Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 9 November 12, 2009

5.25. (a) The marginal densities are

f_{1}(y_{1}) = ∫_{0}^{∞} e^{−y_{1}−y_{2}} dy_{2} = −e^{−y_{1}−y_{2}} |_{0}^{∞} = e^{−y_{1}}

if y_{1} > 0 and zero otherwise. Similarly,

f_{2}(y_{2}) = ∫_{0}^{∞} e^{−y_{1}−y_{2}} dy_{1} = −e^{−y_{1}−y_{2}} |_{0}^{∞} = e^{−y_{2}}

if y_{2} > 0 and zero otherwise. These are both exponential densities with β = 1.

(b) P (1 < Y_{i} < 2.5) = ∫_{1}^{2.5} e^{−y_{i}} dy_{i} = e^{−1} − e^{−2.5} for i = 1, 2.

(c) The conditional density is only defined for y_{2} > 0 (since we must have f_{2}(y_{2}) > 0 in Definition 5.7).

(d) The conditional density f(y_{1}|y_{2}) = \frac{e^{−y_{1}−y_{2}}}{e^{−y_{2}}} = e^{−y_{1}}.

(e) Similarly, f(y_{1}|y_{2}) = e^{−y_{2}}.

(f) The conditional and marginal densities for both variables are equal.

(g) The marginal and conditional probabilities in any interval will also be equal.

5.26. (a) The marginal densities are

f_{1}(y_{1}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{2} = 2y_{1}, and similarly
f_{2}(y_{2}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{1} = 2y_{2}.

Looking ahead to the next section, note that these satisfy the criterion of Theorem 5.4. So Y_{1} and Y_{2} are independent.

(b) To compute the denominator in the conditional probability, we can use the marginal density from part a:

P (Y_{1} ≤ 1/2|Y_{2} ≥ 3/4) = \frac{P (Y_{1} ≤ 1/2, Y_{2} ≥ 3/4)}{P (Y_{2} ≥ 3/4)}

= \frac{∫_{0}^{1/2} ∫_{3/4}^{1} 4y_{1}y_{2} dy_{2} dy_{1}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= \frac{∫_{0}^{1/2} 2y_{1} dy_{1} ∫_{3/4}^{1} 2y_{2} dy_{2}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= ∫_{0}^{1/2} 2y_{1} dy_{1}

= 1/4.

(The cancellation works this way because of independence.)

5.54. No – Y_{1}, Y_{2} cannot be independent when the interval of y_{2} values where the density is nonzero depends on y_{1}. (You can also see this directly by computing f_{1}(y_{1}) and f_{2}(y_{2}). The region where the joint density is nonzero is the triangle with vertices at (0, 0), (2, 0), (2, 1). So

f_{1}(y_{1}) = ∫_{0}^{y_{1}/2} 1 dy_{2} = \frac{y_{1}}{2}

(when 0 ≤ y_{1} ≤ 2; 0 otherwise). Similarly,

f_{2}(y_{2}) = ∫_{2y_{2}}^{2} 1 dy_{1} = 2 − 2y_{2}

(when 0 ≤ y_{2} ≤ 1; 0 otherwise). The product f_{1}(y_{1})f_{2}(y_{2}) is not the same as f(y_{1}, y_{2}).)

5.59. No – same reason as 5.54. The marginal densities are

f_{1}(y_{1}) = ∫_{0}^{y_{1}} e^{−y_{1}} dy_{2} = y_{1}e^{−y_{1}}

and

f_{2}(y_{2}) = ∫_{y_{2}}^{∞} e^{−y_{1}} dy_{1} = e^{−y_{2}}.

Note that f(y_{1}, y_{2}) ≠ f_{1}(y_{1})f_{2}(y_{2}).

5.65. (a) The marginal density of Y_{1} is computed as follows. Note that we try to split up the integral so that the integrand is a product of functions of y_{1} and y_{2} alone. That allows factors depending only on y_{1} to be taken outside the integral sign:

f_{1}(y_{1}) = ∫_{0}^{∞} (1 − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})) e^{−y_{1}−y_{2}} dy_{2}
= e^{−y_{1}} ∫_{0}^{∞} e^{−y_{2}} dy_{2} − α(1 − 2e^{−y_{1}})e^{−y_{1}} ∫_{0}^{∞}(1 − 2e^{−y_{2}})e^{−y_{2}} dy_{2}
= e^{−y_{1}} ∫_{0}^{∞} e^{−y_{2}} dy_{2} − −α(1 − 2e^{−y_{1}})e^{−y_{1}} · (∫_{0}^{∞} e^{−y_{2}} dy_{2} − ∫_{0}^{∞} 2e^{−2y_{2}} dy_{2})
= e^{−y_{1}} · 1 − α(1 − 2e^{−y_{1}})e^{−y_{1}}(1 − 1)
= e^{−y_{1}}

(since the integrals in the next to last line are all integrals of exponential densities over the interval where they are nonzero, hence each equals 1). Hence Y_{1} is exponentially distributed with mean β = 1.

(b) We have f_{2}(y_{2}) = e^{−y_{2}} by the symmetry of the joint density. So Y_{2} is also exponential with mean β = 1.

(c) By a general result (Theorem 5.4), Y_{1} and Y_{2} are independent if and only if f(y_{1}, y_{2}) = f_{1}(y_{1})f_{2}(y_{2}). Here, this is true if and only if

e^{−y_{1}−y_{2}} = (1 − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})) e^{−y_{1}−y_{2}} = e^{−y_{1}−y_{2}} − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})e^{−y_{1}−y_{2}}

which is true if and only if α = 0.

5.76. This can be done most easily by using the marginal densities for the Y_{i} (although there are several other valid methods too). Note that f_{1}(y_{1}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{2} = 2y_{1} if 0 ≤ y_{1} ≤ 1, and zero otherwise. Similarly for f_{2}(y_{2}) by the symmetry of the joint density.

(a) E(Y_{1}) = ∫_{0}^{1} y_{1}f_{1}(y_{1}) dy_{1} = ∫_{0}^{1} 2y_{1}^{2} dy_{1} = \frac{2}{3}.

(b) Then the variance is:

V (Y_{1}) = E(Y_{1}^{2}) − (\frac{2}{3})^{2} = ∫_{0}^{1} 2y_{1}^{3} dy_{1} − \frac{4}{9} = \frac{1}{2} − \frac{4}{9} = \frac{1}{18}.

(c) By linearity, E(Y_{1} − 3Y_{2}) = E(Y_{1}) − 3E(Y_{2}) = \frac{2}{3} − 2 = \frac{−4}{3}. This can also be computed, for instance, as the integral:

∫_{0}^{1} ∫_{0}^{1}(y_{1} − 3y_{2}) · 4y_{1}y_{2} dy_{2} dy_{1}.

5.79. To integrate over this region, is probably easiest to integrate with respect to y_{1} first:

E(Y_{1}Y_{2}) = ∫_{0}^{1} ∫_{y_{2}−1}^{−y_{2}+1} y_{1}y_{2} dy_{1} dy_{2}

= ∫_{0}^{1} y_{2} \frac{(−y_{2} + 1)^{2} − (y_{2} − 1)^{2}}{2} dy_{1}

= 0,

since everything cancels out in the integrand.

5.91. From Exercise 5.76 on the last problem set, E(Y_{1}) = E(Y_{2}) = \frac{2}{3}. Then

E(Y_{1}Y_{2}) = ∫_{0}^{1} ∫_{0}^{1} 4y_{1}^{2}y_{2}^{2} dy_{2} dy_{1} = \frac{4}{9}.

Hence Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) − E(Y_{1})E(Y_{2}) = 0. This is not a surprise since we know Y_{1} and Y_{2} are independent (see solution to Exercise 5.26 on the last problem set).

5.94. “Uncorrelated” means that it is given that Cov(Y_{1}, Y_{2}) = 0.

(a) By the bilinearity of the covariance (part c of Theorem 5.12),

Cov(U_{1}, U_{2}) = Cov(Y_{1} + Y_{2}, Y_{1} − Y_{2})
= Cov(Y_{1}, Y_{1}) + Cov(Y_{2}, Y_{1}) − Cov(Y_{1}, Y_{2}) − Cov(Y_{2}, Y_{2})
= V (Y_{1}) + 0 − 0 − V (Y_{2})
= V (Y_{1}) − V (Y_{2}).

(b) ρ = \frac{V (Y_{1})−V (Y_{2})}{V (Y_{1})V (Y_{2})}.

(c) It is possible for Cov(U_{1}, U_{2}) to be zero. This happens if and only if V (Y_{1}) = V (Y_{2}).

5.97. (a) Cov(Y_{1}, Y_{1}) = E(Y_{1}^{2}) − (E(Y_{1}))^{2} = V (Y_{1}) = 2 by the given information. b,c,d) If Cov(Y_{1}, Y_{2}) = 7, the correlation coefficient ρ = \frac{7}{σ_{1}σ_{2}} = \frac{7}{\sqrt{2}·2\sqrt{2}} = 7/4. That is not possible because −1 ≤ ρ ≤ 1 in all cases. The largest and smallest values for Cov(Y_{1}, Y_{2}) are −4, 4, corresponding to cases where Y_{2} is a decreasing or increasing linear function of Y_{1}.

6.1. (a) Letting U_{1} = 2Y − 1, U_{1} ≤ u ⇔ Y ≤ (u + 1)/2. Then the cdf for U_{1} is

F (u) = ∫_{0}^{(u+1)/2} 2(1 − y) dy = −((1 − u)/2)^{2} + 1

(for −1 ≤ u ≤ 1, and zero otherwise). To find the pdf for U, we differentiate

f(u) = F′(u) = (1 − u)/2 if − 1 ≤ u ≤ 1; 0otherwise

(b) Derivation for U_{2} is very similar to part (a). Answer:

f(u) = (1 + u)/2 if − 1 ≤ u ≤ 1, 0 otherwise .

(c) U_{3} = Y^{2}. We have Y^{2} ≤ u ⇔= \sqrt{u} ≤ Y ≤ \sqrt{u}. Only positive values of y occur with nonzero densities, though, so the cdf for U_{3} is

F (u) = ∫_{0}^{√\overline{u}} 2(1 − y) dy = −(1 − \sqrt{u})^{2}

(for 0 ≤ u ≤ 1, and zero otherwise). The pdf for U_{3} is f(u) = F′(u) = \frac{1−\sqrt{u}}{\sqrt{u}} (for 0 < u ≤ 1, zero otherwise).

(d) Using these pdf’s:

E(U_{1}) = ∫_{−1}^{1} u(1 − u)/2 du = −1/3
E(U_{2}) = ∫_{−1}^{1} u(1 + u)/2 du = 1/3
E(U_{3}) = ∫_{0}^{1} u · \frac{1 − \sqrt{u}}{\sqrt{u}} du = ∫_{0}^{1} \sqrt{u} − u du = 1/6

(e) Using the methods from Chapter 4: We know Y has a Beta distribution with α = 1, β = 2 from the form of the pdf for Y. So E(Y) = 1/3 and V (Y) = 1/18. Hence E(U_{1}) = 2(1/3) − 1 = −1/3, E(U_{2}) = 1 − 2(1/3) = 1/3, and E(U_{3}) = E(Y^{2}) = V (Y) + E(Y)^{2} = 1/18 + (1/3)^{2} = 1/6. Everything checks!

6.8. (a) The inequality U < u is equivalent to 1 − Y < u so Y > 1 − u. Now,

F_{U}(u) = P (1 − Y < u) = ∫_{1−u}^{1} \frac{1}{B(α, β)}y^{α−1}(1 − y)^{β−1} dy.

To see what is happening here, make the substitution z = 1 − y, so y = 1 − z, and dz = −dy. We get

F_{U}(u) = ∫_{u}^{0} \frac{1}{B(α, β)}(1 − z)^{α−1}z^{β−1}(−dz) = ∫_{0}^{u} \frac{1}{B(α, β)}(1 − z)^{α−1}z^{β−1} dz.

Differentiating this with respect to u, using the FTC, we get:

f_{U}(u) = \frac{1}{B(α, β)}u^{β−1}(1 − u)^{α−1}

(b) Since the beta function is symmetric in its arguments, this is the form for the density function of a variable with a beta distribution with parameters β and α (i.e. reversed from the parameters of Y).

(c) E(U) = \frac{β}{α+β} = 1 − \frac{α}{α+β} = 1 − E(Y) as we expect from our general facts about linearity of expected values.

(d) We can see V (U) = V (Y) because of the symmetry of \frac{αβ}{(α+β)^{2}(α+β+1)} in α, β.

6.38. We have

m_{U}(t) = E(e^{tU})
= E(e^{t(a_{1}Y_{1}+a_{2}Y_{2})})
= E(e^{(a_{1}t)Y_{1}} · e^{(a_{2}t)Y_{2}})
= E(e^{(a_{1}t)Y_{1}}) · E(e^{(a_{2}t)Y_{2}}) by Theorem 6.2 (independence)
= m_{Y_{1}}(a_{1}t) · m_{Y_{2}}(a_{2}t),

which is what we wanted to show.

6.39. By the preceding exercise, if Y_{i} are the lives of the two components and U = \frac{1}{2}Y_{1} + \frac{1}{2}Y_{2} is the average life, then

m_{U}(t) = m_{Y_{1}}(\frac{1}{2}t) · m_{Y_{2}}(\frac{1}{2}t) = \frac{1}{(1 − \frac{t}{2})^{2}} .

By the uniqueness theorem, this implies that U has a Γ distribution with α = 2 and β = \frac{1}{2}. The density function is the Γ density:

f(u) = \frac{1}{Γ(2)(1/2)^{2}} ue^{− \frac{u}{1/2}} = 4ue^{−2u}

(for u > 0, and 0 otherwise).

6.41. By Theorem 6.3, U has a normal distribution with mean μ_{U} = (a_{1} + ⋯ + a_{n})μ and σ_{U}^{2} = (a_{1}^{2} + ⋯ + a_{n}^{2})σ^{2}. The density function for U is the usual normal density:

f_{U}(u) = \frac{1}{\sqrt{2πσ_{U}^{2}}} e^{\frac{(u−μ_{U})^{2}}{2σ^{2}_{U}}} .

6.42. Using the result from Exercise 6.41, U = Y_{2}−Y_{1} is normally distributed with μ_{U} = −1000 and σ_{U}^{2} = 300^{2} + 400^{2} = 500^{2}. The probability that the elevator is overloaded is found by standardizing and using the table for the standard normal

P (Y_{2} − Y_{1} > 0) = P (\frac{(Y_{2} − Y_{1}) + 1000}{500} > 2) = P (Z > 2) = .0228.

6.46. Using Exercise 6.38 (with a_{2} = 0) we have

m_{W} (t) = m_{Y} (\frac{2}{β} t) = \frac{1}{(1 − β \frac{2}{β} t)^{n/2}} = \frac{1}{(1 − 2t)^{n/2}} .

By the uniqueness theorem, this implies that W has a χ^{2} distribution with n d.f.