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M442 Practice Problems for Final Exam
Calculators will be allowed on the final. Unjustified answers will not receive credit.
- For the ODE system
y′ 1 = y 2 y′ 2 = 8y 1 − 2 y^31 ,
identify each equilibrium point at the linearized level from the following list: orbit, saddle, stable node, unstable node, improper stable node, improper unstable node, stable spiral, unstable spiral.
- Find all equilibrium points for the system
y 1 ′ = y 2 y 2 ′ = y 3 y 3 ′ = y 1 − 1 ,
and determine whether or not each is stable.
- Solve the ODE system
y 1 ′ = y 1 − 2 y 2 ; y 1 (0) = 0 y 2 ′ = 3y 1 + y 2 ; y 2 (0) = 1.
- For the Lotka-Volterra model
x′^ = ax − bxy y′^ = − ry + cxy,
suppose the birth and death rates a and r are known respectively to be .01 and .02, and that (20, 10) is known to be an equilibrium point. Find values for the parameters b and c.
5a. Show that Pr{A ∩ Bc} = Pr{A} − Pr{A ∩ B}.
5b. Use (5a) to establish
Pr{Ac^ ∩ Bc} = 1 − Pr{A} − Pr{B} + Pr{A ∩ B}.
- A bin of 5 electrical components is known to contain exactly 2 that are defective. If the components are tested one at a time in random order, until both defectives are discovered, find the expected number of tests that are made. Keep in mind that you don’t necessarily have to test the defectives. For example, if you draw three working components in a row, you can conclude that the remaining two are defective. Also, compute the variance of the number of tests.
- The covariance of two random variables X and Y is defined to be
Cov(X, Y ) := E[(X − E[X])(Y − E[Y ])].
Show that Cov(X, Y ) = E[XY ] − E[X]E[Y ].
- Show that for any sequence of random variables X 1 , X 2 , ..., Xn (not necessarily indepen- dent or with the same distributions),
Var(
∑^ n
k=
Xk) =
∑^ n
k=
Var(Xk) +
j 6 =k
Cov(Xj , Xk).
- Show that for any two discrete random variables X and Y
E[XY ] = E[Y E[X|Y ]].
- A laboratory blood test is 95 percent effective in detecting a certain disease when the disease is present. However, the test also yields a “false positive” result for 1% of healthy people tested. If .5% of the population actually has the disease, what is the probability a person has the disease given that the test result is positive.
- Suppose a fair coin is flipped until it lands heads on three consecutive flips. What is the expected number of flips required?
- Suppose U 1 and U 2 are both uniform random variables on [0, 1]. Determine whether or not the random variable X = U 1 + U 2 is a uniform random variable on [0, 2].
- Prove that for any random variable X, and for any constants k > 0 and r > 0,
Pr{|X| ≥ k} ≤
kr^
E[|X|r].
Hint. Use Markov’s inequality: Pr{Y ≥ a} ≤ E[ aY ].
14a. Let U 1 and U 2 denote uniform random variables on [− 1 , 1], and note that pairs (U 1 , U 2 ) are contained in a square of area 4 centered at (0, 0). Explain how simulation can be used in this framework to estimate a value for π.
14b. Determine the number of simulations in (14a) that would be required to ensure that the probability of your error’s exceeding .01 is smaller than .05.
Solutions.
- The equilibrium points solve the system of algebraic equations
y 2 = 0 8 y 1 − 2 y^31 = 0,
By convention, we choose v 1 = 1, and the eigenvector is
~v =
i
√ 6 2
Likewise, the eigenvector associated with r = 1 + i
6 is
~v =
−i
√ 6 2
(In fact, it is easy to show that the eigenvectors for complex conjugate eigenvalues are always complex conjugates.) The general solution for this equation is
~y(t) = C 1
i
√ 6 2
e(1−i
√ 6)t (^) + C 2
−i
√ 6 2
e(1+i
√ 6)t.
Before evaluating the initial conditions, let’s recall that we read this as two equations
y 1 (t) = C 1 e(1−i
√ 6)t (^) + C 2 e (1+i √ 6)t
y 2 (t) = C 1 i
e(1−i
√6)t − C 2 i
e(1+i
√6)t .
In order to find C 1 and C 2 , we observe from the initial conditions that
0 = C 1 + C 2 ⇒ C 1 = −C 2
1 = C 1 i
− C 2 i
⇒ C 1 i
6 = 1 ⇒ C 1 = −
i √ 6
We conclude
~y(t) = −
i √ 6
i
√ 6 2
e(1−i
√ 6)t (^) + √i 6
−i
√ 6 2
e(1+i
√ 6)t.
In order to write this solution entirely in terms of sine and cosine, we use Euler’s formula
eiθ^ = cos θ + i sin θ.
Here,
y 1 (t) = −
i √ 6
et(cos(
6 t) − i sin(
6 t)) +
i √ 6
et(cos(
6 t) + i sin(
6 t))
et^ sin(
6 t).
Similarly,
y 2 (t) =
et(cos(
6 t) − i sin(
6 t)) +
et(cos(
6 t) + i sin(
6 t))
= et^ cos(
6 t).
- By definition, the equilibrium point (20, 10) must solve
0 = .01(20) − b(20)(10) 0 = − .02(10) + c(20)(10).
Solving these equations for b and c, we conclude
b =. 001 c =. 001.
5a. See homework practice problems.
5b. Replacing A in (5a) with Ac, we have
Pr{Ac^ ∩ Bc} = Pr{Ac} − Pr{Ac^ ∩ B}.
Clearly Pr{Ac} = 1 − Pr{A}, and if we again use (5a) with the roles of A and B switched, we have Pr{Ac^ ∩ B} = Pr{B} − Pr{A ∩ B}.
Combining these observations, we obtain the conclusion.
- Let the random variable N denote the number of tests that have to be made. Then the expected value is computed from the relation from class E[N] =
n=2 nPr(N^ =^ n), where a single test cannot possibly be conclusive, and we would never require 5. The probability Pr(N = 2), for example, is the probability that one of two defectives is drawn out of five components (2/5) times the probability that a single defective is drawn out of the four remaining components (1/4). For N = 2 and N = 3, the probabilities are (G=”good”, D=”defective”) DGD, GDD, GGG, and respectively DGG, GDG, GGD.
E[N] = 2 ·
Alternatively (and as is recommended), we could compute the probability that 4 draws are required as 1 − Pr{Fewer than four draws required}. The variance is now straightforward:
Var[N] = 1. 52 ·
Cov(X, Y ) = E[(X − E[X])(Y − E[Y ])] = E[XY − E[X]Y − XE[Y ] + E[X]E[Y ]] = E[XY ] − E[X]E[Y ].
- Since U 1 and U 2 are both uniform on [0, 1], we know that for all a, b ∈ [0, 1], a ≤ b, there holds Pr{a ≤ Uk ≤ b} = b − a, k = 1, 2.
Now let c, d ∈ [0, 2], c ≤ d. The question becomes, is it true that
Pr{c ≤ X ≤ d} =
d − c 2
(Keep in mind that if c = 0 and d = 2 the probability must be 1.) Intuitively, we think that X should not be uniform, because there are many more ways to get a sum such as 1 than a sum such as 0. In order to find a counterexample to (1), consider a particular pair of values, say c = 12 and d = 32. We have
Pr{
≤ X ≤
} = Pr{
≤ U 1 + U 2 ≤
≥ Pr{
≤ U 1 ≤ 1 }Pr{U 2 ≤
} + Pr{U 1 ≤
}Pr{
≤ U 2 ≤ 1 }
≤ U 1 ≤
}Pr{
≤ U 2 ≤
If X were uniformly distributed on [0, 2] we would have
Pr{
≤ X ≤
- Consider the positive random variable
Y := |X|r.
According to Markov’s inequality, we must have
Pr{Y ≥ kr^ } ≤
kr^
E[Y ],
or
Pr{|X|r^ ≥ kr^ } ≤
kr^
E[|X|r].
Finally, for k > 0 and r > 0,
Pr{|X|r^ ≥ kr} = Pr{|X| ≥ k},
and this proves the statement.
14a. First, observe that we can introduce π into this problem by considering the area of the inscribed circle; i.e., the circle centered at the origin with radius 1. The area of this circle is π, and so the probability of a randomly generated point (U 1 , U 2 ) appearing in this circle is π
- Let^ X^ be defined as follows:
X =
1 (U 1 , U 2 ) in circle 0 otherwise
Now let X 1 , X 2 ,... , Xn denote n observations of X. We have
π 4
n
∑^ n
k=
Xk.
14b. Using the inequality from the Weak Law of Large Numbers, we have
Pr{|
n
∑^ n
k=
Xk −
π 4
| ≥ k} ≤
σ^2 nk^2
In this case, we take k = .01, and we must find n sufficiently large so that
1 n(.01)^2
≤. 05 ⇒ n ≥
(.01)^2 (.05)
