Mathematics 375 – Probability Theory Solutions for Problem Set 10 December 9, 2011

6.1. (a) Letting U_{1} = 2Y − 1, U_{1} ≤ u ⇔ Y ≤ (u + 1)/2, and the interval of u values corresponding to 0 ≤ y ≤ 1 is −1 ≤ u ≤ 1. Then the cdf for U_{1} is

F (u) = ∫_{0}^{(u+1)/2} 2(1 − y) dy = 1 − ((1 − u)/2)^{2}

(for −1 ≤ u ≤ 1; and 1 for u ≥ 1, 0 for u ≤ −1). To find the pdf for U, we differentiate

f(u) = F′(u) = (1 − u)/2 if − 1 ≤ u ≤ 1; 0 otherwise

(b) Letting U_{2} = 1 − 2Y, U_{2} ≤ u ⇔ Y ≥ (1 − u)/2, and the interval of u values corresponding to 0 ≤ y ≤ 1 is again −1 ≤ u ≤ 1. The cdf for U_{2} is

F (u) = ∫_{(1−u)/2}^{1} 2(1 − y) dy = ((1 + u)/2)^{2}

(for −1 ≤ u ≤ 1; and 1 for u ≤ −1, 0 for u ≥ 1). To find the pdf for U, we differentiate

f(u) = F′(u) = (1 + u)/2 if − 1 ≤ u ≤ 1; 0 otherwise

(c) U_{3} = Y^{2}. We have Y^{2} ≤ u ⇔= \sqrt{u} ≤ Y ≤ \sqrt{u}. Only positive values of y occur with nonzero densities, though, so the cdf for U_{3} is

F (u) = ∫_{0}^{√\overline{u}} 2(1 − y) dy = −(1 − \sqrt{u})^{2}

(for 0 ≤ u ≤ 1, and zero otherwise). The pdf for U_{3} is f(u) = F′(u) = \frac{1−\sqrt{u}}{\sqrt{u}} (for 0 < u ≤ 1, zero otherwise).

(d) Using these pdf’s:

E(U_{1}) = ∫_{−1}^{1} u(1 − u)/2 du = −1/3
E(U_{2}) = ∫_{−1}^{1} u(1 + u)/2 du = 1/3
E(U_{3}) = ∫_{0}^{1} u · \frac{1 − \sqrt{u}}{\sqrt{u}} du = ∫_{0}^{1} \sqrt{u} − u du = 1/6

(e) (Using the methods from Chapter 4.) The expected values above can be computed directly from the density function for Y. But there is also a more clever way. We recognize Y has a Beta distribution with α = 1, β = 2 from the form of the pdf for Y. So E(Y) = 1/3 and V (Y) = 1/18. Hence E(U_{1}) = 2(1/3) − 1 = −1/3, E(U_{2}) = 1 − 2(1/3) = 1/3, and E(U_{3}) = E(Y^{2}) = V (Y) + E(Y)^{2} = 1/18 + (1/3)^{2} = 1/6. Everything checks!

6.5. Note that y = 1 gives u = 5 and y = 5 gives u = 53. The density for U will be nonzero only on the interval [5, 53]. (Note that there was a small typo in the answer to this problem – the correct form should be 5 ≤ u ≤ 53.) Using the method of distribution functions, if u ∈ [5, 53], then since f(y) is nonzero only when 1 ≤ y ≤ 5,

F_{U}(u) = P (U ≤ u)

= P (2Y^{2} + 3 ≤ u)

= P (1 ≤ Y ≤ \sqrt{\frac{u − 3}{2}})

= ∫_{1}^{√ \overline{\frac{u−3}{2}}} \frac{1}{4} dy

= \frac{1}{4}\sqrt{\frac{u − 3}{2}} − \frac{1}{4}.

Therefore the density function for U is the derivative:

f_{U}(u) = \frac{1}{16\sqrt{\frac{u−3}{2}}}

on the interval 5 ≤ u ≤ 53.

6.7. (a) Let Z be a standard normal random variable and U = Z^{2}. Then for any given u > 0 U ≤ u ⇔ −\sqrt{u} ≤ Z ≤ \sqrt{u}. So

F_{U}(u) = P (U ≤ u) = P (−\sqrt{u} < Z ≤ \sqrt{u}) = ∫_{−\sqrt{u}}^{√\overline{u}} \frac{1}{\sqrt{2π}} e^{−y^{2}/2} dy.

We will now differentiate this function of u to find the density function for U, using the Fundamental Theorem of Calculus:

\frac{d}{du} ∫_{−\sqrt{u}}^{√\overline{u}} \frac{1}{\sqrt{2π}} e^{−y^{2}/2} dy = \frac{u^{−1/2}e^{−u/2}}{2\sqrt{2π}} + \frac{u^{−1/2}e^{−u/2}}{2\sqrt{2π}}

= \frac{u^{−1/2}e^{−u/2}}{\sqrt{π} · 2^{1/2}} .

(b) This is the form of a Γ density function with α = 1/2 and β = 2 because Γ(1/2) = \sqrt{π}.

(c) Another name for this is the χ^{2} distribution with one degree of freedom.

6.11. (a) Since the components operate independently, the joint density for the lifetimes is

f(y_{1}, y_{2}) = \begin{cases}e^{−(y_{1}+y_{2})} & y_{1}, y_{2} > 0 \\ 0 & otherwise. \end{cases}

Let U = (Y_{1} + Y_{2})/2, the average. Then

F_{U}(u) = P (U ≤ u)
= P (Y_{1} + Y_{2} ≤ 2u)
= ∫_{0}^{2u} ∫_{0}^{2u−y_{1}} e^{−(y_{1}+y_{2})} dy_{2} dy_{1}
= 1 − e^{−2u} − 2ue^{−2u}.

Hence

f_{U}(u) = 4ue^{−2u}.

That is, U has a Γ(2, 1/2) distribution. (b) U has mean 1 and variance 1/2, using the formulas for Γ-distributed random variables. Checking our work, using the formulas for linear combinations, E(U) = \frac{1}{2}(E(Y_{1}) + E(Y_{2})) = \frac{1+1}{2} = 1 and V (U) = \frac{1}{4}V (Y_{1}) + \frac{1}{4}V (Y_{2}) + 0 = \frac{1}{2}. Check!

6.12. (a) Since c > 0, we have for all u > 0:

F_{U}(u) = P (U ≤ u)

= P (Y ≤ u/c)

= ∫_{0}^{u/c} \frac{y^{α−1}e^{−y/β}}{β^{α}Γ(α)} dy.

We find f_{U}(u) = F′_{U}(u) this time by using the first part of the Fundamental Theorem of Calculus:

f_{U}(u) = \frac{1}{c} \frac{(u/c)^{α−1}e^{−u/(cβ)}}{β^{α}Γ(α)} = \frac{u^{α−1}e^{−u/(cβ)}}{(cβ)^{α}Γ(α)} .

(b) We recognize this as the density for a Γ(α, cβ)-distributed random variable. (c) The shape parameter is unchanged. The scale parameter is multiplied by c.

6.41. By Theorem 6.3, U has a normal distribution with mean μ_{U} = (a_{1} + ⋯ + a_{n})μ and σ_{U}^{2} = (a_{1}^{2} + ⋯ + a_{n}^{2})σ^{2}. The density function for U is the usual normal density:

f_{U}(u) = \frac{1}{\sqrt{2πσ_{U}^{2}}} e^{− \frac{(u−μ_{U})^{2}}{2σ^{2}_{U}}} .

6.42. Using the result from Exercise 6.41, U = Y_{2}−Y_{1} is normally distributed with μ_{U} = −1000 and σ_{U}^{2} = 300^{2} + 400^{2} = 500^{2}. The probability that the elevator is overloaded is found by standardizing and using the table for the standard normal

P (Y_{2} − Y_{1} > 0) = P (\frac{(Y_{2} − Y_{1}) + 1000}{500} > 2) = P (Z > 2) = .0228.

6.43. (a) Since \overline{Y} is normally distributed with mean μ and variance σ^{2}/n, the density function is

\begin{matrix}√ & \overline{n} \\ √ & \overline{2πσ^{2}} \end{matrix} e^{−n \frac{(u−μ)^{2}}{2σ^{2}}}

(b) We standardize with σ for \overline{Y}:

P (|Y − μ| ≤ 1) = P (|Z| ≤ 5/4) = 1 − 2(.1056) = .7888.

(c) Similarly with n = 36:

P (|Y − μ| ≤ 1) = P (|Z| ≤ 6/4) = 1 − 2(.0668) = .8664,

with n = 64:

P (|Y − μ| ≤ 1) = P (|Z| ≤ 8/4) = 1 − 2(.0228) = .9544,

and with n = 81:

P (|Y − μ| ≤ 1) = P (|Z| ≤ 9/4) = 1 − 2(.0122) = .9756.

As n increases, the variance of \overline{Y} decreases, so the probability that Y is within 1 unit of μ increases. (Note: This observation will be fundamental in our study of statistics next semester.)

6.45. From our previous work, we know that U will have a normal distribution with μ = 100 + 70 + 12 = 182 and standard deviation \sqrt{49(.5)^{2} + 9(.2)^{2} + 2 · 7 · 3 · 0} ≐ 3.55 (we’re using the independence assumption to derive this since that gives Cov(Y_{1}, Y_{2}) = 0) Then we want a value for the bid, B such that P (U > B) = .01. This says P (Z > \frac{B−182}{3.55}) = .01, so \frac{B−182}{3.55} = 2.33, and hence B = 190.27. The independence assumption here is questionable because sand is used in construction to make concrete. The amount of sand used and the amount of cement mix used are probably related.