Mathematics 375 – Probability and Statistics 1 Solutions – Problem Set 9 November 25, 2003

5.4/ a) It is clear from the table that p(y_{1}, y_{2}) ≥ 0 and ∑_{y_{1},y_{2}} p(y_{1}, y_{2}) = 1. Hence this is a valid probability function. b) F (1, 2) = 1 since all possible combinations of values of Y_{i} satisfy Y_{1} ≤ 1 and Y_{2} ≤ 2.

5.5/ The region where the joint density is nonzero is a triangle with corners at (0, 0), (1, 0), (1, 1). a) P (Y_{1} ≤ 1/2, Y_{2} ≤ 1/3) is found by integrating the joint density over the region shaded in this diagram:

The simplest way to set up the double integral here is to integrate with respect to y_{1} first (we don’t need to split the region):

∫_{0}^{1/3} ∫_{y_{2}}^{1/2} 3y_{2} dy_{1} dy_{2} = \frac{23}{216} ≐ .1065

b) P (Y_{2} ≤ Y_{1}/2) is found by integrating the joint density over the region shaded in this diagram:

We get

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

5.7/ The region where the joint density is nonzero is a triangle with corners at (0, 0), (1, 1), (0, 1) (above the line y_{1} = y_{2}).

a) We must have

1 = ∫_{0}^{1} ∫_{y_{1}}^{1} k(1 − y_{2}) dy_{2} dy_{1} ⇒ k = 6.

b) P (Y_{1} ≤ 3/4, Y_{2} ≥ 1/2) is found by integrating the joint density over the region shaded in this diagram:

This is one where either way of setting up the iterated integrals leads to a case where you must split the region. For instance, setting up the integrals to integrate with respect to y_{2} first, we get:

∫_{0}^{1/2} ∫_{1/2}^{1} 6(1 − y_{2}) dy_{2} dy_{1} + ∫_{1/2}^{3/4} ∫_{y_{1}}^{1} 6(1 − y_{2}) dy_{2} dy_{1} = \frac{31}{64}.

5.13. The region where the joint density is nonzero is the infinite “wedge” in the first quadrant between the y_{1} axis and the below the line y_{1} = y_{2}. a) P (Y_{1} < 2, Y_{2} > 1 is found by integrating the density over the region shaded in this diagram:

We get

∫_{1}^{2} ∫_{1}^{y_{1}} e^{−y_{1}} dy_{2} dy_{1} = ∫_{1}^{2} ∫_{y_{2}}^{2} e^{−y_{1}} dy_{1} dy_{2}

The second form is easier (don’t need to integrate by parts!):

∫_{1}^{2} ∫_{y_{2}}^{2} e^{−y_{1}} dy_{1} dy_{2} = ∫_{1}^{2} −e^{−2} + e^{−y_{2}} dy_{2} = e^{−1} − 2e^{−2}.

b) P (Y_{1} ≥ 2Y_{2}) is found by integrating the density over the smaller “wedge” between the y_{1}-axis and the line y_{2} = y_{1}/2.

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

c) P (Y_{1} − Y_{2} ≥ 1) is found by integrating the density over the shifted “wedge” in the first quadrant between the y_{1}-axis and the line y_{2} = y_{1} − 1:

∫_{0}^{∞} ∫_{y_{2}+1}^{∞} e^{−y_{1}} dy_{1} dy_{2} = e^{−1}

5.16/The region of integration here is the shifted quadrant bounded by the lines y_{1} = 1 and y_{2} = 1.

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

The inner integral requires integration by parts with u = y_{1} and dv = e^{(−y_{1}+1)/2} the final result is \frac{3}{2e}.

5.20 a) The marginal probabilities are the row and column totals given in the table. b) The conditional probability function p(y_{2}|y_{1} = 0) is

\begin{aligned}P (y_{2} = 0|y_{1} = 0) & = .38/.76 = .5 \\ P (y_{2} = 1|y_{1} = 0) & = .14/.76 = .184 \\ P (y_{2} = 2|y_{1} = 0) & = .24/.76 = .316 \end{aligned}

c) It’s .24/.29 = .8276

5.21/a)

f_{2}(y_{2}) = ∫_{y_{2}}^{1} 3y_{1} dy_{1} = \frac{3}{2}(1 − y_{2}^{2})

if 0 ≤ y_{2} ≤ 1 and 0 otherwise. b) The conditional density f(y_{1}|y_{2}) = f(y_{1}, y_{2})/f_{2}(y_{2}) will be undefined when the denominator is zero – at y_{2} = ±1. c) The conditional density f(y_{2}|y_{1}) = 1/y_{1} so at y_{1} = 3/4 we get f(y_{2}|y_{1} = 3/4) = 4/3 if 0 ≤ y_{2} ≤ 3/4 and zero otherwise (i.e. a uniform density). Then

P (Y_{2} > 1/2|Y_{1} = 3/4) = ∫_{1/2}^{3/4} 4/3 dy_{2} = 1/3

(Note: we stop at the upper limit y_{2} = 3/4 since the conditional density is zero for y_{2} > 3/4.)

5.31/We take the factor depending only on y_{2} outside the integral with respect to y_{1}, then change variables to u = y_{1}/2

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

= \frac{1}{2} ∫_{0}^{∞} ue^{−u} du · e^{−y_{2}/2}

= \frac{1}{2}Γ(2)e^{−y_{2}/2}

= \frac{1}{2}e^{−y_{2}/2}

if y_{2} > 0, and zero otherwise. This gives the marginal density for Y_{2} Then the desired probability is found by integrating

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

(Note that the problem should say the life of the component is in 100’s of hours to match the previous problem this one builds on. So the lower limit of integration is 2 rather than 200.

5.42/ These random variables are not independent, since for instance

p_{1}(0)p_{2}(0) = (.55)(.76) = .418 ≠ p(0, 0) = .38

5.45/ The marginal densities are

f_{1}(y_{1}) = ∫_{y_{1}}^{1} 6(1 − y_{2}) dy_{2} = 3(1 − y_{1})^{2}

(if 0 ≤ y_{1} ≤ 1, and zero otherwise) and

f_{2}(y_{2}) = ∫_{0}^{y_{2}} 6(1 − y_{2}) dy_{1} = 6y_{2} − 6y_{2}^{2}

(if 0 ≤ y_{1} ≤ 1, and zero otherwise). The product is not equal to f(y_{1}, y_{2}), so Y_{1} and Y_{2} are not independent.

5.53/Y_{1} and Y_{2} are independent by the criterion of Theorem 5.5 in the text: The joint density

\frac{1}{8}y_{1}e^{−(y_{1}+y_{2})/2} = \frac{1}{4}y^{1}e^{−y_{1}/2} · \frac{1}{2}e^{−y_{2}/2}

factors into a product of a function of y_{1} times a function of y_{2} and the region where the density is nonzero is a “rectangle”. (Note that Y_{1} has a gamma distribution with α = 2 and Y_{2} is exponential.) It is important to realize that Theorem 5.5 also applies on “infinite rectangles” i.e. a, c can be −∞ and b, d can be +∞ in that statement).