Mathematics 375 – Probability and Statistics 1 Solutions – Problem Set 6 October 29, 2003

4.1/Since Y takes only positive integer values, p(y) =

P (Y = y) = P (Y ≤ y) − P (Y ≤ y − 1) = F (y) − F (y − 1) = \{ \begin{matrix}F (1) & if y = 1 \\ F (y) − F (y − 1) & if y > 1 \end{matrix}

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}}.

4.4/a) To be a legal density function we must have

1 = ∫_{−∞}^{∞} f(y) dy = ∫_{0}^{1} ky(1 − y) dy = k (\frac{y^{2}}{2} − \frac{y^{3}}{3} ) |_{0}^{1} = k · \frac{1}{6}

So k = 6.

b) and c) Since the distribution is continuous, including the endpoints of the intervals or not makes no difference. Both probabilities equal

∫_{.4}^{1} 6y(1 − y) dy = .648.

d) and e) Both conditional probabilities equal

\frac{∫_{0}^{.4} 6y(1 − y) dy}{∫_{0}^{.8} 6y(1 − y) dy} ≐ \frac{.352}{.896} ≐ .393.

4.9/a) We must show f(y) ≥ 0 for all y and ∫_{−∞}^{∞} f(y) dy = 1. The first condition follows immediately since b > 0 and y^{2} > 0 for all y > 0. For the second

\begin{aligned}∫_{−∞}^{∞} f(y) dy & = ∫_{b}^{∞} \frac{b}{y^{2}} dy \\ & = \frac{−b}{y}|_{b}^{∞} = 0 − (−1) = 1. \end{aligned}

b) The cumulative distribution function F is

F (y) = ∫_{−∞}^{y} f(t) dt

= \{ \begin{matrix}0 & if y ≤ 0 \\ ∫_{b}^{y} \frac{b}{t^{2}} dt & if y > 0 \end{matrix}

= \{ \begin{matrix}0 & if y ≤ b \\ 1 − \frac{b}{y} & if y > b \end{matrix}

c) Using part b,

P (Y > b + c) = 1 − F (b + c) = 1 − (1 − \frac{b}{b + c}) = \frac{b}{b + c}.

d) With d > c, we have by the definition of conditional probability:

P (Y > b + d|Y > b + c) = \frac{P (Y > b + d)}{P (Y > b + c)}.

Using part c, this equals

(\frac{b}{b + d}) / (\frac{b}{b + c}) = \frac{b + c}{b + d}.

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

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

So 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}.

4.19/To perform the desired calculations, we need the density function f(y) = F′(y). We have

f(y) = \{ \begin{matrix}1/8 & if 0 < y < 2 \\ y/8 & if 2 < y < 4 \\ 0 & otherwise \end{matrix}

Then we integrate by splitting the integral up into two terms based on the split in the definition of f(y) at y = 2:

E(Y) = ∫_{−∞}^{∞} yf(y) dy
= ∫_{0}^{2} y/8 dy + ∫_{2}^{4} y^{2}/8 dy
= y^{2}/16|_{0}^{2} + y^{3}/24|_{2}^{4}
≐ 2.583

Similarly,

E(Y^{2}) = ∫_{−∞}^{∞} y^{2}f(y) dy
= ∫_{0}^{2} y^{2}/8 dy + ∫_{2}^{4} y^{3}/8 dy
= y^{3}/24|_{0}^{2} + y^{4}/32|_{2}^{4}
≐ 7.833

Then using the formula from 4.18,

V (Y) = E(Y^{2}) − (E(Y))^{2} ≐ 7.833 − (2.583)^{2} ≐ 1.16.

4.20/a) Using Theorem 4.5 (with g_{1}(Y) = aY and g_{2}(Y) = b (constant)), we have

E(aY + b) = aE(Y) + b = aμ + b.

b) From the definition of the variance, V (aY + b) = E(((aY + b) − (aμ + b))^{2}), and cancelling inside the square and factoring out a^{2} from what remains, we have:

= E(a^{2}(Y − μ)^{2})
= a^{2}E((Y − μ)^{2})
= a^{2}V (Y).

4.28/a) The uniform cdf is this anti-derivative of the uniform density (f(y) = 1 if 0 < y < 1 and 0 otherwise):

F (y) = \{ \begin{matrix}0 & if y < 0 \\ y & if 0 ≤ y < 1 \\ 1 & if y ≥ 1 \end{matrix}

b) Since a ≥ 0 and a + b ≤ 1, we can just use the “middle case” of F to get:

P (a ≤ Y ≤ a + b) = ∫_{a}^{a+b} 1 dy = F (a + b) − F (a) = a + b − a = b.

4.29/a) The parachutist is closer to A than B whenever y = the random location satisfies A ≤ y ≤ \frac{A+B}{2} . We get

P (A ≤ Y ≤ \frac{A + B}{2} ) = ∫_{A}^{(A+B)/2} \frac{1}{B − A} dy = \frac{1}{2}

b) The parachutist is at least 3 times farther from A than from B whenever y = the random location satisfies \frac{A+3B}{4} ≤ y ≤ B. (Find the cutoff point by solving for y in y − A = 3(b − y).) We get

P (\frac{A + 3B}{4} ≤ Y ≤ B) = ∫_{(A+3B)/4}^{B} \frac{1}{B − A} dy = \frac{1}{4}

4.35/a) Y = the delivery time is uniform on [1, 5]. So its density function is f(y) = 1/4 for 1 ≤ y ≤ 5 and 0 otherwise. Hence P (Y > 2) = ∫_{2}^{5} 1/4 dy = 3/4.

b) The expected value of the cost is

E(C) = ∫_{1}^{5}(c_{0} + c_{1}y^{2}) · \frac{1}{4} dy = (c_{0}y/4 + c_{1}y^{3}/12)|_{1}^{5} = c_{0} + 31c_{1}/3.

4.46/Using the table of standard normal probabilities in the text,

a) P (0 ≤ Z ≤ 1.2) = .5 − .1151 = .3849.

b) By symmetry, P (−.9 ≤ Z ≤ 0) = P (0 ≤ Z ≤ .9) = .5 − .1841 = .3159.

c) P (.3 ≤ Z ≤ 1.56) = .3821 − .0594 = .3227.

d) By symmetry, P (−1.56 ≤ Z ≤ −.2) = P (.2 ≤ Z ≤ 1.56) = .4207 − .0594 = .3613.

4.57/a)

P (.12 ≤ Y ≤ .13) = P (\frac{.12 − .13}{.005} ≤ \frac{Y − .13}{.005} ≤ \frac{.14 − .13}{.005}) = P (−2 ≤ \frac{Y − .13}{.005} ≤ 2)

From the table for the standard normal Z, this is .9544 (Note \frac{Y −.13}{.005} = \frac{Y −μ}{σ} is a standard normal).

b) The number that meet the specifications is a binomial (discrete) random variable with n = 4 and p = .9544 from part a(!). Hence the probability that all four meet the specifications is \binom{4}{4}(.9544)^{4}(.0456) ≐ .83.

4.58/

a) Convert to a probability involving a standard normal:

P (Y > 11) = P (\frac{Y − 7}{2.6} > \frac{11 − 7}{2.6}) = P (Z > 1.538) = .0618

b) Similarly,

P (Y < 9) = P (\frac{Y − 7}{2.6} < \frac{9 − 7}{2.6}) = P (Z < .77) = 1 − .2206 = .7794.

4.61/Here we will use the standard normal table “backwards”. We want to pick μ (the mean fill amount) so that P (Y > 8) = .01. Equivalently, so that P (\frac{Y −μ}{.3} > \frac{8−μ}{.3}) > .01. The entry in the standard normal table for z = 2.33 gives P (Z > 2.33) = .0099, so we will use that value. \frac{8−μ}{.3} = 2.33 ⇒ μ ≐ 7.3.

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.