Mathematics 375 – Probability Theory Solutions for Problem Set 8 November 11, 2011

5.1 (a) Since the assignments are random, the probabilities are

P (Y_{1} = 0, Y_{2} = 0) = \frac{1}{9}
P (Y_{1} = 0, Y_{2} = 1) = \frac{2}{9}
P (Y_{1} = 0, Y_{2} = 2) = \frac{1}{9}
P (Y_{1} = 1, Y_{2} = 0) = \frac{2}{9}
P (Y_{1} = 1, Y_{2} = 1) = \frac{2}{9}
P (Y_{1} = 1, Y_{2} = 2) = 0
P (Y_{1} = 2, Y_{2} = 0) = \frac{1}{9}
P (Y_{1} = 2, Y_{2} = 1) = 0
P (Y_{1} = 2, Y_{2} = 2) = 0.

(b) F (1, 0) = P (Y_{1} ≤ 1, Y_{2} ≤ 0) = P (Y_{1} = 0, Y_{2} = 0) + P (Y_{1} = 1, Y_{2} = 0) = \frac{1}{3}.

5.4. (a) All the entries are ≥ 0 and they sum to one so this is a legal joint probability function.

(b) F (y_{1}, y_{2}) is the joint cumulative distribution, so F (1, 2) = P (Y_{1} ≤ 1, Y_{2} ≤ 2) = 1. This is the sum of all the entries in the table, which works out to 1.

5.7. The density is only nonzero on the first quadrant of the (y_{1}, y_{2})-plane. (a) The inequalities Y_{1} < 1, Y_{2} > 5 describe an infinite vertical strip.

P (Y_{1} < 1, Y_{2} > 5) = ∫_{0}^{1} ∫_{5}^{∞} e^{−y_{1}−y_{2}} dy_{2} dy_{1}
= ∫_{0}^{1} −e^{−y_{1}−y_{2}} |_{y_{2}=5}^{y_{2}=∞} dy_{1}
= ∫_{0}^{1} e^{−y_{1}−5} dy_{1}
= −e^{−y_{1}−5}|_{0}^{1}
= e^{−5} − e^{−6}
≐ .004259.

(b) The inequality Y_{1} + Y_{2} < 3 describes the region in the first quadrant below the line Y_{2} = −Y_{1} + 3:

P (Y_{1} + Y_{2} < 3) = ∫_{0}^{3} ∫_{0}^{3−y_{1}} e^{−y_{1}−y_{2}} dy_{2} dy_{1}
= ∫_{0}^{3} e^{−y_{1}} − e^{−3} dy_{1}
= 1 − 4e^{−3}
≐ .8009.

5.8. (a) We must have 1 = ∫_{0}^{1} ∫_{0}^{1} ky_{1}y_{2} dy_{2} dy_{1} = \frac{k}{4}. So k = 4.

(b) The joint cdf is

F (y_{1}, y_{2}) = P (Y_{1} ≤ y_{1}, Y_{2} ≤ y_{2}) = \begin{cases}∫_{0}^{y_{1}} ∫_{0}^{y_{2}} 4t_{1}t_{2} dt_{2} dt_{1} = y_{1}^{2}y_{2}^{2} & if 0 ≤ y_{1}, y_{2} ≤ 1 \\ ∫_{0}^{y_{1}} ∫_{0}^{1} 4t_{1}t_{2} dt_{2} dt_{1} = y_{2}^{2} & if y_{1} ≥ 1, 0 ≤ y_{2} ≤ 1 \\ ∫_{0}^{1} ∫_{0}^{y_{2}} 4t_{1}t_{2} dt_{2} dt_{1} = y_{1}^{2} & if y_{2} ≥ 1, 0 ≤ y_{1} ≤ 1 \\ ∫_{0}^{1} ∫_{0}^{1} 4t_{1}t_{2} dt_{2} dt_{1} = 1 & if y_{1}, y_{2} ≥ 1 \\ 0 & otherwise. \end{cases}

c) P (Y_{1} ≤ 1/2, Y_{2} ≤ 3/4) = ∫_{0}^{1/2} ∫_{0}^{3/4} 4y_{1}y_{2} dy_{2} dy_{1} = (1/2)^{2}(3/4)^{2} = 9/64.

5.11. Since Y_{1}, Y_{2} are uniformly distributed on this triangle of area 1, the probability that (Y_{1}, Y_{2}) lies in any region in the triangle is just the area of that region. (Careful – this is definitely false when the distribution is not uniform!)

(a) P (Y_{1} ≤ 3/4, Y_{2} ≤ 3/4 is the area of the remaining region when we “cut off” two small triangles of areas 1/32 and 1/16 from the big triangle. That leaves an area of 1 − 3/32 = 29/32.

(b) P (Y_{1} − Y_{2} ≥ 0). Y_{1} ≥ Y_{2} below the line y_{1} = y_{2} in the plane. This gives the triangle with vertices (0, 0), (1, 0), (1/2, 1/2), whose area is 1/4.

5.12. Since the density is constant on the triangle with corners at (0, 0), (1, 0), (0, 1), we can find the required probabilities by computing areas. (a) P (Y_{1} ≤ 3/4, Y_{2} ≤ 2) is twice the area of the pentagon with corners at (0, 0), (3/4, 0), (3/4, 1/4), (1/4, 3/4), (0, 3/4). This is 2 · \frac{7}{16} = \frac{7}{8}. (b) Twice the area of a square with side \frac{1}{2}, so equal to \frac{1}{2}.

5.16. (a) The region where the joint density is nonzero is the unit square. So the required probability is computed by

P (Y_{1} < 1/2, Y_{2} > 1/4) = ∫_{0}^{1/2} ∫_{1/4}^{1} y_{1} + y_{2} dy_{2} dy_{1}

= ∫_{0}^{1/2} \frac{3}{4}y_{1} + \frac{15}{32} dy_{1}

= \frac{21}{64}.

(b) The required probability is:

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

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

= \frac{1}{3}.

5.18. The region where the joint density is nonzero is the first quadrant. Since the region of integration is an infinite rectangle, the required double integral can be computed by separating into a product of two one-variable integrals. The first one can be done by parts:

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

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

= \frac{6e^{−1/2} · 2e^{−1/2}}{8}

= \frac{3e^{−1}}{2} ≐ .5518.

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 for y_{i} > 0.

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

5.26. (a) The marginal densities are only nonzero if 0 ≤ y_{i} ≤ 1 and then

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.)

(c and d) The conditional densities are nonzero in the same intervals as the marginal densities, and are given there by

f(y_{1}|y_{2}) = \frac{4y_{1}y_{2}}{2y_{2}} = 2y_{1}

and

f(y_{2}|y_{1}) = \frac{4y_{1}y_{2}}{2y_{1}} = 2y_{2}

(e) Using the conditional density,

P (Y_{1} ≤ 3/4 | Y_{2} = 1/2) = ∫_{0}^{3/4} 2y_{1} dy_{1} = y_{1}^{2}|_{0}^{3/4} = \frac{9}{16}.

5.37. The best way to do this is to compute the marginal density for Y_{2} first:

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

for y_{2} > 0. (Note this says Y_{2} has an exponential distribution with mean β = 2 × 100hours.) Hence the probability that Y_{2} is more than 200 hours is

∫_{2}^{∞} \frac{1}{2}e^{−y_{2}/2} dy_{2} = e^{−1} ≐ .3679.