Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 10 November 30, 2009

5.85. Whenver we can, we use the results of Exercise 5.65 above (not the method of integration of the function times the joint density!!). The point of this problem is that some approaches to doing things can be much simpler than others!

(a) Y_{1}, Y_{2} both have exponential distributions with β = 1, so E(Y_{1}) = E(Y_{2}) = 1.

(b) Similarly, V (Y_{1}) = V (Y_{2}) = 1.

(c) E(Y_{1} − Y_{2}) = E(Y_{1}) − E(Y_{2}) = 0.

(d) This is the only part of this question for which no obvious “shortcut” presents itself. We must compute:

E(Y_{1}Y_{2}) = ∫_{0}^{∞} ∫_{0}^{∞} y_{1}y_{2} (1 − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})) e^{−y_{1}−y_{2}} dy_{2} dy_{1}.

The best approach is to multiply out and split the integral into a sum of several terms where each is a product g(y_{1})h(y_{2}):

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

Integrating by parts, we have ∫_{0}^{∞} xe^{−x} dx = 1 and ∫_{0}^{∞} x(1 − e^{−x})e^{−x} dx = \frac{1}{2}. So E(Y_{1}Y_{2}) = 1 − \frac{α}{4}.

(e) This requires the concept of covariance: V (Y_{1} −Y_{2}) = V (Y_{1})−2Cov(Y_{1}, Y_{2})+V (Y_{2}) = 1+ \frac{α}{2}. The only way to estimate an interval is to use Tchebysheff’s theorem.

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}.

5.109. Using Theorem 5.12,

V (30Y_{1} + 25Y_{2}) = 900V (Y_{1}) + 1500Cov(Y_{1}, Y_{2}) + 625V (Y_{2}).

We compute:

E(Y_{1}) = ∫_{0}^{1} ∫_{0}^{1} y_{1}(y_{1} + y_{2}) dy_{1} dy_{2} = \frac{7}{12}
E(Y_{2}) = \frac{7}{12} by symmetry
E(Y_{1}^{2}) = ∫_{0}^{1} ∫_{0}^{1} y_{1}^{2}(y_{1} + y_{2}) dy_{1} dy_{2} = \frac{5}{12}
E(Y_{1}Y_{2}) = ∫_{0}^{1} ∫_{0}^{1} y_{1}y_{2}(y_{1} + y_{2}) dy_{1} dy_{2} = \frac{1}{3}
E(Y_{2}^{2}) = \frac{5}{12} by symmetry .

Then

V (Y_{i}) = E(Y_{i}^{2}) − E(Y_{i})^{2} = \frac{5}{12} − (\frac{7}{12})^{2} = \frac{11}{144}.

Similarly,

Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) − E(Y_{1})E(Y_{2}) = \frac{1}{3} − \frac{49}{144} = \frac{−1}{144}.

So finally,

V (30Y_{1} + 25Y_{2}) = \frac{1525 · 11 − 1500}{144} = \frac{15275}{144} ≐ 106.1.

By Tchebysheff’s Theorem, and interval containing 30Y_{1} + 25Y_{2} at least 75% of the time would be [μ − 2σ, μ + 2σ] ≐ [11.48, 52.68].

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!

6.7. (a) Let Z be a standard normal random variable and U = Z^{2}. Then for any given u > 0 U ≤ u ⇔ −\sqrt{u} ≤ Z ≤ \sqrt{u}. So

F_{U}(u) = P (U ≤ u) = P (−\sqrt{u} < Z ≤ \sqrt{u}) = ∫_{−\sqrt{u}}^{√\overline{u}} \frac{1}{\sqrt{2π}} e^{−y^{2}/2} dy.

We will now differentiate this function of u to find the density function for U, using the Fundamental Theorem of Calculus:

\frac{d}{du} ∫_{−\sqrt{u}}^{√\overline{u}} \frac{1}{\sqrt{2π}} e^{−y^{2}/2} dy = \frac{u^{−1/2}e^{−u/2}}{2\sqrt{2π}} + \frac{u^{−1/2}e^{−u/2}}{2\sqrt{2π}}

= \frac{u^{−1/2}e^{−u/2}}{\sqrt{π} · 2^{1/2}} .

(b) This is the form of a Γ density function with α = 1/2 and β = 2 because Γ(1/2) = \sqrt{π}.

(c) Another name for this is the χ^{2} distribution with one degree of freedom.

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).

6.40. This is a χ^{2} distribution with 2 d.f. by Theorem 6.4.

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.