# 10.4: Problems on Functions of Random Variables

- Page ID
- 10879

Exercise \(\PageIndex{1}\)

Suppose \(X\) is a nonnegative, absolutely continuous random variable. Let \(Z = g(X) = Ce^{-aX}\), where \(a > 0\), \(C > 0\). Then \(0 < Z \le C\). Use properties of the exponential and natural log function to show that

\(F_Z (v) = 1 - F_X (- \dfrac{\text{In } (v/C)}{a})\) for \(0 < v \le C\)

**Answer**-
\(Z = Ce^{-aX} \le v\) iff \(e^{-aX} \le v/C\) iff \(-aX \le \text{In } (v/C)\) iff \(X \ge - \text{In } (v/C)/a\), so that

\(F_Z(v) = P(Z \le v) = P(X \ge -\text{In } (v/C)/a) = 1 - F_X (-\dfrac{\text{In } (v/C)}{a})\)

Exercise \(\PageIndex{2}\)

Use the result of Exercise 10.4.1 to show that if \(X\) ~ exponential \((\lambda)\), then

\(F_Z (v) = (\dfrac{v}{C})^{\lambda/a}\) \(0 < v \le C\)

**Answer**-
\(F_Z (v) = 1 - [1- exp (-\dfrac{\lambda}{a} \cdot \text{In } (v/C))] = (\dfrac{v}{C})^{\lambda/a}\)

Exercise \(\PageIndex{3}\)

Present value of future costs. Suppose money may be invested at an annual rate *a*, compounded continually. Then one dollar in hand now, has a value \(e^{ax}\) at the end of \(x\) years. Hence, one dollar spent \(x\) years in the future has a *present **value**e*\(^{-ax}\). Suppose a device put into operation has time to failure (in years) \(X\) ~ exponential (\(\lambda\)). If the cost of replacement at failure is \(C\) dollars, then the present value of the replacement is \(Z = Ce^{-aX}\). Suppose \(\lambda = 1/10\), \(a = 0.07\), and \(C =\) $1000.

- Use the result of Exercise 10.4.2. to determine the probability \(Z \le 700, 500, 200\).
- Use a discrete approximation for the exponential density to approximate the probabilities in part (a). Truncate \(X\) at 1000 and use 10,000 approximation points.

**Answer**-
\(P(Z \le v) = (\dfrac{v}{1000})^{10/7}\)

v = [700 500 200]; P = (v/1000).^(10/7) P = 0.6008 0.3715 0.1003 tappr Enter matrix [a b] of x-range endpoints [0 1000] Enter number of x approximation points 10000 Enter density as a function of t 0.1*exp(-t/10) Use row matrices X and PX as in the simple case G = 1000*exp(-0.07*t); PM1 = (G<=700)*PX' PM1 = 0.6005 PM2 = (G<=500)*PX' PM2 = 0.3716 PM3 = (G<=200)*PX' PM3 = 0.1003

Exercise \(\PageIndex{4}\)

Optimal stocking of merchandise. A merchant is planning for the Christmas season. He intends to stock *m* units of a certain item at a cost of *c* per unit. Experience indicates demand can be represented by a random variable \(D\) ~ Poisson (\(\mu\)). If units remain in stock at the end of the season, they may be returned with recovery of \(r\) per unit. If demand exceeds the number originally ordered, extra units may be ordered at a cost of *s* each. Units are sold at a price \(p\) per unit. If \(Z = g(D)\) is the gain from the sales, then

- For \(t \le m\), \(g(t) = (p - c) t- (c - r)(m - t) = (p - r)t + (r - c) m\)
- For \(t > m\), \(g(t) = (p - c)m + (t - m) (p - s) = (p - s) t + (s - c)m\)

Let \(M = (-\infty, m]\). Then

\(g(t) = I_M(t) [(p - r) t + (r - c)m] + I_M(t) [(p - s) t + (s - c) m]\)

Suppose \(\mu = 50\) \(m = 50\) \(c = 30\) \(p = 50\) \(r = 20\) \(s = 40\).

Approximate the Poisson random variable \(D\) by truncating at 100. Determine \(P(500 \le Z \le 1100)\).

**Answer**-
mu = 50; D = 0:100; c = 30; p = 50; r = 20; s = 40; m = 50; PD = ipoisson(mu,D); G = (p - s)*D + (s - c)*m +(s - r)*(D - m).*(D <= m); M = (500<=G)&(G<=1100); PM = M*PD' PM = 0.9209 [Z,PZ] = csort(G,PD); % Alternate: use dbn for Z m = (500<=Z)&(Z<=1100); pm = m*PZ' pm = 0.9209

Exercise \(\PageIndex{5}\)

(See Example 2 from "Functions of a Random Variable") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase ten tickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:

- 11-20, $18 each
- 21-30, $16 each
- 31-50, $15 each
- 51-100, $13 each

If the number of purchasers is a random variable \(X\), the total cost (in dollars) is a random quantity \(Z = g(X)\) described by

\(g(X) = 200 + 18 I_{M1} (X) (X - 10) + (16 - 18) I_{M2} (X) (X - 20) +\)

\((15 - 16) I_{M_3} (X) (X - 30) + (13 - 15) I_{M4} (X) (X - 50)\)

where \(M1 = [10, \infty)\), \(M2 = [20, \infty)\), \(M3 = [30, \infty)\), \(M4 = [50, \infty)\)

Suppose \(X\)~ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine \(P(Z \ge 1000)\), \(P(Z \ge 1300)\) and \(P(900 \le Z \le 1400)\).

**Answer**-
X = 0:150; PX = ipoisson(75,X); G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ... (15 - 16)*(X- 30).*(X>=30) + (13 - 15)*(X - 50).*(X>=50); P1 = (G>=1000)*PX' P1 = 0.9288 P2 = (G>=1300)*PX' P2 = 0.1142 P3 = ((900<=G)&(G<=1400))*PX' P3 = 0.9742 [Z,PZ] = csort(G,PX); % Alternate: use dbn for Z p1 = (Z>=1000)*PZ' p1 = 0.9288

Exercise \(\PageIndex{6}\)

(See Exercise 6 from "Problems on Random Vectors and Joint Distributions", and Exercise 1 from "Problems on Independent Classes of Random Variables")) The pair \(\{X, Y\}\) has the joint distribution

(in m-file npr08_06.m):

\(X = \) [-2.3 -0.7 1.1 3.9 5.1] \(Y = \) [1.3 2.5 4.1 5.3]

\(P = \begin{bmatrix} 0.0483 & 0.0357 & 0.0420 & 0.0399 & 0.0441 \\ 0.0437 & 0.0323 & 0.0380 & 0.0361 & 0.0399 \\ 0.0713 & 0.0527 & 0.0620 & 0.0609 & 0.0551 \\ 0.0667 & 0.0493 & 0.0580 & 0.0651 & 0.0589 \end{bmatrix}\)

Determine \(P(\text{max }\{X, Y\} \le 4)\). Let \(Z = 3X^3 + 3X^2 Y - Y^3\).

Determine \(P(Z< 0)\) and \(P(-5 < Z \le 300)\).

**Answer**-
npr08_06 Data are in X, Y, P jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row matrix of VALUES of X X Enter row matrix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P P1 = total((max(t,u)<=4).*P) P1 = 0.4860 P2 = total((abs(t-u)>3).*P) P2 = 0.4516 G = 3*t.^3 + 3*t.^2.*u - u.^3; P3 = total((G<0).*P) P3 = 0.5420 P4 = total(((-5<G)&(G<=300)).*P) P4 = 0.3713 [Z,PZ] = csort(G,P); % Alternate: use dbn for Z p4 = ((-5<Z)&(Z<=300))*PZ' p4 = 0.3713

Exercise \(\PageIndex{7}\)

(See Exercise 2 from "Problems on Independent Classes of Random Variables") The pair \(\{X, Y\}\) has the joint distribution (in m-file npr09_02.m):

\(X = \) [-3.9 -1.7 1.5 2 8 4.1] \(Y = \) [-2 1 2.6 5.1]

\(P = \begin{bmatrix} 0.0589 & 0.0342 & 0.0304 & 0.0456 & 0.0209 \\ 0.0962 & 0.056 & 0.0498 & 0.0744 & 0.0341 \\ 0.0682 & 0.0398 & 0.0350 & 0.0528 & 0.0242 \\ 0.0868 & 0.0504 & 0.0448 & 0.0672 & 0.0308 \end{bmatrix}\)

Determine \(P(\{X + Y \ge 5\} \cup \{Y \le 2\})\), \(P(X^2 + Y^2 \le 10)\).

**Answer**-
npr09_02 Data are in X, Y, P jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row matrix of VALUES of X X Enter row matrix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P M1 = (t+u>=5)|(u<=2); P1 = total(M1.*P) P1 = 0.7054 M2 = t.^2 + u.^2 <= 10; P2 = total(M2.*P) P2 = 0.3282

Exercise \(\PageIndex{8}\)

(See Exercsie 7 from "Problems on Random Vectors and Joint Distributions", and Exercise 3 from "Problems on Independent Classes of Random Variables") The pair has the joint distribution

(in m-file npr08_07.m):

t = | -3.1 | -0.5 | 1.2 | 2.4 | 3.7 | 4.9 |

u = 7.5 | 0.0090 | 0.0396 | 0.0594 | 0.0216 | 0.0440 | 0.0203 |

4.1 | 0.0495 | 0 | 0.1089 | 0.0528 | 0.0363 | 0.0231 |

-2.0 | 0.0405 | 0.1320 | 0.0891 | 0.0324 | 0.0297 | 0.0189 |

-3.8 | 0.0510 | 0.0484 | 0.0726 | 0.0132 | 0 | 0.0077 |

Determine \(P(X^2 - 3X \le 0)\), \(P(X^3 - 3|Y| < 3)\).

**Answer**-
npr08_07 Data are in X, Y, P jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row matrix of VALUES of X X Enter row matrix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P M1 = t.^2 - 3*t <=0; P1 = total(M1.*P) P1 = 0.4500 M2 = t.^3 - 3*abs(u) < 3; P2 = total(M2.*P) P2 = 0.7876

Exercise \(\PageIndex{9}\)

For the pair \(\{X, Y\}\) in Exercise 10.4.8, let \(Z = g(X, Y) = 3X^2 + 2XY - Y^2\). Determine and plot the distribution function for \(Z\).

**Answer**-
G = 3*t.^2 + 2*t.*u - u.^2; % Determine g(X,Y) [Z,PZ] = csort(G,P); % Obtain dbn for Z = g(X,Y) ddbn % Call for plotting m-procedure Enter row matrix of VALUES Z Enter row matrix of PROBABILITIES PZ % Plot not reproduced here

Exercise \(\PageIndex{10}\)

For the pair \(\{X, Y\}\) in Exercise 8, let

Determine and plot the distribution function for \(W\).

**Answer**-
H = t.*(t+u<=4) + 2*u.*(t+u>4); [W,PW] = csort(H,P); ddbn Enter row matrix of VALUES W Enter row matrix of PROBABILITIES PW % Plot not reproduced here

**For the distributions in Exercises 10-15 below**

- Determine analytically the indicated probabilities.
- Use a discrete approximation to calculate the same probablities.'

Exercise \(\PageIndex{11}\)

\(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1+ t\) (see Exercise 15 from "Problems on Random Vectors and Joint Distributions").

Determine \(P(Z \le 2)\)

**Answer**-
\(P(Z \le 2) = P(Z \in Q = Q1M1 \bigvee Q2M2)\), where \(M1 = \{(t, u): 0 \le t \le 1, 0 \le u \le 1 + t\}\)

\(M2 = \{(t, u) : 1 < t \le 2, 0 \le u \le 1 + t\}\)

\(Q1 = \{(t, u) : 0 \le t \le 1/2\}\), \(Q2 = \{(t, u) : u \le 2 - t\}\) (see figure)

\(P = \dfrac{3}{88} \int_{0}^{1/2} \int_{0}^{1 + t} (2t + 3u^2) du\ dt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{2 - t} (2t + 3u^2) du\ dt = \dfrac{563}{5632}\)

tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 3] Enter number of X approximation points 200 Enter number of Y approximation points 300 Enter expression for joint density (3/88)*(2*t + 3*u.^2).*(u<=1+t) Use array operations on X, Y, PX, PY, t, u, and P G = 4*t.*(t<=1) + (t+u).*(t>1); [Z,PZ] = csort(G,P); PZ2 = (Z<=2)*PZ' PZ2 = 0.1010 % Theoretical = 563/5632 = 0.1000

Exercise \(\PageIndex{12}\)

\(f_{XY} (t, u) = \dfrac{24}{11}\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2 - t\}\)(see Exercise 17 from "Problems on Random Vectors and Joint Distributions").

\(Z = I_M(X, Y) \dfrac{1}{2} X + I_{M^c} (X, Y) Y^2\), \(M = \{(t, u) : u > t\}\)

Determine \(P (Z \le 1/4)\).

**Answer**-
\(P(Z \le 1/4) = P((X, Y) \in M_1Q_1 \bigvee M_2Q_2)\), \(M_1 = \{(t, u): 0 \le t \le u \le 1\}\)

\(M_2 = \{(t, u) : 0 \le t \le 2, 0 \le t \le \text{min } (t, 2 - t)\}\)

\(Q_1 = \{(t, u): t \le 1/2\}\) \(Q_2 = \{(t, u): u \le 1/2\}\) (see figure)

\(P = \dfrac{24}{11} \int_{0}^{1/2} \int_{0}^{1} tu \ du\ dt + \dfrac{24}{11} \int_{1/2}^{3/2} \int_{0}^{1/2} tu\ du\ dt + \dfrac{24}{11} \int_{3/2}^{2} \int_{0}^{2 - t} tu\ du\ dt = \dfrac{85}{176}\)

tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 400 Enter number of Y approximation points 200 Enter expression for joint density (24/11)*t.*u.*(u<=min(1,2-t)) Use array operations on X, Y, PX, PY, t, u, and P G = 0.5*t.*(u>t) + u.^2.*(u<t); [Z,PZ] = csort(G,P); pp = (Z<=1/4)*PZ' pp = 0.4844 % Theoretical = 85/176 = 0.4830

Exercise \(\PageIndex{13}\)

\(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\) (see Exercise 18 from "Problems on Random Vectors and Joint Distributions").

\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y)2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)

Determine \(P(Z \le 1)\)

**Answer**-
\(P(Z \le 1) = P((X, Y) \in M_1Q_1 \bigvee M_2Q_2)\), \(M_1 = \{(t, u): 0 \le t \le 1, 0 \le u \le 1 - t\}\)

\(M_2 = \{(t, u) : 1 \le t \le 2, 0 \le u \le t\}\)

\(Q_1 = \{(t, u): u \le 1 - t\}\) \(Q_2 = \{(t, u): u \le 1/2\}\) (see figure)

\(P = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1-t} (t + 2u) \ du\ dt + \dfrac{3}{23} \int_{1}^{2} \int_{0}^{1/2} (t + 2u)\ du\ dt = \dfrac{9}{46}\)

tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 300 Enter number of Y approximation points 300 Enter expression for joint density (3/23)*(t + 2*u).*(u<=max(2-t,t)) Use array operations on X, Y, PX, PY, t, u, and P M = max(t,u) <= 1; G = M.*(t + u) + (1 - M)*2.*u; p = total((G<=1).*P) p = 0.1960 % Theoretical = 9/46 = 0.1957

Exercise \(\PageIndex{14}\)

\(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2, 3 - t\}\) (see Exercise 19 from "Problems on Random Vectors and Joint Distributions").

\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\)

Determine \(P(Z \le 2)\).

**Answer**-
\(P(Z \le 2) = P((X, Y) \in M_1 Q_1 \bigvee (M_2 \bigvee M_3) Q_2)\), \(M_1 = \{(t, u): 0 \le t \le 1, 1 \le u \le 2\}\)

\(M_2 = \{(t, u) : 0 \le t \le 1, 0 \le u \le 1\}\) \(M_3 = \{(t, u): 1 \le t \le 2, 0 \le u \le 3 - t\}\)

\(Q_1 = \{(t, u): u \le 1 - t\}\) \(Q_2 = \{(t, u) : u \le 1/2\}\) (see figure)

\(P = \dfrac{12}{179} \int_{0}^{1} \int_{0}^{2 - t} (3t^2 + u) du\ dt + \dfrac{12}{179} \int_{1}^{2} \int_{0}^{1} (3t^2 + u) du\ dt = \dfrac{119}{179}\)

tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 300 Enter number of Y approximation points 300 Enter expression for joint density (12/179)*(3*t.^2 + u).*(u<=min(2,3-t)) Use array operations on X, Y, PX, PY, t, u, and P M = (t<=1)&(u>=1); Z = M.*(t + u) + (1 - M)*2.*u.^2; G = M.*(t + u) + (1 - M)*2.*u.^2; p = total((G<=2).*P) p = 0.6662 % Theoretical = 119/179 = 0.6648

Exercise \(\PageIndex{15}\)

\(f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1 + t, 2\}\) (see Exercise 20 from "Problems on Random Variables and joint Distributions")

\(Z = I_M (X, Y) X + I_{M^c} (X, Y) \dfrac{Y}{X}\), \(M = \{(t, u): u \le \text{min } (1, 2 - t)\}\)

Determine \(P(Z \le 1)\).

Figure 10.4.3

**Answer**-
\(P(Z \le 1) = P((X, Y) \in M_1 Q_1 \bigvee V_2Q_2)\), \(M_1 = M\), \(M_2 = M^c\)

\(Q_1 = \{(t, u): 0 \le t \le \}\) \(Q_2 = \{(t, u) : u \le t\}\) (see figure)

\(P = \dfrac{12}{227} \int_{0}^{1} \int_{0}^{1} (3t + 2tu) du\ dt + \dfrac{12}{227} \int_{1}^{2} \int_{2 - t}^{t} (3t + 2tu) du\ dt = \dfrac{124}{227}\)

tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 400 Enter number of Y approximation points 400 Enter expression for joint density (12/227)*(3*t+2*t.*u).*(u<=min(1+t,2)) Use array operations on X, Y, PX, PY, t, u, and P Q = (u<=1).*(t<=1) + (t>1).*(u>=2-t).*(u<=t); P = total(Q.*P) P = 0.5478 % Theoretical = 124/227 = 0.5463

Exercise \(\PageIndex{16}\)

The class \(\{X, Y, Z\}\) is independent.

\(X = -2 I_A + I_B + 3I_C\). Minterm probabilities are (in the usual order)

\(Y = I_D + 3I_E + I_F - 3\). The class \(\{D, E, F\}\) is independent with

\(Z\) has distribution

Value | -1.3 | 1.2 | 2.7 | 3.4 | 5.8 |

Probability | 0.12 | 0.24 | 0.43 | 0.13 | 0.08 |

Determine \(P(X^2 + 3XY^2 >3Z)\).

**Answer**-
% file npr10_16.m Data for Exercise 16. cx = [-2 1 3 0]; pmx = 0.001*[255 25 375 45 108 12 162 18]; cy = [1 3 1 -3]; pmy = minprob(0.01*[32 56 40]); Z = [-1.3 1.2 2.7 3.4 5.8]; PZ = 0.01*[12 24 43 13 8]; disp('Data are in cx, pmx, cy, pmy, Z, PZ') npr10_16 % Call for data Data are in cx, pmx, cy, pmy, Z, PZ [X,PX] = canonicf(cx,pmx); [Y,PY] = canonicf(cy,pmy); icalc3 Enter row matrix of X-values X Enter row matrix of Y-values Y Enter row matrix of Z-values Z Enter X probabilities PX Enter Y probabilities PY Enter Z probabilities PZ Use array operations on matrices X, Y, Z, PX, PY, PZ, t, u, v, and P M = t.^2 + 3*t.*u.^2 > 3*v; PM = total(M.*P) PM = 0.3587

Exercise \(\PageIndex{17}\)

The simple random variable *X* has distribution

- Plot the distribution function \(F_X\) and the quantile function \(Q_X\).
- Take a random sample of size \(n =\) 10,000. Compare the relative frequency for each value with the probability that value is taken on.

**Answer**-
X = [-3.1 -0.5 1.2 2.4 3.7 4.9]; PX = 0.01*[15 22 33 12 11 7]; ddbn Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX % Plot not reproduced here dquanplot Enter VALUES for X X Enter PROBABILITIES for X PX % Plot not reproduced here rand('seed',0) % Reset random number generator dsample % for comparison purposes Enter row matrix of VALUES X Enter row matrix of PROBABILITIES PX Sample size n 10000 Value Prob Rel freq -3.1000 0.1500 0.1490 -0.5000 0.2200 0.2164 1.2000 0.3300 0.3340 2.4000 0.1200 0.1184 3.7000 0.1100 0.1070 4.9000 0.0700 0.0752 Sample average ex = 0.8792 Population mean E[X] = 0.859 Sample variance vx = 5.146 Population variance Var[X] = 5.112