Mathematics 375 – Probability and Statistics I Solutions – Problem Set 6 October 21, 2005

4.2/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 y is the largest integer ≤ y. Hence

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

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}}.

4.3/ a) From the table, P (2 < Y < 5) = .967 − .678 = .289. P (2 ≤ Y < 5) = .967 − .376 = .591 b) P (2 < Y ≤ 5) = .994 − .678 = .316. P (2 ≤ Y ≤ 5) = .994 − .376 = .618. c) No – Y is not a continuous random variable (it’s discrete, and the claim does not apply).

4.5/ a) To be a pdf, we must have 1 = c ∫_{0}^{2} y dy = 2c ⇒ c = 1/2. b) The cdf is

F (y) = ∫_{−∞}^{y} f(t) dt = \{ \begin{matrix}0 & if y ≤ 0 \\ y^{2}/4 & if y > 0 \end{matrix} c) graphs omitted. d) F (2)−F (1) = 1−1/4 = 3/4.

e) In geometric terms, P (1 ≤ Y ≤ 2) is represented by the area under z = y/2 for 1 ≤ y ≤ 2. This is the area of a trapezoid, so by the usual formula = (1/2 + 1)/2 · 1 = 3/4.

4.6/ a,b) F (y) ≥ 0 for all y. lim_{y→−∞} F (y) = 0 and lim_{y→+∞} F (y) = 1 are clear. More-

over, f(y) = F′(y) = \{ \begin{matrix}0 & if y ≤ 0 \\ 2ye^{−y^{2}} & if y > 0 \end{matrix}, which is always ≥ 0. Hence F is nondecreasing.

Therefore, F has the properties of a distribution function according to Theorem 4.1. c)

(Note that times are given in 100’s of hours. So, for instance, 200 hours means y = 2.) P (Y ≥ 2) = 1 − F (2) = e^{−4} ≐ .018. d)

P (Y > 1|Y ≤ 2) = \frac{P (1 < Y ≤ 2)}{P (Y ≤ 2)} = \frac{F (2) − F (1)}{F (2)} = \frac{e^{−1} − e^{−4}}{1 − e^{−4}} ≐ .356.

4.11/a) We must have ∫_{−∞}^{∞} f(y) dy = 1. Hence

1 = ∫_{0}^{1} cy^{2} + y dy = \frac{c}{3} + \frac{1}{2} ⇒ c = \frac{3}{2}.

b) The formula for the cdf F (y) can be derived either by taking the indefinite integral of f(y) “piece by piece” and adjusting the constants of integration to make the resulting function continuous and a legal cdf (as we did in an example from class), or else by doing the definite integral F (y) = ∫_{−∞}^{y} f(t) dt as in an example in the book. The result either way is

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

c) Plots omitted. d) F (−1) = F (0) = 0. F (1) = 1.

e) P (Y < 1/2) = ∫_{0}^{1/2} f(y) dy = F (1/2) − F (0) = 3/16.

f) Using the definition of conditional probability:

P (Y > 1/2|Y > 1/4) = P (Y > 1/2)/P (Y > 1/4) = \frac{1 − F (1/2)}{1 − F (1/4)} = \frac{13/16}{123/128} = 104/123.

4.15/Using the formula for f(y) from 4.11 above, we have

μ = E(Y) = ∫_{−∞}^{∞} yf(y) dy

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

= \frac{17}{24}

Then, to get V (Y), we can proceed as follows:

E(Y^{2}) = ∫_{0}^{1} y^{2} (\frac{3}{2}y^{2} + y) dy = 11/20.

Finally,

V (Y) = E(Y^{2}) − (E(Y))^{2} = 11/20 − (17/24)^{2} ≐ .0483

4.18/The proof here is virtually the same as in the discrete case, using Theorem 4.5: Let μ = E(Y). Then

V (Y) = E((Y − μ)^{2}) = E(Y^{2} − 2μY + μ^{2})
= E(Y^{2}) − 2μE(Y) + μ^{2}
= E(Y^{2}) − 2μ^{2} + μ^{2}
= E(Y^{2}) − (E(Y))^{2}.

4.21/ Easiest method here is to use general facts about expected values of linear combinations: W = 5 − .5Y where Y is as in problem 4.15. Hence E(W) = 5 − .5E(Y) = 5 − .5(17/24) ≐ 4.65. V (W) = (.5)^{2}V (Y) ≐ .012.

4.26/ 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: 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.

4.33/ a) With uniform distribution on [20, 25], P (Y < 22) = ∫_{20}^{22} \frac{1}{5} dy = \frac{2}{5}. b) Similarly, P (Y > 24) = 1/5.

4.36/ a) Uniform distribution on [0, 500] so P (Y ≥ 475) = ∫_{475}^{500} \frac{1}{500} dy = \frac{25}{500} = .05. b) is also .05. c) P (Y < 250) = .5.

4.47/ Using the normal curve area table: a) z_{0} = 0 by symmetry of the standard normal pdf. b) z_{0} = 1.10. c) By the symmetry of the standard normal pdf, to get P (−z_{0} < Z < z_{0}) = .90, we must have P (Z ≥ z_{0}) = .05. From the table, z_{0} ≐ 1.645. d) Similar to c. Answer: z_{0} ≐ 2.575.

4.49/ μ = 16 and σ = 1. Hence P (Y > 17) = P (Y − 16 > 17 − 16) = P (Z > 1) where Z is standard normal. P (Z > 1) = .1587 from the table

4.58/ a) P (Y > 11) = P (\frac{Y −7}{2.6} > \frac{11−7}{2.6}) = P (Z > 1.54) = .0618. b) P (Y < 9) = 1−P (Y ≥ 9) = 1 − P (Z ≥ \frac{9−7}{2.6}) = 1 − .2206 = .7794.

4.65/The constant \frac{1}{\sqrt{2πσ^{2}}} in the formula for the normal density is irrelevant for the location of the inflection points, so we will ignore it and consider the following function g(y) and its derivatives:

g(y) = e^{−(y−μ)^{2}/(2σ^{2})}
⇒ g′(y) = −e^{−(y−μ)^{2}/(2σ^{2})} · (y − μ)/σ^{2} chain rule
⇒ g′′(y) = −e^{−(y−μ)^{2}/(2σ^{2})} (\frac{1}{σ^{2}} − \frac{(y − μ)^{2}}{σ^{4}} ) product rule

We have g′′(y) = 0 if and only if

\begin{aligned}0 & = \frac{1}{σ^{2}} − \frac{(y − μ)^{2}}{σ^{4}} \\ ⇒ (y − μ)^{2} & = σ^{2} \\ ⇒ y − μ & = ±σ \\ ⇒ y & = μ ± σ \end{aligned}

which is what we wanted to show.