Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 6 October 29, 2009

4.6. (a) The probability function for the (discrete) geometric random variable is p(y) = q^{y−1}p. Hence if y = i ≥ 1,

F (i) = P (Y ≤ i) = P (Y = 1) + ⋯ + P (Y = i)

= p(1 + q + q^{2} + ⋯ + q^{i−1})

= p · \frac{1 − q^{i}}{1 − q}

= 1 − q^{i}

(where we used the formula for the sum of a finite geometric series to get the next to last line, then simplified using q = 1 − p). Now, if y > 0 is not an integer, we get F (y) = F (i) where i is the largest integer ≤ y. Hence

F (y) = \begin{cases}0 & if y < 0 \\ 1 − q^{i} & if i ≤ y < i + 1 for i = 0, 1, 2, …. \end{cases}

b) lim_{y→−∞} F (y) = 0 by the definition. Then,

lim_{y→+∞} F (y) = lim_{i→∞} 1 − q^{i} = 1

since q < 1. Finally, F (y) is non-decreasing since

i_{1} > i_{2} ⇒ q^{i_{1}} < q^{i_{2}} ⇒ −q^{i_{1}} > −q^{i_{2}} ⇒ 1 − q^{i_{1}} > 1 − q^{i_{2}}.

4.8. (a) We must have ∫_{−∞}^{+∞} f(y) dy = ∫_{0}^{1} ky(1−y) dy = 1. This gives the equation k (\frac{y^{2}}{2} − \frac{y^{3}}{3} )|_{0}^{1} =

1, so \frac{k}{6} = 1, and k = 6.

(b)

P (.4 ≤ Y ≤ 1) = ∫_{.4}^{1} 6y(1 − y) dy
= 3y^{2} − 2y^{3}|_{.4}^{1}
= .648.

(c) This is the same as (b) because Y is a continuous random variable, so including the endpoints or not makes no difference.

(d) By the definition of conditional probability,

P (Y ≤ .4|Y ≤ .8) = \frac{P (Y ≤ .4)}{P (Y ≤ .8)} = \frac{∫_{0}^{.4} 6y(1 − y) dy}{∫_{0}^{.8} 6y(1 − y) dy} ≐ .393.

(e) This is the same as (d) for the same reason as in (c).

4.13. (a) F (y) = ∫_{−∞}^{y} f(z) dz is a continuous antiderivative of f(y). Hence

F (y) = \begin{cases}0 & if y < 0 \\ \frac{y^{2}}{2} & if 0 ≤ y < 1 \\ y − \frac{1}{2} & if 1 ≤ y < 1.5 \\ 1 & if y ≥ 1.5. \end{cases}

(b) By the FTC, P (0 ≤ Y ≤ .5) = F (.5) − F (0) = \frac{1}{8} = .125.

(c) By the FTC, P (.5 ≤ Y ≤ 1.2) = F (1.2) − F (.5) = .7 − .125 = .575.

4.17. (a) We must have 1 = ∫_{0}^{1} cy^{2} + y dy. So 1 = \frac{c}{3} + \frac{1}{2}, and c = \frac{3}{2}. (b) As in 4.13 (b),

F (y) = \begin{cases}0 & if y < 0 \\ \frac{y^{3}}{2} + \frac{y^{2}}{2} & if 0 ≤ y < 1 \\ 1 & if y ≥ 1. \end{cases}

(c) plots omitted.

(d) F (−1) = 0, F (0) = 0, F (1) = 1.

(e) P (Y < 1/2) = F (1/2) = \frac{1}{16} + \frac{1}{8} = \frac{3}{16} = .1875.

(f) P (Y ≥ 1/2|Y ≥ 1/4) = \frac{P (Y ≥1/2)}{P (Y ≥1/4)} = \frac{1−F (1/2)}{1−F (1/4)} ≐ \frac{.8125}{.9609} ≐ .8455.

4.21. The mean is

E(Y) = ∫_{−∞}^{∞} yf(y) dy

= ∫_{0}^{1} \frac{3y^{3}}{2} + y^{2} dy

= \frac{3y^{4}}{8} + \frac{y^{3}}{3} |_{0}^{1}

= \frac{3}{8} + \frac{1}{3} = \frac{17}{24} ≐ .708.

Then the variance is V (Y) = E(Y^{2}) − E(Y)^{2}.

E(Y^{2}) = ∫_{−∞}^{∞} y^{2}f(y) dy

= ∫_{0}^{1} \frac{3y^{4}}{2} + y^{3} dy

= \frac{3y^{5}}{10} + \frac{y^{4}}{4} |_{0}^{1}

= \frac{3}{10} + \frac{1}{4}

= \frac{11}{20}.

So V (Y) = \frac{11}{20} − (\frac{17}{24})^{2} = \frac{139}{2880} ≐ .0483.

4.26. (a) By Theorem 4.5, E(aY + b) = aE(Y) + b = aμ + b.

(b) Then, by the definition of the variance and Theorem 4.5 again,

V (aY + b) = E(((aY + b) − (aμ + b))^{2})
= E(a^{2}(Y^{2} − μ^{2}))
= a^{2}(E(Y^{2}) − μ^{2})
= a^{2}(E(Y^{2}) − E(Y)^{2}) by one of our standard formulas
= a^{2}V (Y).

4.27. Using the result of 4.26, E(W) = 5 − .5E(Y) = 5 − \frac{17}{48} ≐ 4.646 (dollars). Then V (W) = \frac{1}{4}V (Y) = \frac{139}{4·2880} ≐ .0121.

4.32. (a) E(Y) = \frac{3}{64} ∫_{0}^{4} y^{3}(4−y) dy = \frac{3}{64} ∫_{0}^{4} 4y^{3}−y^{4} dy = 2.4. Then E(Y^{2}) = \frac{3}{64} ∫_{0}^{4} 4y^{4}−y^{5} dy = 6.4 and V (Y) = E(Y^{2}) − E(Y)^{2} = .64.

(b) Easiest method here is to use general facts about expected values of linear combinations as in 4.26 above: E(C) = 200(2.4) = $480. V (C) = 200^{2}(.64) = 25600 (units of variance are dollars^{2}), so σ = 160.

(c) P (C > 600) = P (Y > 3) = \frac{3}{64} ∫_{3}^{4} y^{2}(4 − y) dy ≐ .26, so this can certainly happen.

4.33. (a) We have E(Y) = ∫_{5}^{7} \frac{3}{8}y(7 − y)^{2} dy = \frac{11}{2} = 5.5. Then

V (Y) = E(Y^{2}) − μ^{2}

= ∫ \frac{3}{8}y^{2}(7 − y)^{2} dy − (\frac{11}{2})^{2}

= \frac{3}{20} = .15.

(b) This part can be done via Tchebysheff’s Theorem (which also applies for continuous random variables). But we have not talked about this yet. So, another way (yielding a different interval, of course!) is to find a z such that F (z) = .75, where F is the distribution function. We have F (y) = 0 for all y ≤ 5, F (y) = 1 − \frac{1}{8}(7 − y)^{3} for 5 ≤ y ≤ 7, and F (y) = 1 for all y ≥ 7. Hence we have

\frac{3}{4} = 1 − \frac{1}{8}(7 − z)^{3} ⇔ z = 7 − 2^{1/3} ≐ 5.74.

The interval would be 5 ≤ y ≤ 5.74.

(c) P (Y < 5.5) = ∫_{5}^{5.5} \frac{3}{8}y(7 − y)^{2} dy ≐ .5781, so this is actually very common. 4.41. From Theorem 4.6 (and work in class), we know that E(Y) = \frac{θ_{1}+θ_{2}}{2} . Now compute

E(Y^{2}) = ∫_{θ_{1}}^{θ_{2}} y^{2} · \frac{1}{θ_{2} − θ_{1}} dy

= \frac{1}{θ_{2} − θ_{2}} \frac{y^{3}}{3} |_{θ_{1}}^{θ_{2}}

= \frac{θ_{2}^{3} − θ_{1}^{3}}{3(θ_{2} − θ_{1})}

= \frac{θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}}{3} .

Then

V (Y) = E(Y^{2}) − E(Y)^{2}

= \frac{θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}}{3} − (\frac{θ_{1} + θ_{2}}{2} )^{2}

= \frac{4(θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}) − 3(θ_{2}^{2} + 2θ_{2}θ_{1} + θ_{1}^{2})}{12}

= \frac{(θ_{2} − θ_{1})^{2}}{12} ,

which is what we wanted to show.

4.42. We have

.5 = P (Y ≤ ϕ_{.5}) = ∫_{θ_{1}}^{ϕ_{.5}} \frac{1}{θ_{2} − θ_{1}} dy = \frac{ϕ_{.5} − θ_{1}}{θ_{2} − θ_{1}} .

Solving this for ϕ_{.5}, we get ϕ_{.5} = \frac{θ_{1} + θ_{2}}{2} . This is the median, by definition.

4.43. The mean of the area is \overline{A} = E(A) = ∫_{0}^{1} πr^{2} dr = \frac{π}{3} , and the variance is

V (A) = E(A^{2}) − E(A)^{2}

= ∫_{0}^{1}(πr^{2})^{2} dr − (\frac{π}{3} )^{2}

= \frac{π^{2}}{5} − \frac{π^{2}}{9}

= \frac{4π^{2}}{45} .

4.47. The density function for Y is f(y) = \frac{1}{4} for 1 ≤ y ≤ 5, and zero otherwise. Hence (a) P (Y ≥ 2) = ∫_{2}^{5} \frac{1}{4} dy = \frac{3}{4}.

(b) The expected cost is E(C) = E(c_{0} +c_{1}y^{2}) = c_{0} +c_{1}E(Y^{2}). We can compute E(Y^{2}) directly, or else use the computations from 4.41 above: E(Y^{2}) = \frac{25+5+1}{3} = \frac{31}{3}. Therefore, E(C) = c_{0} + \frac{31c_{1}}{3} .

4.51. This is a conditional probability:

P (Y > 65|Y > 55) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.