = P(R1) + P(Rc1
\ R2 \ R3) =
18
38
+

20
38 ·
18
38 ·
18
38

= 0.591¯7.
On the other hand,
P{X = −1} = P(Rc1
\ R2 \ Rc3
) + P(Rc1
\ Rc2\ R3)
=

20
38 ·
18
38 ·
20
38

+

20
38 ·
20
38 ·
18
38

0.2624,
P{X = −3} = P(Rc1
\ Rc2
\ Rc3
) =

20
38
3
0.1458.
In particular, P{X > 0} = P{X = 1} = 0.591¯7.
(b) Are you convinced that the strategy is indeed a “winning” strategy? Explain your
answer.
Solution. One may think at first that this is a winning strategy because P{win} > 0.5.
But it is the expectation that counts, and we will see next that EX < 0. This
means that the strategy is a losing strategy.
(c) Compute EX.
Solution. We have
EX = 0.591¯7 − 0.2624 − (3 × 0.1458) −0.108.
Therefore, every time we play we expect to lose about 10 cents.
2. One of the numbers 1 through 10 is chosen at random. You are to try and guess
the number chosen by asking questions with “yes-no” answers. Compute the expected
number of questions that you need to ask in each of the following two cases:
(a) Your ith question is to be, “Is it i?”, for i = 1, . . . , 10.
Solution. Let Q denote the number of questions asked. It follows that P{Q = i} = 1/10.
Therefore,
E(Q) =
1
10
+
2
10
+ · · · +
10
10
= 5.5
(b) With each question you try to eliminate one-half of the remaining numbers, as
nearly as possible.
Solution. Consider the following strategy: Divide what you have in halves and ask, “is it
greater than or equal to x?” where x is the middle of the numbers. For instance,
your first question is, “is it greater than or equal to 5.5?”. [All other strategies of
this type are similar. But you need to be consistent.] Let X denote the random
number, so that P{X = i} = 1/10 for i = 1, . . . , 10. If X = 1, 2, 3, 6, 7, 8, then
Q = 3. If X = 4, 5, 9, 10 then Q = 4 (check!). Therefore, P{Q = 3} = P{X =
1} + P{X = 2} + P{X = 3} + P{X = 6} + P{X = 7} + P{X = 8} = 0.6, and
P{Q = 4} = 0.4. Thus, E(Q) = (3 × 0.6) + (4 × 0.4) = 3.4.
3. A person tosses a fair coin until a tail appears for the first time. If “tail” appears
on the nth flip, then the person wins 2n dollars. Let X denote the person’s winnings.
Show that EX = 1. This is known as the “St.-Petersbourg paradox.”
Solution. For all n 1,
P{X = n} = P(H1 \ · · · \ Hn−1 \ Tn) =
1
2n .
Else, P{X = n} = 0. Therefore,
EX =
1X
n=1

2n ×
1
2n

= 1.
4. A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If
they are the same color, then you win $1.10; if they are different colors, then you win
−$1.00 (i.e., you lose one dollar). Compute:
(a) the expected value of the amount you win;
Solution. Let Ri denote the event, “red on ith draw.” Then, P(R1 \ R2) = (5/10)(4/9)
and P(Rc1
\ Rc2
) = (5/10)(4/9). Let X denote the win, where X < 0 means loss.
Then, P{X = 1.10} = 2×(5/10)(4/9) = 4/9, and P{X = −1} = 5/9. Therefore,
EX = (1.1 × 4
9 ) − 5
9 = −0.0¯6.
(b) the variance of the amount you win.
Solution. Evidently, E[X2] = (1.1× 4
9 )+ 5
9 = 1.0¯4. Therefore, Var(X) = 1.0¯4 −(−0.0¯6)2 =
1.04.
Theoretical Problems:
1. Let X be such that P{X = 1} = p and P{X = −1} = 1 − p. Find a constant c 6= 1
such that E[cX] = 1.
Solution. First, for all real c,
E[cX] = cp + c−1(1 − p).
Set this equal to one to find that cp + (1 − p)/c = 1. I.e., pc2 − c + (1 − p) = 0.
This is a quadratic equation (in c) and has two roots:
c =
1 ±
p
1 − 4p(1 − p)
2p
.
But 1 − 4p(1 − p) = 1 − 4p + 4p2 = (1 − 2p)2. Therefore,
c =
1 ± (1 − 2p)
2p
= 1 and
1 − p
p
.
So the answer is c = (1 − p)/p.
2. Prove that if X is a Poisson random variable with parameter , then for all n 1,
E[Xn] = E[(X + 1)n−1]. (1)
Use this to compute EX, E[X2], VarX, and E[X3].
Solution. We calculate directly:
E[Xn] =
1X
k=0
knP{X = k}
=
1X
k=0
kn e− k
k!
=
1X
k=1
kn e− k
k!
=
1X
k=1
kn−1 e− k−1
(k − 1)!
=
1X
j=0
(j + 1)n−1 e− j
j!
= E[(X + 1)n−1].
Therefore,
EX = E[(1 + X)0] = ,
E[X2] = E[(1 + X)] = (1 + ) = + 2,
VarX = E[X2] − (EX)2 = ,
E[X3] = E[(1 + X)2] =

E

1 + 2X + X2
=

1 + 2 + + 2
= + 3 2 + 3.

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Math 678. Homework 3 Solutions.

Math 678. Homework 3 Solutions.
http://math.gmu.edu/~memelian/teaching/Fall11/math678/hw/hw3sol.pdf
#1
We need to derive the formula for the solution of the IVP
ut 􀀀 u + cu = f: in Rn (0;1)
u = g; on Rn ft = 0g
Some observations: (1) the solution to the non-homogeneous problem can be
obtained from the solution to the homogeneous problem via Duhamel's principle;
(2) if the term cu is not present, we know the exact solution to this IVP, (3)
ut + cu = 0 is equivalent to ectut + cectu = 0 which converts to (ectu)t = 0 and
leads to u = Ce􀀀ct as a solution (should remind you of an integrating factor
technique).
From the above observations, we see that a good way to proceed is to mul-
tiply both sides by the function ect. This gives us:
ectut + cectu 􀀀 ect u = ectf
Since (ectu) = ect u, the above can be written as
(ectu)t 􀀀 (ectu) = ectf
and hence this converts into a regular heat equation formulation in terms of ectu,
with non-homogeneous right hand side. The initial condition is unchanged, since
(ectu)(x; 0) = u(x; 0) = g(x). So now we can use Duhamel's principle and write
the exact solution for this modi ed IVP as:
ectu(x; t) =
Z
Rn
(x 􀀀 y; t)g(y)dy +
Z t
0
Z
Rn
(x 􀀀 y; t 􀀀 s)ecsf(y; s)dyds
The solution to the original IVP then follows:
u(x; t) = e􀀀ct
h Z
Rn
(x 􀀀 y; t)g(y)dy +
Z t
0
Z
Rn
(x 􀀀 y; t 􀀀 s)ecsf(y; s)dyds
i
#2
Now let us consider the usual heat equation ut = uxx in 1d with u(x; t) being a
solution.
(a) Take v(x; t) = u(x􀀀y; t). How does one show this is also a solution? The
easiest way is to plug it into the equation. First we need to use chain rule and
compute partial derivatives: vt = (u(x􀀀y; t))t = ut(x􀀀y; t), vx = ux(x􀀀y; t),
relying on the fact that (x 􀀀 y)x = 1. Since ut(x; t) = uxx(x; t) for any x 2 R,
we have vt = vxx. Notice that if we had an IVP originally with u(x; 0) = g(x),
the initial condition for v would be hy(x) = g(x 􀀀 y), and the formula for the
solution using fundamental function (x; t) would give us:
u(x 􀀀 y; t) =
Z
R
(x 􀀀 y 􀀀 s; t)hy(s)ds
1
where s0 = y + s
(b) For any derivative D , where is a multiindex, and the derivative is
taken either in t or in x, we have (D u)t = D ut = D uxx = (D u)xx since
derivatives commute. it follows that any derivative of the solution of the heat
equation is again a solution.
(c) (au + bv)t = aut + bvt = auxx + bvxx = (au + bv)xx, hence au + bv is
again a solution, if u; v are solutions.
(d) This follows from the fact that the operations of integration and di er-
entiation are commutative:
@
@t
Z x
0
u(y; t)dy

=
Z x
0
ut(y; t)dy =
Z x
0
uxx(y; t)dy = @2
@x2
Z x
0
u(y; t)dy

(e) Let v(x; t) = u(
p
ax; at). Then vt(x; t) = av(x; t); vx(x; t) =
p
av(x; t); vxx =
av(x; t). It follows that vt = vxx.
#3
The easiest example of a Dirichlet problem with no solution can be constructed
as follows. Let UT = U (0; T) and = U
T 􀀀 UT be the boundary of this
cylinder including the top, bottom and the sides, while denoting 􀀀T to be the
parabolic boundary comprised of the bottom and vertical sides only. Consider
ut 􀀀 u = 0; in UT
u = f; on
Suppose u 2 C2
1 (UT ) \ C( U
T ) solves this IVP. Then it should satisfy the
weak maximum principle, namely
max
U
T
u(x; t) = max
􀀀T
u(x; t)
Since u(x; t) = f(x) on 􀀀T , we need the solution to satisfy max
U
T
u = max
􀀀T
f.
In particular, f(x; t = T) < max
􀀀T
f, which does not hold for all continuous
functions.
If we allow t = T to be part of the cylinder UT = U (0; T], we need to
construct a more elaborate example of a Dirichlet problem with no solution.
One such example is a ball in R3 with deformable surface. We can push in a
sharp spike at some point on this surface and assume that near the tip of the
spike the surface takes the form of a conical surface obtained by rotating the
curve
y =

e􀀀1=x: for x > 0
0: for x = 0
about the x-axis. Then we can consider heat conduction on the interior of the
deformed ball de ned this way, called
. If the temperature distribution on @

is given by a continuous function f which is equal to zero at points of the spike
2
and is equal to a large positive constant temperature T at points away from the
spike, the seady state temperature u(x) should be close to T for all x in
. But
this is impossible, since u(x) won't be able to approach the zero temperature
as x approaches the spike from within of
. Basically, the spike doe not have
enough surface area to keep the temperature at surrounding points close to zero,
hence the solution fails to be continuous in the closure
. More details about
the subject are available in Helms "Introduction to potential theory" (1975).
#4
Consider v(x; t) = k(x; t)u(x=t;􀀀1=t); t > 0. To see that this solved the heat
equation, let us compute its partial derivatives, using chain rule and product
rule:
vt(x; t) = kt(x; t)u(x
t
;􀀀
1
t
) + k(x; t)(
1
t2 )ut(x
t
;􀀀
1
t
) 􀀀 k(x; t)( x
t2 )ux(x
t
;􀀀
1
t
)
vx(x; t) = kx(x; t)u(x
t
;􀀀
1
t
) + k(x; t)(
1
t
)ux(x
t
;􀀀
1
t
)
vxx(x; t) = kxx(x; t)u(x
t
;􀀀
1
t
) + 2kx(x; t)(
1
t
)ux(x
t
;􀀀
1
t
) + k(x; t)(
1
t2 )uxx(x
t
;􀀀
1
t
)
Now using the fact that kt = kxx; ut = uxx, we can simplify this to
vt 􀀀 vxx = 􀀀
1
t
xk(x; t)
t
+ 2kx(x; t))

ux(x
t
;􀀀
1
t
)
Using the de nition of k(x; t), we can easily see that kx(x; t) = 􀀀(x=2t)k(x; t),
so that xk(x; t) + 2tkx(x; t) = 0. This con rms the claim that v(x; t) solves
the heat equation. Since u(x; t) was de ned for all t < 0, s = 􀀀1=t covers the
domain (0;1). In other words, v(x; t) is a solution for all t > 0.
3

IASTATE Math 266 Exams Solutions

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