Mathematics 375 – Probability and Statistics 1 Solutions – Problem Set 10 November 30, 2005

5.87/ By linearity of expected values, E(3Y_{1} + 4Y_{1} − 6Y_{3}) = 3E(Y_{1}) + 4E(Y_{2}) − 6E(Y_{3}) = 6 − 4 − 24 = −22. We use Theorem 5.12b to find

V (3Y_{1} + 4Y_{1} − 6Y_{3}) = 9V (Y_{1}) + 16V (Y_{2}) + 36V (Y_{3})
+ 24Cov(Y_{1}, Y_{2}) − 18Cov(Y_{1}, Y_{3}) − 48Cov(Y_{2}, Y_{3})

= 36 + 96 + 288 + 24 + 36 + 0 = 480.

5.92/ By computing marginal densities here, we see Y_{1} has a Gamma distribution with α = 2 and β = 1. Y_{2} is exponentially distributed. So E(Y_{1}) = 2, E(Y_{2}) = 1. So E(Y_{1} − Y_{2}) = 1. An alternate way to see this: integrate against the joint density:

E(Y_{1} − Y_{2}) = ∫_{0}^{∞} ∫_{0}^{y_{1}}(y_{1} − y_{2})e^{−y_{1}} dy_{1} dy_{2}

= \frac{1}{2} ∫_{0}^{∞} y_{1}^{2}e^{−y_{1}} dy_{1}

= \frac{1}{2}Γ(3)

= 1

Then by Theorem 5.12 b,

V (Y_{1} − Y_{2}) = V (Y_{1}) + V (Y_{2}) − 2Cov(Y_{1}, Y_{2})
= 2 + 1 − 2(E(Y_{1}Y_{2}) − 2 · 1)
= 7 − 2 ∫_{0}^{∞} ∫_{0}^{y_{1}} y_{1}y_{2}e^{−y_{1}} dy_{2} dy_{1}
= 7 − ∫_{0}^{∞} y_{1}^{3}e^{−y_{1}} dy_{1}
= 7 − Γ(4)
= 7 − 6 = 1.

It is unlikely that a customer spends 4 minutes or more at the window, since 4 = μ + 3σ (three standard deviations above mean).

6.1/ a) Letting U_{1} = 2Y − 1, U_{1} ≤ u ⇔ Y ≤ (u + 1)/2. Then the cdf for U_{1} is

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

(for −1 ≤ u ≤ 1, and zero otherwise). To find the pdf for U, we differentiate

f(u) = F′(u) = \{ \begin{matrix}(1 − u)/2 & if −1 ≤ u ≤ 1 \\ 0 & otherwise \end{matrix}

b) Derivation for U_{2} is very similar to part a. Answer:

f(u) = \{ \begin{matrix}(1 + u)/2 & if −1 ≤ u ≤ 1 \\ 0 & otherwise \end{matrix} .

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:

\begin{aligned}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 \end{aligned}

e) Using the methods from Chapter 4: We know 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!

6.3/ a) Since U = 10Y − 4, U ≤ u ⇔ Y ≤ (u + 4)/10. So

F (u) = ∫_{−∞}^{(u+10)/4} f(y) dy = \{ \begin{matrix}0 & if u < −4 \\ \frac{1}{2}((u + 4)/10)^{2} & if −4 ≤ u ≤ 6 \\ −\frac{1}{2} + (u + 4)/10 & if 6 ≤ u ≤ 11 \\ 1 & if u ≥ 11 \end{matrix}

(Note: this is found by integrating each “piece”, then adjusting constants of integration to make the whole function F continuous.) Finally, to get the pdf, we take

f(u) = F′(u) == \{ \begin{matrix}0 & if u < −4 \\ (u + 4)/100 & if −4 ≤ u ≤ 6 \\ 1/10 & if 6 ≤ u ≤ 11 \\ 0 & if u ≥ 11 \end{matrix}

b) Integrate “piecewise”:

E(U) = ∫_{−4}^{6} u · (u + 4)/100 du + ∫_{6}^{11} u · (1/10) du = 67/12.

c) “Using the methods of Chapter 4” means finding E(Y) then using linearity to get E(U):

E(Y) = ∫_{0}^{1} y · y dy + ∫_{1}^{3/2} y dy = 23/24.

So E(U) = 10E(Y) − 4 = (230/24) − 4 = 134/24 = 67/12. Check!

6.8/ Note U = Y_{1} − Y_{2} ≤ u ⇔ Y_{2} ≥ Y_{1} − u. The region where the joint density is nonzero is the infinite “wedge” bounded by y_{2} = 0, y_{2} = y_{1} and y_{1} ≥ 0. Note that y_{2} = y_{1} − u is a line parallel to the slanted edge of the wedge. To find F_{U}(u) = P (U ≤ u) we want to integrate over the portion of the wedge above the line y_{2} = y_{1} − u. For this, it is simplest to set up the iterated integral with dy_{1} “inside” (since the left and right boundaries are given by the same formulas all the way, unlike the bottom edge if we integrate with dy_{2} inside – that way works too, of course, but it’s significantly more complicated!).

F_{U}(u) = ∫_{0}^{∞} ∫_{y_{2}}^{y_{2}+u} e^{−y_{1}} dy_{1} dy_{2}
= ∫_{0}^{∞} −e^{−(y_{2}+u)} + e^{−y_{2}} dy_{2}
= e^{−(y_{2}+u)} − e^{−y_{2}} |_{0}^{∞}
= 1 − e^{−u}

So f_{U}(u) = F′_{U}(u) = e^{−u} (exponential with β = 1.)

6.22/ a) Let U = Y^{m} ⇔ Y = U^{1/m} = h^{−1}(U) Then \frac{dh^{−1}(u)}{du} = \frac{1}{m}u^{1/m−1} and

f_{U}(u) = f_{Y} (h^{−1}(u)) |\frac{dh^{−1}(u)}{du}|

= \frac{1}{α}m (u^{1/m})^{m−1} e^{−u/α} · \frac{1}{m}u^{1/m−1}

= \frac{1}{α}e^{−u/α}

(after some simplification).

b) If U = Y^{k} (k ≠ m), the same transformation gives f_{U}(u) = \frac{m}{α}u^{(m−k)/k}e^{−u^{m/k}/α}. So

E(U) = ∫_{0}^{∞} \frac{m}{α}u^{m/k}e^{−u^{m/k}/α} du

The substitution v = u^{m/k}, dv = \frac{m}{k}u^{(m−k)/k} takes this to the form

E(U) = \frac{1}{α} ∫_{0}^{∞} v^{k/m}e^{−v/α} dv = α^{k/m}Γ (\frac{k}{m} + 1) .

6.24/ U = −2 ln Y ⇔ Y = e^{−U/2} = h^{−1}(U). Hence \frac{dh^{−1}(u)}{du} = −\frac{1}{2}e^{−u/2}. So f(u) =

1 · |\frac{dh^{−1}(u)}{du} | = \frac{1}{2}e^{−y/2}. (That is, U is exponentially distributed with β = 1/2.)

6.33/ U = Y_{1}/2 + Y_{2}/2 where each Y_{i} is exponentially distributed and Y_{1}, Y_{2} are independent. Hence by Theorem 6.2, the mgf for U is

m_{U}(t) = m_{Y_{1}}(t/2) · m_{Y_{2}}(t/2) = \frac{1}{(1 − t/2)^{2}} .

Comparing with the table inside the back cover of the text, we see that U has a Gamma distribution with α = 2 and β = 1/2.

6.34/ Y_{i} are N(0, 1), so by Example 6.11 Y_{i}^{2} are χ^{2}(1). Then by Theorem 6.4, U = Y_{1}^{2}+Y_{2}^{2} has a χ^{2}(2) distribution. Therefore f_{U}(u) = \frac{1}{2}e^{−u/2}. (Note: this is also same as an exponential pdf with β = 2.)

6.35/ By Theorem 6.3, U is normally distributed with μ_{U} = (∑_{i} a_{i})μ and σ_{U}^{2} = (∑_{i} a_{i}^{2})σ^{2}. Hence

f_{U}(u) = \frac{1}{\sqrt{2πσ_{U}^{2}}} e^{−(u−μ_{U})^{2}/(2σ_{U}^{2} )}

6.36/ Using the result from Exercise 6.35, U = Y_{1} − Y_{2} 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_{1} − Y_{2} > 0) = P (\frac{(Y_{1} − Y_{2}) + 1000}{500} > 2) = P (Z > 2) = .0228.