Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 6 October 22, 2009

3.145. By the binomial theorem,

m(t) = E(e^{tY}) = ∑_{y=0}^{n} e^{ty}p^{y}q^{n−y}

= ∑_{y=0}^{n}(e^{t}p)^{y}q^{n−y}

= (pe^{t} + q)^{n}.

This is what we wanted to show.

3.146. The idea of this problem is to use the moment generating function for a binomial Y to give a different derivation of the formulas we know for E(Y) and V (Y). By the chain rule for derivatives, m′(t) = n(pe^{t} + q)^{n−1}pe^{t}. So E(Y) = m′(0) = np. Then

m′′(t) = n(n − 1)(pe^{t} + q)^{n−2}p^{2}e^{2t} + n(pe^{t} + q)^{n−1}pe^{t}

so E(Y^{2}) = m′′(0) = n(n − 1)p^{2} + np. Then

V (Y) = E(Y^{2}) − E(Y)^{2} = n(n − 1)p^{2} + np − n^{2}p^{2} = −np^{2} + np = np(1 − p) = npq.

3.150. The uniqueness theorem says that if two random variables have the same moment generating function, then they have the same distribution. This given formula has the form of the moment generating function for a geometric random variable with p = .3 (this was proved in class; also see the indicated problem 3.147). So the random variable is geometric with p = .3.

3.153. By the Uniqueness Theorem for the moment generating function, and the formulas from class and the text for moment generating functions for standard types of random variables,

(a) Binomial, n = 5, p = 1/3.

(b) Geometric, p = 1/2.

(c) Poisson, λ = 2.

3.158. Since e^{bt} is not random (it’s a fixed function of t),

m_{aY +b}(t) = E(e^{(aY +b)t}) = E(e^{bt}e^{(at)Y}) = e^{bt}E(e^{(at)Y}) = e^{bt}m_{Y} (at).

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}.

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}.

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.