14.1 Problems on conditional expectation, regression (Page 5/5)

Applied probability Page 5 / 5

Suppose $X \sim$ uniform on 1 through n and $Y \sim$ conditionally binomial $(i, p)$ , given $X = i$ .

Determine $E [Y]$ from $E [Y | X = k]$ .
Determine the joint distribution for ${X, Y}$ for $n = 50$ and $p = 0.3$ . Use jcalc to determine $E [Y]$ ; compare with the theoretical value.

$E [Y | X = i] = i p$ , so
$E [Y] = \sum_{i = 1}^{n} E [Y | X = i] P (X = i) = \frac{p}{n} \sum_{i = 1}^{n} i = \frac{p (n + 1)}{2}$
$P (X = i) = 1 / n, 1 \leq i \leq n$ , $P (Y = k | X = i) = i b i n o m (i, p, 0 : i), 0 \leq k \leq i$ . n = 50; p = 0.3; X = 1:n; Y = 0:n; P0 = zeros(n,n+1); % Could use randbernfor i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i);end P = rot90(P0);jcalc: X Y P - - - - - - - - - - -EY = dot(Y,PY) EY = 7.6500 % Comparison with part (a): 0.3*51/2 = 0.765

A number X is selected randomly from the integers 1 through 100. A pair of dice is thrown X times. Let Y be the number of sevens thrown on the X tosses. Determine the joint distribution for ${X, Y}$ and then determine $E [Y]$ .

$P (X = i) = 1 / n$ , $E [Y | X = i] = i / 6$ , so
$E [Y] = \frac{1}{6} \sum_{i = 0}^{n} i / n = \frac{(n + 1)}{12}$
n = 100; p = 1/6; X = 1:n; Y = 0:n; PX = (1/n)*ones(1,n); P0 = zeros(n,n+1); % Could use randbernfor i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i);end P = rot90(P0);jcalc EY = dot(Y,PY)EY = 8.4167 % Comparison with part (a): 101/12 = 8.4167

Got questions? Get instant answers now!

A number X is selected randomly from the integers 1 through 100. Each of two people draw X times, independently and randomly, a number from 1 to 10. Let Y be the number of matches (i.e., both draw ones, both draw twos, etc.). Determine thejoint distribution and then determine $E [Y]$ .

Same as [link] , except $p = 1 / 10$ . $E [Y] = (n + 1) / 20$ .

n = 100; p = 0.1; X = 1:n; Y = 0:n; PX = (1/n)*ones(1,n);
P0 = zeros(n,n+1);         % Could use randbernfor i = 1:n
  P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i);end
P = rot90(P0);jcalc
- - - - - - - - - -EY = dot(Y,PY)
EY =  5.0500                  % Comparison with part (a): EY = 101/20 = 5.05

Got questions? Get instant answers now!

$E [Y | X = t] = 10 t$ and X has density function $f_{X} (t) = 4 - 2 t$ for $1 \leq t \leq 2$ . Determine $E [Y]$ .

E [Y] = \int E [Y | X = t] f_{X} (t) d t = \int_{1}^{2} 10 t (4 - 2 t) d t = 40 / 3

Got questions? Get instant answers now!

$E [Y | X = t] = \frac{2}{3} (1 - t)$ for $0 \leq t < 1$ and X has density function $f_{X} (t) = 30 t^{2} {(1 - t)}^{2}$ for $0 \leq t \leq 1$ . Determine $E [Y]$ .

E [Y] = \int E [Y | X = t] f_{X} (t) d t = \int_{0}^{1} 20 t^{2} {(1 - t)}^{3} d t = 1 / 3

Got questions? Get instant answers now!

$E [Y | X = t] = \frac{2}{3} (2 - t)$ and X has density function $f_{X} (t) = \frac{15}{16} t^{2} {(2 - t)}^{2} 0 \leq t < 2$ . Determine $E [Y]$ .

E [Y] = \int E [Y | X = t] f_{X} (t) d t = \frac{5}{8} \int_{0}^{2} t^{2} {(2 - t)}^{3} d t = 2 / 3

Got questions? Get instant answers now!

Suppose the pair ${X, Y}$ is independent, with $X \sim$ Poisson $(μ)$ and $Y \sim$ Poisson $(λ)$ . Show that X is conditionally binomial $(n, μ / (μ + λ))$ , given $X + Y = n$ . That is, show that

P (X = k | X + Y = n) = C (n, k) p^{k} {(1 - p)}^{n - k}, 0 \leq k \leq n, for p = μ / (μ + λ)

$X \sim$ Poisson $(μ)$ , $Y \sim$ Poisson $(λ)$ , Use of property (T1) and generating functions shows that $X + Y \sim$ Poisson $(μ + λ)$

P (X = k | X + Y = n) = \frac{P (X = k, X + Y = n)}{P (X + Y = n)} = \frac{P (X = k, Y = n - k)}{P (X + Y = n)}

= \frac{e^{- μ} \frac{μ^{k}}{k!} e^{- λ} \frac{λ^{n - k}}{(n - k)!}}{e^{- (μ + λ)} \frac{{(μ + λ)}^{n}}{n!}} = \frac{n!}{k! (n - k)!} \frac{μ^{k} λ^{n - k}}{{(μ + λ)}^{n}}

Put $p = μ / (μ + λ)$ and $q = 1 - p = λ / (μ + λ)$ to get the desired result.

Got questions? Get instant answers now!

Use the fact that $g (X, Y) = g^{*} (X, Y, Z)$ , where $g^{*} (t, u, v)$ does not vary with v . Extend property (CE10) to show

E [g (X, Y) | X = t, Z = v] = E [g (t, Y) | X = t, Z = v] a . s . [P_{X Z}]

E [g (X, Y) | X = t, Z = v] = E [g^{*} (X, Z, Y) | (X, Z) = (t, v)] = E [g^{*} (t, v, Y) | (X, Z) = (t, v)]

= E [g (t, Y) | X = t, Z = v] a . s . [P_{X Z}] by (CE10)

Got questions? Get instant answers now!

Use the result of [link] and properties (CE9a) and (CE10) to show that

E [g (X, Y) | Z = v] = \int E [g (t, Y) | X = t, Z = v] F_{X | Z} (d t | v) a . s . [P_{Z}]

By (CE9) , $E [g (X, Y) | Z] = E {E [g (X, Y) | X, Z] | Z} = E [e (X, Z) | Z] a . s .$

By (CE10) ,

E [e (X, Z) | Z = v] = E [e (X, v) | Z = v] =

\int e (t, v) F_{X | Z} (d t | v) a . s .

By [link] ,

\int E [g (X, Y) | X = t, Z = v] F_{X | Z} (d t | v) =

\int E [g (t, Y) | X = t, Z = v] F_{X | Z} (d t | v) a . s . [P_{Z}]

Got questions? Get instant answers now!

A shop which works past closing time to complete jobs on hand tends to speed up service on any job received during the last hour before closing. Suppose the arrivaltime of a job in hours before closing time is a random variable $T \sim$ uniform $[0, 1]$ . Service time Y for a unit received in that period is conditionally exponential $β (2 - u)$ , given $T = u$ . Determine the distribution function for Y .

F_{Y} (v) = \int F_{Y | T} (v | u) f_{T} (u) d u = \int_{0}^{1} (1 - e^{- β (2 - u) v}) d u =

1 - e^{- 2 β v} \frac{e^{β v} - 1}{β v} = 1 - e^{- β v} [\frac{1 - e^{- β v}}{β v}], 0 < v

Got questions? Get instant answers now!

Time to failure X of a manufactured unit has an exponential distribution. The parameter is dependent upon the manufacturing process. Suppose the parameter isthe value of random variable $H \sim$ uniform on[0.005, 0.01], and X is conditionally exponential $(u)$ , given $H = u$ . Determine $P (X > 150)$ . Determine $E [X | H = u]$ and use this to determine $E [X]$ .

F_{X | H} (t | u) = 1 - e^{- u t} f_{H} (u) = \frac{1}{0.005} = 200, 0.005 \leq u \leq 0.01

F_{X} (t) = 1 - 200 \int_{0.005}^{0.01} e^{- u t} d u = 1 - \frac{200}{t} [e^{- 0.005 t} - e^{- 0.01 t}]

P (X > 150) = \frac{200}{150} [e^{- 0.75} - e^{- 1.5}] \approx 0.3323

E [X | H = u] = 1 / u E [X] = 200 \int_{0.005}^{0.01} \frac{d u}{u} = 200 ln 2

Got questions? Get instant answers now!

A system has n components. Time to failure of the i th component is X _i and the class

${X_{i} : 1 \leq i \leq n}$ is iid exponential $(λ)$ . The system fails if any one or more of the components fails. Let W be the time to system failure. What is the probability the failure is due to the i th component?

Suggestion . Note that $W = X_{i}$ iff $X_{j} > X_{i}$ for all $j \neq i$ . Thus

{W = X_{i}} = {(X_{1}, X_{2}, \dots, X_{n}) \in Q}, Q = {(t_{1}, t_{2}, \dots t_{n}) : t_{k} > t_{i}, \forall k \neq i}

P (W = X_{i}) = E [I_{Q} (X_{1}, X_{2}, \dots, X_{n})] = E {E [I_{Q} (X_{1}, X_{2}, \dots, X_{n}) | X_{i}]}

Let $Q = {(t_{1}, t_{2}, \dots, t_{n}) : t_{k} > t_{i}, k \neq i}$ . Then

P (W = X_{i}) = E [I_{Q} (X_{1}, X_{2}, \dots, X_{n})] = E {E [I_{Q} (X_{1}, X_{2}, \dots, X_{n}) | X_{i}]}

= \int E [I_{Q} (X_{1}, X_{2}, \dots, t_{i}, \dots X_{n})] F_{X} (d t)

E [I_{Q} (X_{1}, X_{2}, \dots, t_{i}, \dots X_{n})] = \prod_{k \neq i} P (X_{k} > t) = {[1 - F_{X} (t)]}^{n - 1}

If F _X is continuous, strictly increasing, zero for $t < 0$ , put $u = F_{X} (t)$ , $d u = f_{X} (t) d t$ . $t = 0 \sim u = 0$ , $t = \infty \sim u = 1$ . Then

P (W = X_{i}) = \int_{0}^{1} {(1 - u)}^{n - 1} d u = \int_{0}^{1} u^{n - 1} d u = 1 / n

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied probability' conversation and receive update notifications?

Ask

	Introduction to sociology 2e By OpenStax Read Online Course
	2 Psychology MCQ 2009 1 Exam By John Gabrieli Start Exam
	10 Neuroanatomy 10 Corticospinal Tract By Stephen Voron Start Quiz
	4 CDL Quiz - General Knowledge Part 2 By Jazzycazz Jackson Start Quiz
	6 Microeconomics 06 Elasticity By OpenStax Start Flashcards
	13 AP 13 Nervous System Essay By OpenStax Start Flashcards
	6 Microbiology Midterm Practice By Madison Christian Start Test
	Anthropology Environment, Adaption Subsistence By Richley Crapo Start Assignment
©flickr:	Spanish Test By Tess Armstrong Start Quiz
	1 Business Law MCQ 1 By Maureen Miller Start Exam