Mathematics 375 – Probability and Statistics 1 Solutions – Problem Set 10 December 10, 2003

General Comment: A lot of work can be saved on this assignment by using results of previous problems. Also, in general it is not necessary, or even an especially good idea in many cases, to compute the marginal densities and use those to get variances, etc. Using the joint density is usually easier.

57. 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).

b) We have f_{2}(y_{2}) = e^{−y_{2}} by the symmetry of the joint density.

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.

62. Since we are given that Y_{1} has a binomial distribution, E(Y_{1}) = np = 2/3, and V (Y_{1}) = npq = 4/9. Since the probability function table is symmetric, it follows that Y_{2} has the same distribution, and E(Y_{1} − Y_{2}) = E(Y_{1}) − E(Y_{2}) = 0.

66. Here Y_{1} and Y_{2} have a uniform joint distribution over the triangle with vertices (0, 0), (2, 0), (0, 1). So

E(Y_{1}) = ∫_{0}^{2} ∫_{0}^{y_{1}/2} y_{1} dy_{2} dy_{1} = 4/3

and

E(Y_{2}) = ∫_{0}^{2} ∫_{0}^{y_{1}/2} y_{2} dy_{2} dy_{1} = 1/3

b) Also,

E(Y_{1}^{2}) = ∫_{0}^{2} ∫_{0}^{y_{1}/2} y_{1}^{2} dy_{2} dy_{1} = 2,

so V (Y_{1}) = E(Y_{1}^{2}) − (E(Y_{1}))^{2} = 2 − (4/3)^{2} = 2/9. Similarly,

E(Y_{2}^{2}) = ∫_{0}^{2} ∫_{0}^{y_{1}/2} y_{2}^{2} dy_{2} dy_{1} = 1/6,

so V (Y_{1}) = E(Y_{1}^{2}) − (E(Y_{1}))^{2} = 1/6 − (1/3)^{2} = 1/18.

c) By part a, E(Y_{1} − Y_{2}) = E(Y_{1}) − E(Y_{2}) = 4/3 − 1/3 = 1.

d) Probably the easiest way to compute V (Y_{1} − Y_{2}) is to compute

E((Y_{1} − Y_{2})^{2}) = ∫_{0}^{2} ∫_{0}^{y_{1}/2}(y_{1} − y_{2})^{2} dy_{2} dy_{1} = 7/6

Then V (Y_{1} − Y_{2}) = 7/6 − (1)^{2} = 1/6. The last question in part d is asking for an interval which is guaranteed to contain Y_{1} − Y_{2} with high probability. By the empirical rule or Tchebysheff’s theorem, one such interval would be (μ−2σ, μ+2σ) = (1−2/\sqrt{6}, 1+2/\sqrt{6}).

69. For this problem and several later ones as well, we will use the fact from 5.53 (on Problem Set 9) that with this joint density, Y_{1} and Y_{2} are independent, and moreover,

• Y_{1} has a gamma distribution with α = 2 and β = 2 (i.e. f_{1}(y_{1}) = \frac{1}{4}y_{1}e^{−y_{1}/2}).

• Y_{2} has an exponential distribution with β = 2 (i.e. f_{2}(y_{2}) = \frac{1}{2}e^{−y_{2}/2}). Then by independence,

E(Y_{2}/Y_{1}) = E(Y_{2})E(1/Y_{1}) = 2E(1/Y_{1})

In addition, letting u = y_{1}/2,

E(1/Y_{1}) = ∫_{0}^{∞} \frac{1}{y_{1}} \frac{1}{4}y_{1}e^{−y_{1}/2} dy_{1} = \frac{1}{2} ∫_{0}^{∞} e^{−u} du = \frac{1}{2}Γ(1) = \frac{1}{2}

So E(Y_{2}/Y_{1}) = 2/2 = 1.

70. Instead of splitting up the expected value or computing marginal densities, the fastest way to compute this is simply to take

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

73. Using the results shown in problem 57 above saves a lot of work here! a) Since Y_{1} and Y_{2} are both exponentially distributed with β = 1, E(Y_{1}) = E(Y_{2}) = 1. b) For the same reason, V (Y_{1}) = V (Y_{2}) = 1. (We’re using the general formulas for exponential random variables from Theorem 4.10 or the table inside the back cover of the text.) c)

E(Y_{1} − Y_{2}) = 1 − 1 = 0 by part a. d) Here we do have to work a bit. But note that we can try to split up the double integral into terms where the integrand is a product of functions of y_{1} and y_{2} alone. That allows us to evaluate a double integral as product of two one-variable integrals:

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

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

= Γ(2)Γ(2) − α(Γ(2) − Γ(2)/2)(Γ(2) − Γ(2)/2)
= 1 − α/4

e) Using parts b and d, and a formula from later, we have

V (Y_{1} − Y_{2}) = V (Y_{1}) + V (Y_{2}) − 2Cov(Y_{1}, Y_{2}) = 2 − 2(E(Y_{1}Y_{2}) − E(Y_{1})E(Y_{2})) = 2 + α/2

The last question is similar to the last part of 66. The interval (μ−2σ, μ+2σ) contains Y_{1}− Y_{2} with probability ·.95 by the empirical rule, or with probability ≥ .75 by Tchebysheff’s theorem.

83. 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}.

85. a) By the results of 73a,d, we have

Cov(Y_{1}Y_{2}) = E(Y_{1}Y_{2}) − E(Y_{1})E(Y_{2}) = (1 − α/4) − 1 = −α/4

b) From this formula, it is clear that Cov(Y_{1}, Y_{2}) = 0 ⇔ α = 0. c) Since σ_{1} = σ_{2} = 1 it is also true that ρ = 0 ⇔ α = 0. If Y_{1}, Y_{2} are independent, then Cov(Y_{1}, Y_{2}) = 0 ⇒ ρ = 0. On the other hand, if ρ = 0, then α = 0, and hence Y_{1}, Y_{2} are independent by the result of 57, part c. Note that it is not true in general that Cov(Y_{1}, Y_{2}) = 0 ⇒ Y_{1}, Y_{2} independent, though. Example 5.24 in the text gives a counterexample.

91. We begin by computing as follows:

\begin{aligned}E(Y_{1}) & = ∫_{0}^{1} ∫_{0}^{1−y_{1}} y_{1} dy_{2} dy_{1} = 1/3 \\ E(Y_{2}) & = ∫_{0}^{1} ∫_{0}^{1−y_{1}} y_{2} dy_{2} dy_{1} = 1/3 \\ E(Y_{1}^{2}) & = ∫_{0}^{1} ∫_{0}^{1−y_{1}} y_{1}^{2} dy_{2} dy_{1} = 1/6 \\ E(Y_{1}Y_{2}) & = ∫_{0}^{1} ∫_{0}^{1−y_{1}} y_{1}y_{2} dy_{2} dy_{1} = 1/12 \\ E(Y_{2}^{2}) & = ∫_{0}^{1} ∫_{0}^{1−y_{1}} y_{2}^{2} dy_{2} dy_{1} = 1/6 \end{aligned}

Then we can put these pieces together to get all the desired information for the problem. E(Y_{i}) are already done, so E(Y_{1} + Y_{2}) = 1/3 + 1/3 = 2/3. Then

V (Y_{1} + Y_{2}) = V (Y_{1}) + V (Y_{2}) + 2Cov(Y_{1}, Y_{2}) = 2 · (\frac{1}{6} − (\frac{1}{3})^{2}) + 2 · (\frac{1}{12} − (\frac{1}{3})^{2}) = \frac{1}{18}

94. Once again, using the results from previous problems simplifies life immensely in this problem! a) From 5.53 on Problem Set 9, we know Y_{1} and Y_{2} are independent, and moreover,

• Y_{1} has a gamma distribution with α = 2 and β = 2.

• Y_{2} has an exponential distribution with β = 2. So a) E(50 + 2Y_{1} + 4Y_{2}) = 50 + 2 · 4 + 4 · 2 = 66. Then for b), noting that the variance of a constant is zero,

V (50 + 2Y_{1} + 4Y_{2}) = 4V (Y_{1}) + 16V (Y_{2}) + 16Cov(Y_{1}, Y_{2}) = 4 · 8 + 16 · 4 + 16 · 0 = 96

(Recall Y_{1}, Y_{2} independent ⇒ Cov(Y_{1}, Y_{2}) = 0.)

99. This is a multinomial situation:

a) \frac{6!}{3!2!1!}(1/3)^{6}

b) Y_{1} is binomial with p = 1/3, so E(Y_{1}) = n/3 and V (Y_{1}) = npq = 2n/9.

c) Cov(Y_{2}, Y_{3}) = −n(1/3)(1/3) = −n/9

d) E(Y_{2} − Y_{3}) = 0, and

V (Y_{2} − Y_{3}) = V (Y_{2}) + V (Y_{3}) − 2Cov(Y_{2}, Y_{3}) = 2n/9 + 2n/9 − 2(−n/9) = 2n/3

103. Also multinomial. The probability is

\frac{4!}{2!1!1!}(.73)^{2}(.2)^{1}(.07)^{1}