Mathematics 375 – Probability Theory Solutions for Problem Set 9 December 2, 2011

5.48. No Y_{1} and Y_{2} are not independent. If they were, then for instance P (Y_{1} = 0, Y_{2} = 0) = P (Y_{1} = 0)P (Y_{2} = 0) = (.76)(.55) = .418. However, that does not match the given value.

5.51. (a) By Theorem 5.4, Y_{1} and Y_{2} are independent here. Each has an exponential distribution with mean β = 1 and the joint density is the product e^{−(y_{1}+y_{2})} = e^{−y_{1}} · e^{−y_{2}}. (b) Yes, this explains the results in 5.25.

5.55. No they are not independent. “Uniformly distributed” just means that the joint density is constant on the triangle in the figure from 5.11. But computing marginal densities we have

f_{1}(y_{1}) = \begin{cases}y_{1} + 1 & −1 ≤ y_{1} ≤ 0 \\ −y_{1} + 1 & 0 ≤ y_{1} ≤ 0 \\ 0 & otherwise. \end{cases}

(These are separate cases based on the equation of the top edge of the triangle on different intervals of y_{1}-values; they are not added together.)

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

for 0 ≤ y_{2} ≤ 1. The product of the marginal densities is not equal to the constant joint density.

5.61. Yes they are independent by Theorem 5.5: take g(y_{1}) = \frac{1}{4}y_{1}e^{−y_{1}/2} and h(y_{2}) = \frac{1}{2}e^{−y_{2}/2}. (These are the marginal densities in fact – Y_{1} has a Γ(2, 2) distribution and Y_{2} has an exponential distribution with β = 2.)

5.70. As in the Hint, let Y_{1} = bus arrival time and Y_{2} = passenger arrival time, both uniformly distributed on [0, 1]. We want

P (Y_{2} ≤ Y_{1} ≤ Y_{2} + 1/4).

This can be set up as a double integral, but the most direct way is to think in terms of areas. The area we want is what is left after two triangles are cut out of the unit square. The first triangle is the one above the line y_{2} = y_{1}, so its area is \frac{1}{2}. The second triangle is below the line y_{2} = y_{1} − \frac{1}{4}, so it has area \frac{1}{2} · (\frac{3}{4})^{2} = \frac{9}{32}. The area we want is then

1 − \frac{1}{2} − \frac{9}{32} = \frac{7}{32}.

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} − Y_{2}) = E(Y_{1}) − E(Y_{2}) = \frac{2}{3} − \frac{2}{3} = 0. This can also be computed, for instance, as the integral:

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

5.81. There are again a multitude of different correct ways to approach this. The most direct way is to compute as follows:

E(Y_{1}/Y_{2}) = ∫_{0}^{∞} ∫_{0}^{∞} \frac{y_{2}}{y_{1}} · \frac{1}{8}y_{1}e^{−(y_{1}+y_{2})/2} dy_{2}dy_{1}

= ∫_{0}^{∞} ∫_{0}^{∞} \frac{1}{8}y_{2}e^{−(y_{1}+y_{2})/2} dy_{2}dy_{1}

= \frac{1}{8} ∫_{0}^{∞} e^{−y_{1}/2} dy_{1} · ∫_{0}^{∞} y_{2}e^{−y_{2}/2} dy_{2}

= \frac{1}{8} · 2 · Γ(1) · 4 · Γ(2)

= 1.

To derive the values of the integrals, I substituted u = y_{1}/2 in the first and recognized that the value was just 2Γ(1). Similarly, the second integral is evaluated by setting t = y_{2}/2.

Another, somewhat more clever, way is as follows. Theorem 5.12 implies that Y_{1} and Y_{2} are independent. Note that independence of Y_{1} and Y_{2} implies that E(g(Y_{1})h(Y_{2})) = E(g(Y_{1}))E(h(Y_{2})) for any functions g, h (why?). Hence E(Y_{2}/Y_{1}) = E(Y_{2})E(1/Y_{1}). By examining the joint density and factoring, we can see Y_{2} has an exponential distribution with β = 2, so E(Y_{2}) = 2. We also see Y_{1} has a Γ(2, 2) distribution. By Exercise 4.111 from a previous problem set,

E(1/Y_{1}) = \frac{1}{2(2 − 1)} = \frac{1}{2}.

Therefore, E(Y_{2}/Y_{1}) = 1 as before(!)

5.92. We have

E(Y_{1}) = ∫_{0}^{1} ∫_{y_{1}}^{1} 6y_{1}(1 − y_{2}) dy_{2} dy_{1} = \frac{1}{4}

and

E(Y_{2}) = ∫_{0}^{1} ∫_{y_{1}}^{1} 6y_{2}(1 − y_{2}) dy_{2} dy_{1} = \frac{1}{2}.

Therefore

Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) − \frac{1}{8}

= ∫_{0}^{1} ∫_{y_{1}}^{1} 6y_{1}y_{2}(1 − y_{2}) dy_{2} dy_{1} − \frac{1}{8}

= \frac{3}{20} − \frac{1}{8}

= \frac{1}{40}.

Since Cov(Y_{1}, Y_{2}) ≠ 0, they are not independent.

5.93. (a) We have

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

and

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

Therefore

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

(b) In Exercise 5.55 earlier, we saw that Y_{1} and Y_{2} are not independent. (This example shows that the implication in Theorem 5.11 goes only one way.) (c) The correlation coefficient is also 0. (d) No, it is possible for two dependent random variables to have zero covariance.

5.96. (a) By the definition,

Cov(Y_{1}, Y_{2}) = E((Y_{1} − μ_{1})(Y_{2} − μ_{2})) = E((Y_{2} − μ_{2})(Y_{1} − μ_{1})) = Cov(Y_{2}, Y_{1}).

(b) Similarly,

Cov(Y_{1}, Y_{1}) = E((Y_{1} − μ_{1})^{2}) = V (Y_{1}) = σ_{1}^{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}.

5.103. By the general formulas for expected value and variance of a linear combination of random variables,

E(3Y_{1} + 4Y_{2} − 6Y_{3}) = 3E(Y_{1}) + 4E(Y_{2}) − 6E(Y_{3}) = (3)(2) + (4)(−1) − (6)(4) = −22

and

V (3Y_{1} + 4Y_{2} − 6Y_{3}) = 9V (Y_{1}) + 16V (Y_{2}) + 36V (Y_{3})
+ 24Cov(Y_{1}, Y_{2}) − 36Cov(Y_{1}, Y_{3}) − 48Cov(Y_{2}, Y_{3})

= 36 + 96 + 288 + 24 + 36 + 0 = 480.

5.105. From Exercise 5.76, we know V (Y_{i}) = \frac{1}{18}. So we have V (Y_{1} − Y_{2}) = V (Y_{1}) + V (Y_{2}) − 2Cov(Y_{1}, Y_{2}) = \frac{1}{9}.

5.114. (a) We recognize that Y_{1} has a Γ(4, 1) distribution and Y_{2} has an exponential distribution with β = 2. Hence from our general formulas we know E(Y_{1}) = 4 and E(Y_{2}) = 2. So E(U) = E(Y_{1}) − E(Y_{2}) = 2. (b) Similarly, V (Y_{1}) = 4 and V (Y_{2}) = 2, so V (U) = V (Y_{1}) + V (Y_{2}) + 2Cov(Y_{1}, Y_{2}) = 6, since independence implies the covariance is zero. (c) 0 is a bit more than one standard deviation below the mean for U. The daily profit could drop below zero, but it would not be extremely common.